Probability
Question 1 |
For n > 2, let a {0,1}n be a non-zero vector. Suppose that x is chosen uniformly at random from {0,1}n.
0.5 |
Question 1 Explanation:
Question 2 |
Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is __________.
0.021 | |
0.022 | |
0.023 | |
0.024 |
Question 2 Explanation:
When two identical dice are rolled
⇾ A person wins who gets lower number compared to other person.
⇾ There could be “tie”, if they get same number.
Favorable cases = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Probability (tie) = 6/36 (when two dice are thrown, sample space = 6 × 6 = 36)
= 1/6
“Find the probability that one of them wins in the third attempt"
⇾ Which means, first & second time it should be tie and third time it should not be tie
⇾ P (tie) * P (tie) * P (not tie)
⇒ 1/6* 1/6 * (1 - 1/6)
⇒ (5/36×6) = 0.138/6 = 0.023
⇾ A person wins who gets lower number compared to other person.
⇾ There could be “tie”, if they get same number.
Favorable cases = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Probability (tie) = 6/36 (when two dice are thrown, sample space = 6 × 6 = 36)
= 1/6
“Find the probability that one of them wins in the third attempt"
⇾ Which means, first & second time it should be tie and third time it should not be tie
⇾ P (tie) * P (tie) * P (not tie)
⇒ 1/6* 1/6 * (1 - 1/6)
⇒ (5/36×6) = 0.138/6 = 0.023
Question 3 |
Let X be a Gaussian random variable with mean 0 and variance σ2. Let Y = max(X, 0) where max(a,b) is the maximum of a and b. The median of Y is __________.
0 | |
1 | |
2 | |
3 |
Question 3 Explanation:
Given that, Y = max(X,0),which means, Y is 0 for negative values of X and Y is X for positive values of X.
Median is a point, where the probability of getting less than median is 1/2 and probability of getting greater than median is 1/2.
From the given details, we can simply conclude that, median is 0. (0 lies exactly between positive and negative values)
Median is a point, where the probability of getting less than median is 1/2 and probability of getting greater than median is 1/2.
From the given details, we can simply conclude that, median is 0. (0 lies exactly between positive and negative values)
Question 4 |
Nβ (1 - β) | |
Nβ | |
N (1 - β) | |
Not expressible in terms of N and β alone |
Question 4 Explanation:
For a discrete random variable X,
Given gy (z) = (1 - β + βz)N ⇾ it is a binomial distribution like (x+y)n
Expectation (i.e., mean) of a binomial distribution will be np.
The polynomial function
,
given
Mean of Binomial distribution of b(xj,n,p)=
The probability Mass function,
Given:
Given gy (z) = (1 - β + βz)N ⇾ it is a binomial distribution like (x+y)n
Expectation (i.e., mean) of a binomial distribution will be np.
The polynomial function
,given
Mean of Binomial distribution of b(xj,n,p)=
The probability Mass function,
Given:
Question 5 |
P and Q are considering to apply for job. The probability that p applies for job is 1/4. The probability that P applies for job given that Q applies for the job 1/2 and The probability that Q applies for job given that P applies for the job 1/3.The probability that P does not apply for job given that Q does not apply for the job
 | |
 | |
 | |
 |
Question 5 Explanation:
Probability that ‘P’ applies for a job = 1/4 = P(p) ⇾ (1)
Probability that ‘P’ applies for the job given that Q applies for the job = P(p/q) = 1/2 ⇾ (2)
Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job = P(p/q) = 1/3 ⇾ (3)
Bayes Theorem:
(P(A/B) = (P(B/A)∙P(A))/P(B) ; P(A/B) = P(A∩B)/P(B))
⇒ P(p/q) = (P(q/p)∙P(p))/p(q)
⇒ 1/2 = (1/3×1/4)/p(q)
p(q) = 1/12×2 = 1/(6) (P(q) = 1/6) ⇾ (4)
From Bayes,
P(p/q) = (P(p∩q))/(P(q))
1/2 = P(p∩q)/(1⁄6)
(p(p∩q) = 1/12)
We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job = P(p'/q')
From Bayes theorem,
P(p'/q') = (P(p'∩q'))/P(q') ⇾ (5)
We know,
p(A∩B) = P(A) + P(B) - P(A∪B)
also (P(A'∩B') = 1 - P(A∪B))
P(p'∩q') = 1 - P(p∪q)
= 1 - (P(p) + P(q) - P(p∩q))
= 1 - (P(p) + P(q) - P(p) ∙ P(q))
= 1 - (1/4 + 1/6 - 1/12)
= 1 - (10/24 - 2/24)
= 1 - (8/24)
= 2/3
(P(p'∩q') = 2/3) ⇾ (6)
Substitute in (5),
P(p'⁄q') = (2⁄3)/(1-P(q)) = (2⁄3)/(1-1/6) = (2⁄3)/(5⁄6) = 4/5
(P(p'/q') = 4/5)
Probability that ‘P’ applies for the job given that Q applies for the job = P(p/q) = 1/2 ⇾ (2)
Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job = P(p/q) = 1/3 ⇾ (3)
Bayes Theorem:
(P(A/B) = (P(B/A)∙P(A))/P(B) ; P(A/B) = P(A∩B)/P(B))
⇒ P(p/q) = (P(q/p)∙P(p))/p(q)
⇒ 1/2 = (1/3×1/4)/p(q)
p(q) = 1/12×2 = 1/(6) (P(q) = 1/6) ⇾ (4)
From Bayes,
P(p/q) = (P(p∩q))/(P(q))
1/2 = P(p∩q)/(1⁄6)
(p(p∩q) = 1/12)
We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job = P(p'/q')
From Bayes theorem,
P(p'/q') = (P(p'∩q'))/P(q') ⇾ (5)
We know,
p(A∩B) = P(A) + P(B) - P(A∪B)
also (P(A'∩B') = 1 - P(A∪B))
P(p'∩q') = 1 - P(p∪q)
= 1 - (P(p) + P(q) - P(p∩q))
= 1 - (P(p) + P(q) - P(p) ∙ P(q))
= 1 - (1/4 + 1/6 - 1/12)
= 1 - (10/24 - 2/24)
= 1 - (8/24)
= 2/3
(P(p'∩q') = 2/3) ⇾ (6)
Substitute in (5),
P(p'⁄q') = (2⁄3)/(1-P(q)) = (2⁄3)/(1-1/6) = (2⁄3)/(5⁄6) = 4/5
(P(p'/q') = 4/5)
Question 6 |
If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)2] equals _________.
