Test: Probability- 2

# Test: Probability- 2

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## 30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2023 Mock Test Series | Test: Probability- 2

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*Answer can only contain numeric values
Test: Probability- 2 - Question 1

### Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is  . The value of X is ______

Detailed Solution for Test: Probability- 2 - Question 1

There are only two possible sets whose elements sum to 22 : {6,6,6,4}, {6,6,5,5}
Number of permutations for 1st set : 4!/3! = 4
Number of permutations for 2nd set : 4!/(2!*2!) = 6
So total number of ways to sum 22 = 10
So X =10.

Test: Probability- 2 - Question 2

### 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) and respectively are

Detailed Solution for Test: Probability- 2 - Question 2

It immediately follows from the monotonicity property that ,
0 <= P(E) <= 1 ,
probability of atleast one means union of the probability of events , i.e.
P(A U B) = P(A) + P(B) - P(A⋂B)
here , P(AUB) = 1 , because it can not be more than 1 and if atleast one of event have probability 1 (here , P(A) = 1) , then union of both should be 1 .
so , P(A U B) = P(A) + P(B) - P(A⋂B)
1 = 1 + 1/2 - P(A⋂B) ,
P(A⋂B) = 1/2 ,
now , P(A/B) = P(A⋂B) / P(B) = (1/2) / (1/2) = 1 ,
P(B/A) = P(A⋂B) / P(A) = (1/2) / (1) = 1/2 .
hence option (D) ,
NOTE :- if atleast one of the two events has/have the probability 1 . then both events should be independent events but vise - versa not true .

Test: Probability- 2 - Question 3

### You have to play three games with opponents A and B in a specified sequence. You win the series if you win two consecutive games. A is a stronger player than B. Which sequence maximizes your chance of winning the series?

Detailed Solution for Test: Probability- 2 - Question 3

Let the probability of winning against player A be  and the probability of winning against player B be b.

be the probability of winning the series in which the games played are against x,y and z in order.

We can see that not all probabilities are equal, so option E is not correct.
We can also see that options A and D result in the same value, so they are not correct either.
Comparing option B and option C.

Hence, option B is the correct answer.

Test: Probability- 2 - Question 4

Three dice are rolled independently. What is the probability that the highest and the lowest value differ by 4?

Detailed Solution for Test: Probability- 2 - Question 4

Case 1:largest is 5 , smallest 1 and middle is 2 or 3 or 4 : 3*3!
Case 2.largest is 5 , smallest 1 and middle is 1 or 5 : 3!*2/2!
Case 3:largest is 6 , smallest 2 and middle is 3 or 4 or 5 : 3*3!
Case 4:largest is 6 , smallest 2 and middle is 6 or 2: 3!*2/2!
So probability the highest and the lowest value differ by 4 =( 3*3!+3!*2/2!+3*3!+3!*2/2!)/63 =2/9

Test: Probability- 2 - Question 5

The probability of throwing six perfect dices and getting six different faces is

Detailed Solution for Test: Probability- 2 - Question 5

B) for all the 6 different faces 1,2,3,4,5,6 the probability is
1/66 and then for
6! different permutations :
6! / 66

Test: Probability- 2 - Question 6

There are 1000 balls in a bag, of which 900 are black and  are white. I randomly draw 100 balls from the bag. What is the probability that the 100st ball will be black?

