Probability
Question 1 
0.021  
0.022  
0.023  
0.024 
⇾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 them coins 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 2 
0  
1  
2  
3 
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 3 
Nβ(1β)  
Nβ  
N(1β)  
Not expressible in terms of N and β alone 
Given g_{y} (z)=(1β+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(x_{j},n,p)=
The probability Mass function,
Given:
Question 4 
4/5  
5/6  
7/8  
11/12 
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')=1P(A∪B))
P(p'∩q')=1P(p∪q)
=1(P(p)+P(q)P(p∩q))
=1(P(p)+P(q)P(p)∙P(q))
=1(1/4+1/61/12)
=1(10/242/24)
=1(8/24)
=2/3
(P(p'∩q')=2/3) ⇾⑥
Substitute in ⑤,
P(p'⁄q')=(2⁄3)/(1P(q))=(2⁄3)/(11/6)=(2⁄3)/(5⁄6)=4/5
(P(p'/q')=4/5)
Question 5 
54  
55  
56  
57 
Mean = Variance
E(X)=E(X^{2})  (E(X))^{2} = 5
E(X^{2}) = 5+(E(X))^{2} = 5+25 = 30
So, E[(X+2)^{2}] = E[X^{2} + 4 +4X]
= E(X^{2})+4+4E(X) = 30+4+4×5
= 54
Question 6 
0.7  
0.6  
0.5  
0.8 
or
where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x^{2} , a≤x≤1
The area under curve,
1 + 1/a = 1
1/a = 2
a = 0.5
Question 7 
0.33  
0.34  
0.35  
0.36 
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)/(11/4)
=1/3
=0.33
Question 8 
0.55  
0.56  
0.57  
0.58 
The bulbs of Type 1, Type 2 are same in number.
So the probability to choose a type is ½.
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 9 
0.75  
0.76  
0.77  
0.78 
Pr(Y=0/ X_{3}=0) = Pr(Y=0 ∩ X_{3}=0)/ Pr(X_{3}=0)
= 3/8 / 4/8 = 3/4 = 0.75
Question 10 
10  
11  
12  
13 
To get ‘22’ as Sum of four outcomes
x_{1} + x_{2} + x_{3} + x_{4} = 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+x_{1}+x_{2} = 22
x_{1}+x_{2} = 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 11 
11.90  
11.91  
11.92  
11.93 
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’ nonworking 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 nonworking 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 12 
3.9  
4.0  
4.1  
4.2 
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 13 
0.259 to 0.261  
0.260 to 0.262  
0.261 to 0.263  
0.262 to 0.264 
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)=1P(A∪B∪C)=0.26
Question 14 
0.25  
0.26  
0.27  
0.28 
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 15 
8/(2e^{3})  
9/(2e^{3})  
17/(2e^{3})  
26/(2e^{3}) 
P(x:λ)=(e^{λ} λ^{x})/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} 3^{0})/0!+(e^{3} 3^{1})/1!+(e^{3} 3^{2})/2!
=e^{3}+e^{3}∙3+(e^{3})∙9)/2
=(17e^{3<})/2
=17/(2e^{3} )
Question 16 
0 and 0.5  
0 and 1  
0.5 and 1  
0.25 and 0.75 
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 17 
10/21  
5/12  
2/3  
1/6 
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 18 
R = 0  
R < 0  
R ≥ 0  
R > 0 
So the answer will be R≥0.
Question 19 
1/3  
1/4  
1/2  
2/3 
Question 20 
1/5  
4/25  
1/4  
2/5 
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
^{5}C_{1} × ^{4}C_{1} = 20
So probability = 4/20 = 1/5
Question 21 
pq + (1  p)(1  q)
 
(1  q)p
 
(1  p)q
 
pq 
A faulty product can be correctly assembled &resulted correctly with (AT) case, where the probability is p*q.
A faulty product can be assembled faulty & resulted wrongly which is A’T’ case, where probability = assembly fail probability* computer testing fail probability = (1p)*(1q)
Total probability = pq +(1p)*(1q).
Question 22 
1/625  
4/625  
12/625
 
16/625 
We can write 10^{99} as 10^{96}×10^{3}
So, (10^{99})/(10^{96}) to be a whole number, [10^{96}×10^{3}/10^{96}]➝ (1)
We can observe that every divisor of 10^{3} is a multiple of 10^{96}
So number of divisor of 10^{3} to be found first
⇒ 10^{3}=(5×2)^{3}=2^{3}×5^{3}
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 10^{99} are 10^{99}=2^{99}×5^{99}=100×100=10000
Probability that divisor of 10^{99} is a multiple of 10^{96} is
⇒16/10,000
Question 23 
0.453
 
0.468
 
0.485  
0.492

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 24 
0.24
 
0.36  
0.4  
0.6

(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 25 
3  
2  
√2  
1 
We can compare their values using standard normal distributions.
The above equation satisfies when σ_{y} will be equal to 3.
Question 26 
11/12  
10/12  
9/12  
8/12 
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+43/ 6
= 5/6
= 10/12
Question 27 
Question 28 
1/2  
1/10  
9!/20!  
None of these