54 | |
55 | |
56 | |
57 |
Question 6 Explanation:
In Poisson distribution:
Mean = Variance
E(X) = E(X2) - (E(X))2 = 5
E(X2) = 5 + (E(X))2 = 5 + 25 = 30
So, E[(X + 2)2] = E[X2 + 4 + 4X]
= E(X2) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Mean = Variance
E(X) = E(X2) - (E(X))2 = 5
E(X2) = 5 + (E(X))2 = 5 + 25 = 30
So, E[(X + 2)2] = E[X2 + 4 + 4X]
= E(X2) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Question 7 |
A probability density function on the interval [a,1] is given by 1/x2 and outside this interval the value of the function is zero. The value of a is _________.
0.7 | |
0.6 | |
0.5 | |
0.8 |
Question 7 Explanation:
The property of probability density function is area under curve = 1
or
where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x2 , a≤x≤1
The area under curve,
- 1 + 1/a = 1
1/a = 2
a = 0.5
or
where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x2 , a≤x≤1
The area under curve,
- 1 + 1/a = 1
1/a = 2
a = 0.5
Question 8 |
Consider the following experiment.
-
Step1. Flip a fair coin twice.
Step2. If the outcomes are (TAILS, HEADS) then output Y and stop.
Step3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop.
Step4. If the outcomes are (TAILS, TAILS), then go to Step 1.
The probability that the output of the experiment is Y is (up to two decimal places) ________.
0.33 | |
0.34 | |
0.35 | |
0.36 |
Question 8 Explanation:
If a coin is flipped twice, the possible outcomes {HH, HT, TH, TT}
Stop conditions:
If outcome = TH then Stop [output 4] --------------- (1)
else
outcome = HH/ HT then Stop [output N] -------------- (2)
We get ‘y’ when we have (1) i.e., ‘TH’ is output.
(1) can be preceded by ‘TT’ also, as ‘TT’ will reset (1) again
Probability of getting y = TH + (TT)(TH) + (TT)(TT)(TH) + …
= 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + ...
= (1/4)/(1-1/4)
= 1/3
= 0.33
Stop conditions:
If outcome = TH then Stop [output 4] --------------- (1)
else
outcome = HH/ HT then Stop [output N] -------------- (2)
We get ‘y’ when we have (1) i.e., ‘TH’ is output.
(1) can be preceded by ‘TT’ also, as ‘TT’ will reset (1) again
Probability of getting y = TH + (TT)(TH) + (TT)(TT)(TH) + …
= 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + ...
= (1/4)/(1-1/4)
= 1/3
= 0.33
Question 9 |
Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is _________.
0.55 | |
0.56 | |
0.57 | |
0.58 |
Question 9 Explanation:
The bulbs of Type 1, Type 2 are same in number.
So, the probability to choose a type is 1/2.
The probability to choose quadrant ‘A’ in diagram is
P(last more than 100 hours/ type1) = 1/2 × 0.7
P(last more than 100 hours/ type2) = 1/2 × 0.4
Total probability = 1/2 × 0.7 + 1/2 × 0.4 = 0.55
Question 10 |
Suppose Xi for i = 1, 2, 3 are independent and identically distributed random variables whose probability mass functions are Pr[Xi = 0] = Pr[Xi = 1]=1/2 for i = 1, 2, 3. Define another random variable Y = X1X2 ⊕ X3, where ⊕ denotes XOR. Then Pr[Y = 0|X3 = 0] = _______________.
0.75 | |
0.76 | |
0.77 | |
0.78 |
Question 10 Explanation:
Pr(Y=0/ X3=0) = Pr(Y=0 ∩ X3=0)/ Pr(X3=0)
= 3/8 / 4/8 = 3/4 = 0.75
Question 11 |
Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is X⁄1296. The value of X is ___________.
