Question 1

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities

Accepted Answer

Let E_1, E_2, and E_3 be the events that the candidate travels by bus, scooter, and car, respectively. The probabilities of choosing these modes of transport are: P(E_1) = 2 5 P(E_2) = 1 5 P(E_3) = 2 5 Let A be the event that the candidate reaches the examination centre late. The conditional probabilities of reaching late are: P(A|E_1) = 1 5 P(A|E_2) = 1 3 P(A|E_3) = 1 4 We need to find the probability that the candidate travelled by bus given that they reached late, which is P(E_1|A). Using Bay

Question 2

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :

Accepted Answer

The probability that the first pair drawn consists of one blue and one green ball is: P_1 = ^ 6 C_ 1 ^ 6 C_ 1 ^ 12 C_ 2 = 36 66 = 6 11 Since the balls are drawn without replacement, 5 blue and 5 green balls remain. The probability that the second pair consists of one blue and one green ball is: P_2 = ^ 5 C_ 1 ^ 5 C_ 1 ^ 10 C_ 2 = 25 45 = 5 9 Continuing this process for all 6 pairs, the probabilities are: P_3 = ^ 4 C_ 1 ^ 4 C_ 1 ^ 8 C_ 2 = 16 28 = 4 7 P_4 = ^ 3 C_ 1 ^ 3 C_ 1 ^ 6 C_ 2 = 9 15 = 3 5

Question 3

A bag contains (N+1) coins - N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is 9 16 , then N is equal to:

Accepted Answer

Let E_1 be the event of selecting a fair coin and E_2 be the event of selecting the two-headed coin. P(E_1) = N N+1 P(E_2) = 1 N+1 Let H be the event of getting a Head. P(H|E_1) = 1 2 P(H|E_2) = 1 Using the theorem of total probability: P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2) P(H) = ( N N+1 ) ( 1 2 ) + ( 1 N+1 )(1) P(H) = N+2 2(N+1) Given P(H) = 9 16 : N+2 2(N+1) = 9 16 N+2 N+1 = 9 8 8N + 16 = 9N + 9 N = 7 Answer: 7

Question 4

The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tou

Accepted Answer

Let A be the event that player A is selected as captain, and B the event that player B is selected as captain. Let W be the event that the team wins the tournament. Given: P(A) = 0.6, P(B) = 0.4 P(W A) = 0.8, P(W B) = 0.7 By the law of total probability: P(W) = P(A) P(W A) + P(B) P(W B) P(W) = (0.6)(0.8) + (0.4)(0.7) P(W) = 0.48 + 0.28 = 0.76 Hence, the correct option is (2)\ 0.76.

Question 5

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

Accepted Answer

Let E_1 be the event that the letter came from KANPUR and E_2 be the event that the letter came from ANANTPUR. Since the letter is equally likely to come from either city, P(E_1) = P(E_2) = 1 2 . Let A be the event that the two consecutive visible letters are AN. In the word KANPUR, there are 6 letters, so there are 5 pairs of consecutive letters: KA, AN, NP, PU, UR. The pair AN appears exactly once. Thus, P(A|E_1) = 1 5 . In the word ANANTPUR, there are 8 letters, so there are 7 pairs of consec

Question 6

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to a b , where a,b N and (a,b)=1, then a+b is equal to ______

Accepted Answer

The total number of ways to select 3 different dates from a month of 31 days is given by choosing 3 days out of 31: ^ 31 C_ 3 = 31 30 29 3 2 1 = 4495 Let the three selected dates be x, y, z such that x 2y = x + z For y to be an integer, the sum x + z must be an even number. This is only possible if both x and z are either odd or even. Once x and z are chosen, y is uniquely determined as their exact average. In a month of 31 days, the number of odd dates (1, 3, 5, , 31) is 16, and the number of e

Question 7

A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is p, then 96p is equal to _____.

Accepted Answer

Let the outcomes of the 8 tosses be denoted by X_1, X_2, X_3, X_4, X_5, X_6, X_7, X_8. Let A be the number of heads in the first 3 tosses (X_1, X_2, X_3). Let B be the number of heads in the next 3 tosses (X_4, X_5, X_6). Let C be the number of heads in the last 2 tosses (X_7, X_8). According to the given conditions: Number of heads in the first six tosses: A + B = 4 Number of heads in the last five tosses: B + C = 3 Since C is the number of heads in 2 tosses, C 2. Thus, B = 3 - C 1. Also, B 3.

Question 8

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is m n , (m, n) = 1, then m + n i

Accepted Answer

Let h be the number of heads and t be the number of tails obtained. The total points obtained is 10h + 5t. For the man to get exactly 30 points, we must have: 10h + 5t = 30 2h + t = 6 Since h and t are non-negative integers, the possible pairs of (h, t) are (3, 0), (2, 2), (1, 4), and (0, 6). For each pair, the number of possible sequences of heads and tails is given by ^ h+t C_ h , and the probability of any specific sequence is ( 1 2 )^ h+t . The total probability is the sum of the probabiliti

Question 9

The probability distribution of a random variable X is given below : \( array |c|c|c|c|c|c|c|c|c| X & 4k & 30 7 k & 32 7 k & 34 7 k & 36 7 k & 38 7 k & 40 7 k & 6k \ P(X) & 2 15 & 1 15 & 2 15 & 1 5 &