Detailed Solution for Test: Probability- 2 - Question 6

Here we are having a total of 1000 Balls, out of which we firstly draw 100 balls , and then 101st ball..
Firstly we have to find expected number of white and black balls in drawn 100 balls , as both can occur in 100 balls..
We have a situation like this:

Expected number of white balls = n*W/N = 100*(100/1000) = 10
Expected number of black balls = n*B/N = 100*(900/1000) = 90
So,  we have drawn 100 balls(90 black, 10 white)
Left balls = (810 Black , 90 white) = 900 total
Now,

probability for 101st ball to be black = 810/900 = 9/10
So, option (A) is Correct

Test: Probability- 2 - Question 7

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

Detailed Solution for Test: Probability- 2 - Question 7

there n binary bits that can differ but only d should differ in this case
ways of choosing these d bits = nCd probability of d bits differ but, n - d bits do not differ = (1/2)d(1/2)(n - d)
no of ways = nCd/2n

Test: Probability- 2 - Question 8

A biased coin is tossed repeatedly. Assume that the outcomes of different tosses are independent and probability of heads is 2/3 in each toss. What is the probability of obtaining an even number of heads in  tosses, zero being treated as an evennumber?

Detailed Solution for Test: Probability- 2 - Question 8

probability of obtaining an even number of heads in 5 tosses, zero being treated as an even number number of event = 0 head or 2 head or 4 head
Probability of tail = 1/3
Probability =
= 121/243

Test: Probability- 2 - Question 9

You are lost in the National park of Kabrastan. The park population consists of tourists and Kabrastanis. Tourists comprise two-thirds of the population the park, and give a correct answer to requests for directions with probability 3/4. The air of Kabrastan has an amnesaic quality however, and so the answers to repeated questions to tourists are independent, even if the question and the person are the same. If u ask a Kabrastani for directions, the answer is always wrong.
Suppose you ask a randomly chosen passer by whether the exit from the park is East or West. The answer is East. You then ask the same person again, and the reply is again East. What is the probability of East being correct?

Detailed Solution for Test: Probability- 2 - Question 9

Tourists are 2/3 of population of the park
Among them probability of correct answer is 3/4
so, 2/3*3/4=1/2

Test: Probability- 2 - Question 10

Doctors A and B perform surgery on patients in stages III and IV of a disease. Doctor A has performed a 100 surgeries (on 80 stage III and 20 stage  patients) and 80 out of her 100 patients have survived (78 stage III and 2 stage IV survivors). Doctor B has also performed 100 surgeries (on 50 stage III and 50 stage IV patients). Her success rate is  (49 stage III survivors and 11 stage IV survivors). A patient has been advised that she is equally likely to besuffering from stage III or stage IV of this disease. Which doctor would you recommend to this patient and why?

Detailed Solution for Test: Probability- 2 - Question 10

As, %of Doctor A successful for stage III 78/80*100=97.5
for stage IV 2/20*100=10
As, %of Doctor B successful for stage III 49/50*100=98
stage IV 11/50*100=22

Test: Probability- 2 - Question 11

An unbiased die is thrown n times. The probability that the product of numbers would be even is

Detailed Solution for Test: Probability- 2 - Question 11

Even number = 2,4,6
odd number = 1,3,5
Product will come even when even one time even number comes odd product will be if every time odd comes

Test: Probability- 2 - Question 12

If a fair coin is tossed four times. What is the probability that two heads and two tails will result?

Detailed Solution for Test: Probability- 2 - Question 12

out of 4 times 2 times head should be present
ways of selecting these 2 places 4C2 probability of getting 2 heads and 2 tails = (1/22)(1/22)
probability = 4C2/24 = 3/8

Test: Probability- 2 - Question 13

Consider a -sided die with all sides not necessarily equally likely such that probability of an even number is , probability of a multiple of    and probability of . Given the above conditions, choose the strongest (most stringent) condition of the following that must always hold about , the probability of 5.

Detailed Solution for Test: Probability- 2 - Question 13

P{3,5} = 1 - P{2,4,6} - P{1} = 1/2 - 1/6 = 1/3
Can P{3} = 0? then P{6} = 1/3 and P{2,4} = 1/2 - 1/3 = 1/6. And P{5} = 1/3. Possible. So, option D.

Test: Probability- 2 - Question 14

Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is

Detailed Solution for Test: Probability- 2 - Question 14

probability of getting all heads = 1/16
probability of getting all tails = 1/16
probability of getting at least one head and one tail = 1 - 1/16 - 1/16 = 7/8

Test: Probability- 2 - Question 15

The probability of three consecutive heads in four tosses of a fair coin is.