10 | |
11 | |
12 | |
13 |
Question 11 Explanation:
Each dice can result from {1, 2, 3, 4, 5, 6}
To get ‘22’ as Sum of four outcomes
x1 + x2 + x3 + x4 = 22
The maximum Sum = 6+6+6+6 = 24 which is near to 22
So, keeping three 6’s, 6+6+6+x = 22
x = 4 combination① = 6 6 6 4
Keeping two 6’s, 6+6+x1+x2 = 22
x1+x2 = 10 possible x’s (5, 5) only
combination② = 6 6 5 5
No. of permutation with 6664 = 4!/ 3! = 4
“ “ “ 6655 = 4!/ 2!2! = 6
Total = 4+6 = 10 ways out of 6×6×6×6 = 1296
Pnb (22) = 10/1296 ⇒ x = 10
To get ‘22’ as Sum of four outcomes
x1 + x2 + x3 + x4 = 22
The maximum Sum = 6+6+6+6 = 24 which is near to 22
So, keeping three 6’s, 6+6+6+x = 22
x = 4 combination① = 6 6 6 4
Keeping two 6’s, 6+6+x1+x2 = 22
x1+x2 = 10 possible x’s (5, 5) only
combination② = 6 6 5 5
No. of permutation with 6664 = 4!/ 3! = 4
“ “ “ 6655 = 4!/ 2!2! = 6
Total = 4+6 = 10 ways out of 6×6×6×6 = 1296
Pnb (22) = 10/1296 ⇒ x = 10
Question 12 |
The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p =_____________.
11.90 | |
11.91 | |
11.92 | |
11.93 |
Question 12 Explanation:
There are 10 systems.
Out of which ‘4’ are chosen at random without replaced and created.
If at least three of them are working then system is deemed functional
i.e., there should be only ‘one’ non-working system in set of ‘4’.
It is possible with combination
W – W – W – N,
W – W – N – W,
W – N – W – W,
N – W – W – W.
For W – W – W – N, the probability = (choosing working out of 10) × (choosing working out of 9) × (choosing working out of 8) × (choosing non-working out of 7)
=4/10×3/9×2/8×1/7
where 4/10 ⇒ 4 working out of 10
3/9 ⇒ 3 working are remaining out of ‘9’ as ‘1’ is already taken
For ‘4’ Sum combinations
Total probability =4×[4/10×3/9×2/8×1/7]=600/5040
We need 100p ⇒100×600/5040=11.90
Out of which ‘4’ are chosen at random without replaced and created.
If at least three of them are working then system is deemed functional
i.e., there should be only ‘one’ non-working system in set of ‘4’.
It is possible with combination
W – W – W – N,
W – W – N – W,
W – N – W – W,
N – W – W – W.
For W – W – W – N, the probability = (choosing working out of 10) × (choosing working out of 9) × (choosing working out of 8) × (choosing non-working out of 7)
=4/10×3/9×2/8×1/7
where 4/10 ⇒ 4 working out of 10
3/9 ⇒ 3 working are remaining out of ‘9’ as ‘1’ is already taken
For ‘4’ Sum combinations
Total probability =4×[4/10×3/9×2/8×1/7]=600/5040
We need 100p ⇒100×600/5040=11.90
Question 13 |
Each of the nine words in the sentence "The quick brown fox jumps over the lazy dog" is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)
3.9 | |
4.0 | |
4.1 | |
4.2 |
Question 13 Explanation:
"The quick brown fox jumps over the lazy dog"
There are ‘9’ words in this sentence.
No. of characters in each word
The (3)
quick (5)
brown (5)
fox (3)
jumps (5)
over (4)
the (3)
lazy (4)
dog (3)
Each word has equal probability.
So expected length = 3×1/9+5×1/9+5×1/9+3×1/9+5×1/9+ 4×1/9+3×1/9+4×1/9+3×1/9
=35/9
=3.9
There are ‘9’ words in this sentence.
No. of characters in each word
The (3)
quick (5)
brown (5)
fox (3)
jumps (5)
over (4)
the (3)
lazy (4)
dog (3)
Each word has equal probability.
So expected length = 3×1/9+5×1/9+5×1/9+3×1/9+5×1/9+ 4×1/9+3×1/9+4×1/9+3×1/9
=35/9
=3.9
Question 14 |
The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______ .
0.259 to 0.261 | |
0.260 to 0.262 | |
0.261 to 0.263 | |
0.262 to 0.264 |
Question 14 Explanation:
Let A = divisible by 2, B = divisible by 3 and C = divisible by 5, then
n(A)=50, n(B)=33, n(C)=20
n(A∩B)=16, n(B∩C)=6, n(A∩C)=10
n(A∩B∩C)=3
P(A∪B∪C)=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C) -P(A∩C)+P(A∩B∩C)=74/100
∴ Required probability is P(A∩B∩C)=1-P(A∪B∪C)=0.26
n(A)=50, n(B)=33, n(C)=20
n(A∩B)=16, n(B∩C)=6, n(A∩C)=10
n(A∩B∩C)=3
P(A∪B∪C)=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C) -P(A∩C)+P(A∩B∩C)=74/100
∴ Required probability is P(A∩B∩C)=1-P(A∪B∪C)=0.26
Question 15 |
Let S be a sample space and two mutually exclusive events A and B be such that A∪B = S. If P(∙) denotes the probability of the event, the maximum value of P(A)P(B) is __________.
0.25 | |
0.26 | |
0.27 | |
0.28 |
Question 15 Explanation:
We know that
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2≥√(P(A)∙P(B))
1/2≥√(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4≥P(A)∙P(B)
P(A)∙P(B)≤1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2≥√(P(A)∙P(B))
1/2≥√(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4≥P(A)∙P(B)
P(A)∙P(B)≤1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
Question 16 |
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
8/(2e3) | |
9/(2e3) | |
17/(2e3) | |
26/(2e3) |
Question 16 Explanation:
The formula for the Poisson probability mean function
P(x:λ)=(e-λ λx)/x! for x = 0,1,2….