Accepted Answer

E(X) = X_i P(X_i) = 526k 15 7 = 263 15 k = 7 2 Substituting k = 7 2 , the values of X become: X: 14, 15, 16, 17, 18, 19, 20, 21 P(X): 2 15 , 1 15 , 2 15 , 1 5 , 1 15 , 2 15 , 1 5 , 1 15 P(X = 2 15 + 1 15 + 2 15 + 1 5 + 1 15 + 2 15 = 11 15

Question 10

A bag contains 10 balls out of which k are red and (10-k) are black, where 0 k 10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability th

Accepted Answer

Using Bayes' theorem: P(k=1| all black ) = P( all black |k=1) P(k=1) P( all black ) where P(k) = 1 11 for each k \ 0,1,...,10\ . When k=1: P( all 3 black |k=1) = 9 3 10 3 = 84 120 = 7 10 . Calculating P( all black ) = 1 11 _ k=0 ^ 10 10-k 3 120 . Numerator is 120 + 84 + 56 + 35 + 20 + 10 + 4 + 1 = 330. So P( all black ) = 1 11 330 120 = 1 4 . Therefore P(k=1| all black ) = 7 10 1 11 1 4 = 7 10 4 11 = 28 110 = 14 55

Question 11

Let S be a set of 5 elements and P ( S ) denote the power set of S. Let E be an event of choosing an ordered pair ( A , B ) from the set P ( S ) P ( S ) such that A B = . If the probability of the eve

Accepted Answer

S has 5 elements, so P(S) has 2^5 = 32 subsets. Total ordered pairs from P(S) P(S) is 32^2 = 1024 For pairs (A, B) with A B = : each of the 5 elements has 3 choices - in A only, in B only, or in neither Number of favorable outcomes = 3^5 = 243 Probability = 243 1024 = 3^5 2^ 10 With p = 5 and q = 10, we get p + q = 15

Question 12

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

Accepted Answer

Binomial distribution with n=8, p= 1 10 , q= 9 10 . P(X 7) = P(X=7) + P(X=8) P(X=7) = 8 7 ( 1 10 )^7 ( 9 10 ) = 72 10^8 P(X=8) = ( 1 10 )^8 = 1 10^8 P(X 7) = 73 10^8

Question 13

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randoml

Accepted Answer

Bag A: 9W, 8B. Bag B: 6W, 4B. Case 1: White transferred from B (prob = 6 10 ). A becomes 10W, 8B. P( white from A ) = 10 18 = 5 9 . Case 2: Black transferred from B (prob = 4 10 ). A becomes 9W, 9B. P( white from A ) = 9 18 = 1 2 . Total probability = 6 10 5 9 + 4 10 1 2 = 1 3 + 1 5 = 8 15 . p = 8, q = 15, (8,15) = 1. So p + q = 23.

Question 14

From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that a-b 10 is m n , gcd (m, n)=1, then m+n is equal to \_\_\_\_.

Accepted Answer

Total ordered selections = 100 99 = 9900. Favorable (a - b 10): for each b from 1 to 90, a ranges from b+10 to 100, giving 91 - b choices. Count = _ b=1 ^ 90 (91-b) = _ k=1 ^ 90 k = 90 91 2 = 4095. Probability = 4095 9900 = 91 220 (after dividing by 45). (91, 220) = 1, so m = 91, n = 220. m + n = 311.

Question 15

If a random variable x has the probability distribution array |c|c|c|c|c|c|c|c|c| x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \ P(x) & 0 & 2k & k & 3k & 2k^ 2 & 2k & k^ 2 +k & 7k^ 2 \ array then P(3 < x 6) is

Accepted Answer

Sum of all probabilities = 1 0 + 2k + k + 3k + 2k^2 + 2k + (k^2+k) + 7k^2 = 1 10k^2 + 9k - 1 = 0. (10k-1)(k+1) = 0 k = 1 10 (rejecting k = -1). P(3 = 2k^2 + 2k + k^2 + k = 3k^2 + 3k = 3(0.01) + 3(0.1) = 0.33.

Question 16

Two distinct numbers a and b are selected at random from 1,2,3, , 50. The probability, that their product a b is divisible by 3, is

Accepted Answer

Total ways = 50 2 = 1225. ab is divisible by 3 iff at least one of a, b is divisible by 3. Multiples of 3 in \ 1, ,50\ : 3,6, ,48 16 numbers. Non-multiples of 3: 34. P( neither divisible by 3 ) = 34 2 50 2 = 561 1225 . P(ab divisible by 3 ) = 1 - 561 1225 = 664 1225 .

Question 17

A random variable X takes values 0,1,2,3 with probabilities 2 a+1 30 , 8 a-1 30 , 4 a+1 30 , b respectively, where a , b R . Let and respectively be the mean and standard deviation of X such that ^ 2

Accepted Answer

Sum of probabilities = 1: 2a+1+8a-1+4a+1 30 + b = 1 b = 29-14a 30 . ^2 + ^2 = E(X^2) = 2. E(X^2) = 8a-1 30 + 4(4a+1) 30 + 9(29-14a) 30 = -102a + 264 30 = 2. -102a + 264 = 60 a = 2. b = 29 - 28 30 = 1 30 . a b = 2 1/30 = 60.

Probability — Previous Year Questions & Solutions