Detailed Solution for Test: Probability- 2 - Question 15

Let the 4 tosses be named P,Q,R and S

Thus, the probability of getting 3 consecutive heads is given by:

Hence, option D is the correct answer.
Another way of looking at it is:

*Answer can only contain numeric values
Test: Probability- 2 - Question 16

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 _________.

Detailed Solution for Test: Probability- 2 - Question 16

We know that the sum of all the probabilities is 1.
Therefore on integrating    with limits a to 1, the result should be 1.

Hence a = 0.5

Integrate 1/x2 in the limits a to 1.. and equate it to 1.. solving we get a=0.5

*Answer can only contain numeric values
Test: Probability- 2 - Question 17

Consider the following experiment.
Step 1. Flip a fair coin twice.
Step 2. If the outcomes are (TAILS, HEADS) then output Y and stop.
Step 3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop.
Step 4. 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) ____________.

Detailed Solution for Test: Probability- 2 - Question 17

1st time it is 0.25(1/4), when tail tail comes, entire process gets repeated, so next time probability of Y to happen is 0.25*0.25 ((1/4)*(1/4)), likewise it goes on as infinite GP
Sum of infinite GP = a/(1-r)
here a= 1/4 and r = 1/4
so answer becomes 1/3 i.e 0.33

P(HH + HT)=1/2
now if TT comes then toss again,
So, P(TTTH)= 1/16 . and so on.... P(TH+TTTH+........) = 1/4 + 1/16+... = 1/3

Test: Probability- 2 - Question 18

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 same. 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?

Detailed Solution for Test: Probability- 2 - Question 18

Given that probability of getting 2 or  4 or 6 is same.

Test: Probability- 2 - Question 19

A man has three cats. At least one is male. What is the probability that all three are male?

Detailed Solution for Test: Probability- 2 - Question 19

Given that A man has three cats and  At least one is male.
Possible combination for atleast one cat is male=(M,F,F),(F,M,F),(F,F,M),(M,F,M),(M,M,F),(F,M,M),(M,M,M)

Probability that all three are male = 1/7
Hence,Option(B)
1/7 is the correct Choice.

Test: Probability- 2 - Question 20

The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is:

Detailed Solution for Test: Probability- 2 - Question 20

First digit can be chosen in 8 ways from 1−9 excluding 7
Second digit can be chosen in 9 ways from 0−9 excluding 7 and similarly the third digit in 9 ways.
So, total no. of ways excluding 7 = 8∗9∗9
Total no. of ways including 7 = 9∗10∗10
So, ans = (8∗9∗9)/(9∗10∗10)=18/25

Test: Probability- 2 - Question 21

The 12 houses on one side of a street are numbered with even numbers starting at 2 and going up to 24. A free newspaper is delivered on Monday to 3 different houses chosen at random from these 12. Find the probability that at least 2 of these newspapers are delivered to houses with numbers strictly greater than 14.

Detailed Solution for Test: Probability- 2 - Question 21

There are  12 houses on one side of a street are numbered with even numbers.
In which 5 houses are strictly greater than Number 14.
And remaining 7 houses are numbered smaller than 14 (i.e. including 14)
No of way of choosing at least 2 of these newspapers are delivered to houses with numbers strictly greater than 14.
5C3  +5C2 * 7C1 =80
Total way of choosing 3 houses=12C3 =220
So Required probability=80/220=4/11

Test: Probability- 2 - Question 22

You arrive at a snack bar and you can’t decide whether to order a lime juice or a lassi. You decide to throw a fair 6-sided die to make the choice, as follows.

• If you throw 2 or 6 you order a lime juice.
• If you throw a 4, you order a lassi.
• Otherwise, you throw the die again and follow the same algorithm.