‘λ’ is the average number (mean)
Given that mean = λ = 3
The probability of observing fewer than three cars is
P(zero car) + P(one car) + P(two cars)
=(e-3 30)/0!+(e-3 31)/1!+(e-3 32)/2!
=e-3+e-3∙3+(e-3)∙9)/2
=(17e-3<)/2
=17/(2e3 )
P(x:λ)=(e-λ λx)/x! for x = 0,1,2….
‘λ’ is the average number (mean)
Given that mean = λ = 3
The probability of observing fewer than three cars is
P(zero car) + P(one car) + P(two cars)
=(e-3 30)/0!+(e-3 31)/1!+(e-3 32)/2!
=e-3+e-3∙3+(e-3)∙9)/2
=(17e-3<)/2
=17/(2e3 )
Question 17 |
Consider a random variable X that takes values +1 and −1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = −1 and +1 are
0 and 0.5 | |
0 and 1 | |
0.5 and 1 | |
0.25 and 0.75 |
Question 17 Explanation:
Cumulative probability is sum of probabilities of all other points till the given value.
The sum of probabilities at x=1, x=-1 itself is 0.5+0.5 =1. It is evident that, there is no probability for any other values.
The F(x=-1) is 0.5 as per given probabilities and
F(x=1) = sum of F(x=-1) +F(x=0)=...f(X=1) = 0.5 +0.5 =1
The sum of probabilities at x=1, x=-1 itself is 0.5+0.5 =1. It is evident that, there is no probability for any other values.
The F(x=-1) is 0.5 as per given probabilities and
F(x=1) = sum of F(x=-1) +F(x=0)=...f(X=1) = 0.5 +0.5 =1
Question 18 |
Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
10/21 | |
5/12 | |
2/3 | |
1/6 |
Question 18 Explanation:
The value on the die for first time rolling= {1, 2, 3}
The value on second time can be {1, 2, 3, 4, 5, 6}
So the Sum can be
We have Sample space = 36
The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}
So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①
If we get ‘6’ in the first time itself, then we do not go for rolling die again.
So, its probability = 1/6
Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12
The value on second time can be {1, 2, 3, 4, 5, 6}
So the Sum can be
We have Sample space = 36
The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}
So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①
If we get ‘6’ in the first time itself, then we do not go for rolling die again.
So, its probability = 1/6
Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12
Question 19 |
If the difference between the expectation of the square of a random variable (E[X2]) and the square of the expectation of the random variable (E[X])2 is denoted by R, then
R = 0 | |
R < 0 | |
R ≥ 0 | |
R > 0 |
Question 19 Explanation:
We know that difference of E(X2) and E(X))2 is nothing but variance or V(X) which is always greater than or equal to zero.
So the answer will be R≥0.
So the answer will be R≥0.
Question 20 |
If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads?
1/3 | |
1/4 | |
1/2 | |
2/3 |
Question 20 Explanation:
Sample space = {HH, HT, TH}
Required probability = 1/3
Question 21 |
A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second.
1/5 | |
4/25 | |
1/4 | |
2/5 |
Question 21 Explanation:
The possible events are
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
5C1 × 4C1 = 20
So probability = 4/20 = 1/5
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
5C1 × 4C1 = 20
So probability = 4/20 = 1/5
Question 22 |
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?
pq + (1 - p)(1 - q)
| |
(1 - q)p
| |
(1 - p)q
| |
pq |
Question 22 Explanation:
The probability of the computer being declared faulty is,
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of tetsing process giving incorrect result × Probability that computer is not faulty
= p × q + (1 - p) (1 - q)
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of tetsing process giving incorrect result × Probability that computer is not faulty
= p × q + (1 - p) (1 - q)
Question 23 |
What is the probability that divisor of 1099 is a multiple of 1096?
1/625 | |
4/625 | |
12/625
| |
16/625 |
Question 23 Explanation:
Probability that divisor of 1099 is a multiple of 1096
We can write 1099 as 1096×103
So, (1099)/(1096) to be a whole number, [1096×103/1096]➝ (1)
We can observe that every divisor of 103 is a multiple of 1096
So number of divisor of 103 to be found first
⇒ 103=(5×2)3=23×53
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 1099 are 1099=299×599=100×100=10000
Probability that divisor of 1099 is a multiple of 1096 is
⇒16/10,000
We can write 1099 as 1096×103
So, (1099)/(1096) to be a whole number, [1096×103/1096]➝ (1)
We can observe that every divisor of 103 is a multiple of 1096
So number of divisor of 103 to be found first
⇒ 103=(5×2)3=23×53
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 1099 are 1099=299×599=100×100=10000
Probability that divisor of 1099 is a multiple of 1096 is
⇒16/10,000
Question 24 |
An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the
If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3?
0.453
| |
0.468
| |
0.485 | |
0.492
|
Question 24 Explanation:
P(0) = Probability of getting odd no. face.
P(e) = Probability of getting even no. face.
It is given that,
P(0) = 0.9 P(e) ----- (I)
Also we know that,
P(0) + P(e) = 1 ----- (II)
Solving equation (I) and (II) we get,
P(e) = 0.52
Also even no. can be 2 or 4 or 6.
And given in question that P(2) = P(4) = P(6).