→ 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 29 
7/8  
1/2  
7/16  
5/32 
The probability of obtaining heads
= (1/2)(5/8) + (1/2)(1/4)
= (5/16) + (1/8)
= 7/16
Question 30 
5  
6  
7  
10 
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 31 
In a multiuser 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 × 10^{6} × e^{20}  
1.02 × 10^{6} × e^{20}  
6.9 × 10^{3} × e^{20}  
1.02 × 10^{3} × e^{20} 
So, λ=15
Question 32 
1/n  
2  
√n  
n 
The given probability P_{i} is for selection of each item independently with probability 1/2.
Now, Probability for x_{1} to be smallest in S = 1/2
Now, Probability for x_{2} to be smallest in S = Probability of x_{1} not being in S × Probability of x_{2} being in S
= 1/2 × 1/2
Similarly, Probability x_{i} to be smallest = (1/2)^{i}
Now the Expected value is
Question 33 
No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nC_{n}*(1/2)^{n}*(1/2)^{n}=2nC_{n}/4^{n}
Question 34 
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 
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 35 
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/(1p)  
1/p^{2}  
1/(1p^{2}) 
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 36 
f(b  a)
 
f(b)  f(a)  
Then the probablity be area of the corresponding curve i.e.,
Question 37 
1/2^{n}
 
1  1/n
 
1/n!
 
1(1/2^{n})

Hence Probability = (2^{n}  1) /2^{n} = 1  1/2^{n}
Question 38 
4/19
 
5/19
 
2/9  
19/30 
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 39 
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 
3  
4  
5  
6 
Question 41 
3/8  
1/2  
5/8  
3/4 
Then total number of possibilities = 2^{4} = 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 42 
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 
Probability of selecting a wrong answer = 3/4
For correct answer +1, for wrong answer0.25;
Expected marks for each question = (1/4) × 1 + (3/4) (0.25)
= 1/4 + (3/16)
= 43/16
= 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 43 
^{n}C_{d} /2^{n}
 
^{n}C_{d} / 2^{d}
 
d/2^{n}
 
1/2^{d}

Total no. of cases where n positions have any binary bit = 2^{n}
The probability of 'd' bits differ = ^{n}C_{d} / 2^{n}
Question 44 
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) = 1/10 = 1
Y goes from 0 to 2, so PDF(y) = 1/20 = 1/2
Now we evaluate,
Question 45 
Question 46 
44  
96  
120  
3125 
Question 47 
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 48 
→ They representing the probability density functions of time taken to execute.
→ f_{1} can be executed in 'x' time.
f_{2} can be executed in 'tx' time.
→ The probability density function =
Question 49 
Atleast one head = 1/16
Atleast one tail = 1/16
Probability of getting one head and one tail is = 1  1/16  1/16 = 1611/16 = 14/16 = 7/8
Question 50 
0  
1 
Pr(E_{1} ∪ E_{2}) = 1
E_{1} and E_{2} are Independent.
P(E_{1} ∪ E_{2}) = Pr(E_{1}) + Pr(E_{2})  Pr(E_{1} ∩ E_{2})
1 = a + a  Pr(E_{1})⋅Pr(E_{2})
1 = a + a  a⋅a
2a  a^{2} = 1
a^{2}  2a + 1 = 0
(a  1)^{2} = 0
a = 1 Pr(E_{1}) = 1
Answer: Option (D)
Question 51 
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 
E(X) = x_{1}P_{1} + x_{2}P_{2} + ... + x_{n}P_{n}
In question, E(X) is given as 5.
E(X) = 5, 0≤P_{i}≤1
P_{1} + P_{2} + ... + P_{n} = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the x_{i} value is greater than 5.
Question 52 
Events E_{1} and E_{2} are independent  
Events E_{1} and E_{2} are not independent  
Question 53 
Total no. of possibilities are 8.
Probability of getting exactly one odd = 3/8
Question 54 
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 
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 55 
1  (5/6 × 5/6) = 1  (25/36) = 11/36
Question 56 
E_{2} : Last card being ace
Note that E_{1} and E_{2} 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(E_{2} / E_{1}) = 3/51
∴ P(E_{1} and E_{2}) = P(E_{1}) ⋅ P(E_{2} / E_{1}) = 4/52 * 3/51
Question 57 
Question 58 
Probability of first ball black and second one white is,
Question 59 
P(A∩B) = P(A)P(B)  
P(A∪B) = P(A) + P(B)  
P(AB) = P(A∩B)P(B)  
P(A∪B) ≤ P(A) + P(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 60 
P_{2} + P_{3} 
P_{3} = P(A)  P_{2}
P(A) = P_{2} + P_{3}
Question 61 
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 
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 62 
Such as total no. of days in a week = 7
Total probability = 7 × 1/7^{7} = 1/7^{6}
Question 63 
For poisson distribution, the mean is twice the variance.  
In queuing theory, if arrivals occur according to poisson distribution, then the interarrival 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). 
Option C is also False, because waiting time is dependent on the service discipline.