What is the probability that you end up ordering a lime juice?

Detailed Solution for Test: Probability- 2 - Question 22

If we want lime juice then we need to throw {2 or 6} . And If we don't get {2 or 6} in first go, then we need to throw {1 or 3 or 5} in first go and again we will try to get {2 or 6} in 2nd throw and so on.
Note - If we throw 4 in any go then we will end up getting lassi. But we want lime juice.
So probability of getting lime juice = (2/6)  + (3/6) * (2/6) + (3/6)2 * (2/6) + .......  to infinity.

Test: Probability- 2 - Question 23

Consider the  function on three bits, maj : where maj   if and only if  be the probability that the output is 1 when each input is set to 1 independently with probability α. What is

Detailed Solution for Test: Probability- 2 - Question 23

Test: Probability- 2 - Question 24

If P and Q are two random events, then the following is TRUE:

Detailed Solution for Test: Probability- 2 - Question 24

here i check option---> i let P = probability,  P and Q is replaced by A and B.
A) it  is saying independent means P(A ^ B) =  P(A)* P(B)  so option A is false
B) P(AÜ B) =  P(A) + P(B) - P(A ^ B)
hence P(AÜB) >= P(A) + P(B)  its false option B is false
C) if A and B is mutually exclusive means P(A ^ B) =  0 then its independent ........its false  there is no relation independent and mutually exclusive
D)  P(A ^ B) <= P(A)

its true

hence option D is correct

Test: Probability- 2 - Question 25

Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?

Detailed Solution for Test: Probability- 2 - Question 25

for every car accident we can pick a day in 7 ways
total number of ways in which accidents can be assigned to days = 77
probability of accidents happening on a particular day = 1/77
we can choose a day in 7 ways
hence probability = 7/77= 1/76

Test: Probability- 2 - Question 26

E1 and E2 are events in a probability space satisfying the following constraints:

• Pr(E1) = Pr(E2)
• Pr(E1 ∪ E2) = 1
• E1 and E2 are independent

The value of Pr(E1), the probability of the event E1, is

Detailed Solution for Test: Probability- 2 - Question 26

let probability of Event E1 = x = prob of E2
prob(E1 union E2) = prob(E1) + prob(E2) - prob(E1 intersect E2)
1 = x + x -x2 (prob(E1 intersect E2) = prob(E1) * prob(E2) as events are independent)
x = 1

Test: Probability- 2 - Question 27

The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from same suit is

Detailed Solution for Test: Probability- 2 - Question 27

There are 4 sets of cards. So, up till 8 cards there is a chance that no more than 2 cards are from a given set. But, once we pick the 9th one, it should make  3 cards from any one of the sets. So, (C) is the answer.

Test: Probability- 2 - Question 28

A bag contains  white balls and  black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is:

Detailed Solution for Test: Probability- 2 - Question 28

probability of first ball white and second one black = (10/25)x(15/24)
probability of first ball black and second one white = (15/25)x(10/24)
probabilty = sum of above two probabilities = 1/2

*Answer can only contain numeric values
Test: Probability- 2 - Question 29

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______ .

Detailed Solution for Test: Probability- 2 - Question 29

no of integers divisible by 2 = 50
no of integers divisible by 3 = 33
no of integers divisible by 5 = 20
no of integers divisible by 2 and 3 = 16
no of integers divisible by 2 and 5 = 10
no of integers divisible by 3 and 5 = 6
no of integers divisible by 2 and 3 and 5 = 3
total numbers divisible by 2 or 3 or 5 = 50 + 33 + 20 -16 -10 - 6 + 3 = 74
total number not divisible by 2 or 3 or 5 = 26
probability = 0.26 [EDIT]

Test: Probability- 2 - Question 30

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?

Detailed Solution for Test: Probability- 2 - Question 30

prob(at least one head) = 3/4
using bayes' theorem = (1/4)/(3/4) = 1/3

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