So, 3 × P(2) = 0.52
P(2) = 0.175
So, P(2) = P(4) = P(6) = 0.175
Also in question it is given that,
P(e/>3) = 0.75
P(even no. greater than 3)/ P(no. greater than 3) = 0.75
P(4,6)/P(>3) = 0.75
(0.175×2)/P(>3) = 0.75
P(>3) = 0.35/0.75 = 0.467
P(e) = Probability of getting even no. face.
It is given that,
P(0) = 0.9 P(e) ----- (I)
Also we know that,
P(0) + P(e) = 1 ----- (II)
Solving equation (I) and (II) we get,
P(e) = 0.52
Also even no. can be 2 or 4 or 6.
And given in question that P(2) = P(4) = P(6).
So, 3 × P(2) = 0.52
P(2) = 0.175
So, P(2) = P(4) = P(6) = 0.175
Also in question it is given that,
P(e/>3) = 0.75
P(even no. greater than 3)/ P(no. greater than 3) = 0.75
P(4,6)/P(>3) = 0.75
(0.175×2)/P(>3) = 0.75
P(>3) = 0.35/0.75 = 0.467
Question 25 |
Aishwarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
0.24
| |
0.36 | |
0.4 | |
0.6
|
Question 25 Explanation:
→ Aishwarya studied CS on Monday. Then we have two possibilities to she study computer science on wednesday.
(i) She study Mathematics on Tuesday and computer science on wednesday.
⇒ 0.6×0.4
⇒ 0.24
(ii) She study computer science on Tuesday and computer science on wednesday.
⇒ 0.4×0.4
⇒ 0.16
→ The probability that she study computer science on wednesday is
0.24+0.16=0.40
(i) She study Mathematics on Tuesday and computer science on wednesday.
⇒ 0.6×0.4
⇒ 0.24
(ii) She study computer science on Tuesday and computer science on wednesday.
⇒ 0.4×0.4
⇒ 0.16
→ The probability that she study computer science on wednesday is
0.24+0.16=0.40
Question 26 |
Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean -1 and variance unknown. If P(X ≤ -1) = P(Y ≥ 2), the standard deviation of Y is
3 | |
2 | |
√2 | |
1 |
Question 26 Explanation:
P(X ≤ -1) = P(Y ≥ 2)
We can compare their values using standard normal distributions.
The above equation satisfies when σy will be equal to 3.
We can compare their values using standard normal distributions.
The above equation satisfies when σy will be equal to 3.
Question 27 |
A sample space has two events A and B such that probabilities P(A ∩ B) = 1/2, P(A') = 1/3, P(B') = 1/3. What is P(A ∪ B)?
11/12 | |
10/12 | |
9/12 | |
8/12 |
Question 27 Explanation:
P(A ∩ B) = 1/2
P(A') = 1/3; P(A) = 2/3
P(B') = 1/3; P(B) = 2/3
P(A ∪ B) = P(A) +P(B) - P(A ∩ B)
= 2/3 + 2/3 - 1/2
= 4+4-3/ 6
= 5/6
= 10/12
P(A') = 1/3; P(A) = 2/3
P(B') = 1/3; P(B) = 2/3
P(A ∪ B) = P(A) +P(B) - P(A ∩ B)
= 2/3 + 2/3 - 1/2
= 4+4-3/ 6
= 5/6
= 10/12
Question 28 |
What is the probability that in a randomly chosen group of r people at least three people have the same birthday?
 | |
 | |
 | |
 |
Question 28 Explanation:
Question 29 |
Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1, 2, 3,….., 20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?
1/2 | |
1/10 | |
9!/20! | |
None of these
|
Question 29 Explanation:
1, 2, 3, 4, …….20
→ Total no. of possible even number = 10
→ Here we are not considering odd number.
→ The probability that 2 appears at an earlier position than any other even number is =1/10
→ Total no. of possible even number = 10
→ Here we are not considering odd number.
→ The probability that 2 appears at an earlier position than any other even number is =1/10
Question 30 |
Suppose there are two coins. The first coin gives heads with probability 5/8 when tossed, while the second coin gives heads with probability 1/4. One of the two coins is picked up at random with equal probability and tossed. What is the probability of obtaining heads ?
7/8 | |
1/2 | |
7/16 | |
5/32 |
Question 30 Explanation:
Each coin has equal probability to pick i.e., 1/2.
The probability of obtaining heads
= (1/2)(5/8) + (1/2)(1/4)
= (5/16) + (1/8)
= 7/16
The probability of obtaining heads
= (1/2)(5/8) + (1/2)(1/4)
= (5/16) + (1/8)
= 7/16
Question 31 |
Consider a hash function that distributes keys uniformly. The hash table size is 20. After hashing of how many keys will the probability that any new key hashed collides with an existing one exceed 0.5.
5 | |
6 | |
7 | |
10 |
Question 31 Explanation:
Total spaces = 20, no. of entries = 1
Probability of collision for each entry = 1/20
After inserting X values then probability becomes 1/2
i.e., (1/20)X = 1/2
X = b
Probability of collision for each entry = 1/20
After inserting X values then probability becomes 1/2
i.e., (1/20)X = 1/2
X = b
Question 32 |
In a multi-user operating system on an average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in 45 minutes is given by :
6.9 × 106 × e-20 | |
1.02 × 106 × e-20 | |
6.9 × 103 × e-20 | |
1.02 × 103 × e-20 |
Question 32 Explanation:
q0 request in 1 hour. So we can expect 15 requests in 45 minutes.
So, λ=15
So, λ=15
Question 33 |
We are given a set X = {x1, .... xn} where xi = 2i. A sample S ⊆ X is drawn by selecting each xi independently with probability pi = 1/2. The expected value of the smallest number in sample S is:
1/n | |
2 | |
√n | |
n |
Question 33 Explanation:
The smallest element in sample S would be xi for which i is the smallest.
The given probability Pi is for selection of each item independently with probability 1/2.
Now, Probability for x1 to be smallest in S = 1/2
Now, Probability for x2 to be smallest in S = Probability of x1 not being in S × Probability of x2 being in S
= 1/2 × 1/2
Similarly, Probability xi to be smallest = (1/2)i
Now the Expected value is
The given probability Pi is for selection of each item independently with probability 1/2.
Now, Probability for x1 to be smallest in S = 1/2
Now, Probability for x2 to be smallest in S = Probability of x1 not being in S × Probability of x2 being in S
= 1/2 × 1/2
Similarly, Probability xi to be smallest = (1/2)i
Now the Expected value is
Question 34 |
For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is:
 | |
 | |
 | |
 |
Question 34 Explanation:
Total number of coins = 2n
No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nCn*(1/2)n*(1/2)n=2nCn/4n
No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nCn*(1/2)n*(1/2)n=2nCn/4n
Question 35 |
In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?
0.4 | |
0.6 | |
0.8 | |
0.9 |
Question 35 Explanation:
Probability rain in afternoon = (0.5×probability of rain when temp≤25) + (0.5×Probability of rain when temp>25)
0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)
0.6 = (2) + (0.5×P(rain at temp>25)
P(rain at temp>25) = 0.8
0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)
0.6 = (2) + (0.5×P(rain at temp>25)
P(rain at temp>25) = 0.8
Question 36 |
When a coin is tossed, the probability of getting a Head is p, 0<p<1. Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is
1/p | |
1/(1-p) | |
1/p2 | |
1/(1-p2) |
Question 36 Explanation:
E = 1 × P + (2 × (1 - p) p) + (3 × ( 1 - p) (1 - p) p) + ......
Multiply both sides with (1 - p) and subtract,
E - (1 - p) E = 1 × p + (1 - p) p + (1 - p) (1 - p) p + ......
E - (1 - p) E = p/(1 - (1 - p))
(1 - 1 + p) E = 1
pE = 1
E = 1/p
Multiply both sides with (1 - p) and subtract,
E - (1 - p) E = 1 × p + (1 - p) p + (1 - p) (1 - p) p + ......
E - (1 - p) E = p/(1 - (1 - p))
(1 - 1 + p) E = 1
pE = 1
E = 1/p
Question 37 |
Let f(x) be the continuous probability density function of a random variable X. The probability that a < X ≤ b, is:
f(b - a)
| |
f(b) - f(a) | |
 | |
 |
Question 37 Explanation:
f(x) be the continuous probability density function of random variable X.
Then the probablity be area of the corresponding curve i.e.,
Then the probablity be area of the corresponding curve i.e.,
Question 38 |
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is:
1/2n
| |
1 - 1/n
| |
1/n!
| |
1-(1/2n)
|
Question 38 Explanation:
Total combinations of strings that can be generated are 2n. We will get one such string in the first experiment. So, favourable cases for the second string are 2n - 1, so that it doesn't match with the previous generated string.
Hence Probability = (2n - 1) /2n = 1 - 1/2n
Hence Probability = (2n - 1) /2n = 1 - 1/2n
Question 39 |
Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows:
(i) Select a box
(ii) Choose a ball from the selected box such that each ball in
the box is equally likely to be chosen. The probabilities of
selecting boxes P and Q are (1/3) and (2/3), respectively.
Given that a ball selected in the above process is a red ball, the probability that it came from the box P is4/19
| |
5/19
| |
2/9 | |
19/30 |
Question 39 Explanation:
P → 2 red, 3 blue
Q → 3 red, 1 blue
The probability of selecting a red ball is
(1/3)(2/5) + (2/3)(3/4)
2/15 + 1/2 = 19/30
The probability of selecting a red ball from P
(1/3) * (2/5) = 2/15
→ The colour of ball is selected is to be red and that is taken from the box P.
⇒ Probability of selecting a red ball from P/Probability of selecting a red ball
⇒ (2/15)/(19/30)
⇒ 4/19
Q → 3 red, 1 blue
The probability of selecting a red ball is
(1/3)(2/5) + (2/3)(3/4)
2/15 + 1/2 = 19/30
The probability of selecting a red ball from P
(1/3) * (2/5) = 2/15
→ The colour of ball is selected is to be red and that is taken from the box P.
⇒ Probability of selecting a red ball from P/Probability of selecting a red ball
⇒ (2/15)/(19/30)
⇒ 4/19
Question 40 |
A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is
1/36 | |
1/6 | |
1/4 | |
1/3 |
Question 40 Explanation:
Question 41 |
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is
3 | |
4 | |
5 | |
6 |
Question 41 Explanation:
Generally which uses a recursion strategy to solve this problem.
Other than using recursion we can solve using a formula which we know
The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)
The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)
The expected number of coin flips for getting n consecutive same tosses is (2^(n+1) -2) / 2.
where n = 2,
which is (2^(3+1) - 2) / 2 = 3
The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)
The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)
The expected number of coin flips for getting n consecutive same tosses is (2^(n+1) -2) / 2.
where n = 2,
which is (2^(3+1) - 2) / 2 = 3
Question 42 |
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?
3/8 | |
1/2 | |
5/8 | |
3/4 |
Question 42 Explanation:
A fair coin is tossed 4 times.
Then total number of possibilities = 24 = 16
No. of possibilities getting 2 heads and 2 tails is
HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6
Probability of getting 2 heads and 2 tails is
= No. of possibilities/Total no. of possibilities = 6/16 = 3/8
Then total number of possibilities = 24 = 16
No. of possibilities getting 2 heads and 2 tails is
HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6
Probability of getting 2 heads and 2 tails is
= No. of possibilities/Total no. of possibilities = 6/16 = 3/8
Question 43 |
An examination paper has 150 multiple choice questions of one mark each, with each question having four choices. Each incorrect answer fetches -0.25 marks. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is
0 | |
2550 | |
7525 | |
9375 |
Question 43 Explanation:
Probability of choosing a correct answer = 1/4
Probability of selecting a wrong answer = 3/4
For correct answer +1, for wrong answer-0.25;
Expected marks for each question = (1/4) × 1 + (3/4) -(0.25)
= 1/4 + (-3/16)
= 4-3/16
= 1/16
= 0.0625
Expected marks for 150 questions = 150 × 0.625 = 9.375
The sum total of expected marks obtained by 1000 students is = 1000×9.375 = 9375
Probability of selecting a wrong answer = 3/4
For correct answer +1, for wrong answer-0.25;
Expected marks for each question = (1/4) × 1 + (3/4) -(0.25)
= 1/4 + (-3/16)
= 4-3/16
= 1/16
= 0.0625
Expected marks for 150 questions = 150 × 0.625 = 9.375
The sum total of expected marks obtained by 1000 students is = 1000×9.375 = 9375
Question 44 |
Two n bit binary strings, S1 and S2, are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is
nCd /2n
| |
nCd / 2d
| |
d/2n
| |
1/2d
|
Question 44 Explanation:
n binary bits with difference 'd' then no. of favourable cases = nCd
Total no. of cases where n positions have any binary bit = 2n
The probability of 'd' bits differ = nCd / 2n
Total no. of cases where n positions have any binary bit = 2n
The probability of 'd' bits differ = nCd / 2n
Question 45 |
A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0,0), (1,0), (1,2) and (0,2). If p is the length of the position vector of the point, the expected value of p2 is
 | |
 | |
 | |
 |
Question 45 Explanation:
Above diagram shows the scenario of our question.
The length p of our position vector (x,y) is
Now we need to calculate the probability density function of X and Y.
Since distribution is uniform,
X goes from 0 to 1, so PDF(x) = 1/1-0 = 1
Y goes from 0 to 2, so PDF(y) = 1/2-0 = 1/2
Now we evaluate,
Question 46 |
Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z = min(X,Y), then the mean of Z is given by
 | |
 | |
 | |
 |
Question 47 |
In how many ways can we distribute 5 distinct balls, B1, B2, …, B5 in 5 distinct cells, C1, C2, …, C5 such that Ball B, is not in cell Ci, ∀i = 1, 2, …, 5 and each cell contains exactly one ball?
44 | |
96 | |
120 | |
3125 |
Question 47 Explanation:
Just apply inclusion exclusion principle,
Question 48 |
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A | B) and P(B | A) respectively are
 | |
 | |
 | |
 |
Question 48 Explanation:
P(A)=1, P(B)=1/2
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
Question 49 |
A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by
 | |
 | |
 | |
 |
Question 49 Explanation:
f1(t) and f2(t) are executed sequentially.
→ They representing the probability density functions of time taken to execute.
→ f1 can be executed in 'x' time.
f2 can be executed in 't-x' time.
→ The probability density function =
→ They representing the probability density functions of time taken to execute.
→ f1 can be executed in 'x' time.
f2 can be executed in 't-x' time.
→ The probability density function =
Question 50 |
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is
 | |
 | |
 | |
 |
Question 50 Explanation:
Total possibilities = 42 = 16
Atleast one head = 1/16
Atleast one tail = 1/16
Probability of getting one head and one tail is = 1 - 1/16 - 1/16 = 16-1-1/16 = 14/16 = 7/8
Atleast one head = 1/16
Atleast one tail = 1/16
Probability of getting one head and one tail is = 1 - 1/16 - 1/16 = 16-1-1/16 = 14/16 = 7/8
Question 51 |
0 | |
 | |
 | |
1 |
Question 51 Explanation:
Pr(E1) = Pr(E2) = a(say)
Pr(E1 ∪ E2) = 1
E1 and E2 are Independent.
P(E1 ∪ E2) = Pr(E1) + Pr(E2) - Pr(E1 ∩ E2)
1 = a + a - Pr(E1)⋅Pr(E2)
1 = a + a - a⋅a
2a - a2 = 1
a2 - 2a + 1 = 0
(a - 1)2 = 0
a = 1 Pr(E1) = 1
Answer: Option (D)
Pr(E1 ∪ E2) = 1
E1 and E2 are Independent.
P(E1 ∪ E2) = Pr(E1) + Pr(E2) - Pr(E1 ∩ E2)
1 = a + a - Pr(E1)⋅Pr(E2)
1 = a + a - a⋅a
2a - a2 = 1
a2 - 2a + 1 = 0
(a - 1)2 = 0
a = 1 Pr(E1) = 1
Answer: Option (D)
Question 52 |
Suppose that the expectation of a random variable X is 5. Which of the following statements is true?
There is a sample point at which X has the value 5. | |
There is a sample point at which X has value greater than 5. | |
There is a sample point at which X has a value greater than or equal to 5. | |
None of the above |
Question 52 Explanation:
Expectation of discrete random variable
E(X) = x1P1 + x2P2 + ... + xnPn
In question, E(X) is given as 5.
E(X) = 5, 0≤Pi≤1
P1 + P2 + ... + Pn = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the xi value is greater than 5.
E(X) = x1P1 + x2P2 + ... + xnPn
In question, E(X) is given as 5.
E(X) = 5, 0≤Pi≤1
P1 + P2 + ... + Pn = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the xi value is greater than 5.
Question 53 |
 | |
Events E1 and E2 are independent | |
Events E1 and E2 are not independent | |
 |
Question 53 Explanation:
Question 54 |
A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is
 | |
 | |
 | |
 |
Question 54 Explanation:
The possibilities of getting exactly one odd number is {OEE, EOE, EEO}.
Total no. of possibilities are 8.
Probability of getting exactly one odd = 3/8
Total no. of possibilities are 8.
Probability of getting exactly one odd = 3/8
Question 55 |
The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow?
0.3 | |
0.25 | |
0.35 | |
0.4 |
Question 55 Explanation:
P(Today) = 0.5
P(Tomorrow) = 0.6
P(T∪To) = 0.7
P(T∩To) = P(T) + P(To) - P(T∪To)
= 0.5 + 0.6 - 0.07
= 1.1 - 0.7
= 0.4
P(Tomorrow) = 0.6
P(T∪To) = 0.7
P(T∩To) = P(T) + P(To) - P(T∪To)
= 0.5 + 0.6 - 0.07
= 1.1 - 0.7
= 0.4
Question 56 |
Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is
 | |
 | |
 | |
 |
Question 56 Explanation:
1 - no. 6 on both dice
1 - (5/6 × 5/6) = 1 - (25/36) = 11/36
1 - (5/6 × 5/6) = 1 - (25/36) = 11/36
Question 57 |
The probability that top and bottom cards of a randomly shuffled deck are both aces in
 | |
 | |
| |
 |
Question 57 Explanation:
E1 : First card being ace
E2 : Last card being ace
Note that E1 and E2 are dependent events, i.e., probability of last card being ace if first is ace will be lesser than the probability of last card being ace if first card is not ace.
So, probability of first card being ace = 4/52
Probability of last card being ace given that first card is ace is,
P(E2 / E1) = 3/51
∴ P(E1 and E2) = P(E1) ⋅ P(E2 / E1) = 4/52 * 3/51
E2 : Last card being ace
Note that E1 and E2 are dependent events, i.e., probability of last card being ace if first is ace will be lesser than the probability of last card being ace if first card is not ace.
So, probability of first card being ace = 4/52
Probability of last card being ace given that first card is ace is,
P(E2 / E1) = 3/51
∴ P(E1 and E2) = P(E1) ⋅ P(E2 / E1) = 4/52 * 3/51
Question 58 |
The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is:
 | |
 | |
 | |
 |
Question 58 Explanation:
Question 59 |
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is:
 | |
 | |
 | |
 |
Question 59 Explanation:
Probability of first ball white and second one black is,
Probability of first ball black and second one white is,
Probability of first ball black and second one white is,
Question 60 |
Let A and B be any two arbitrary events, then, which one of the following is true?
P(A∩B) = P(A)P(B) | |
P(A∪B) = P(A) + P(B) | |
P(A|B) = P(A∩B)P(B) | |
P(A∪B) ≤ P(A) + P(B) |
Question 60 Explanation:
(A) Happens when A and B are independent.
(B) Happens when A and B are mutually exclusive.
(C) Not happens.
(D) P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).
(B) Happens when A and B are mutually exclusive.
(C) Not happens.
(D) P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).
Question 61 |
P2 + P3 |
Question 61 Explanation:
P(A∩B') = P(A) - P(A∩B)
P3 = P(A) - P2
P(A) = P2 + P3
P3 = P(A) - P2
P(A) = P2 + P3
Question 62 |
Let A, B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is __________
0.93 |
Question 62 Explanation:
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) + P(A∩B∩C)
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 - 0.4 - 0.5 - 0.24 + 0.12
= 0.93
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 - 0.4 - 0.5 - 0.24 + 0.12
= 0.93
Question 63 |
Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?
 | |
| |
 | |
 |
Question 63 Explanation:
Probability of all accidents on sunday = 1/77
Such as total no. of days in a week = 7
Total probability = 7 × 1/77 = 1/76
Such as total no. of days in a week = 7
Total probability = 7 × 1/77 = 1/76
Question 64 |
Which of the following statements are FALSE?
For poisson distribution, the mean is twice the variance. | |
In queuing theory, if arrivals occur according to poisson distribution, then the inter-arrival time is exponentially distributed. | |
The distribution of waiting time is independent of the service discipline used in selecting the waiting customers for service. | |
If the time between successive arrivals is exponential, then the time between the occurences of every third arrival is also exponential. | |
Both (A) and (C). |
Question 64 Explanation:
In Poisson distribution, the mean is not equal to the twice the variance.
Option C is also False, because waiting time is dependent on the service discipline.
Option C is also False, because waiting time is dependent on the service discipline.
There are 64 questions to complete.