## Question

### Solution

Correct option is

4/9

Let E1 denote the event that an ace occurs and E2 the event that it does not occur. Let A denote the event that the person reports that it is an ace. Then P(E1) = 1/6, P(E2) = 5/6,  By Bayes’ theorem, #### SIMILAR QUESTIONS

Q1

A group of 6 boys and 6 girls is randomly divided into two equal groups. The probability that each group contains 3 boys and 3 girls is

Q2

In a hurdle race, a runner has probability p of jumping over a specific hurdle. Given that in 5 trials, the runner succeeded 3 times, the conditional probability that the runner had succeeded in the first trial is

Q3

Three integers are chosen at random without replacement from the first 20 integers. The probability that their product is even 2/19.

Q4

A box contains tickets numbered 1 to Nn tickets are drawn from the box with replacement. The probability that the largest number on the tickets is is

Q5

A box contain N coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, when a baised coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

Q6

Given that AB and C are events such that P(A) = P(B) = P(C) = 1/5, P(A B) = P(B ∩ C) = 0 and P(A ∩ C) = 1=10. The probability that at least one of the events AB or C occurs is …….

Q7

Let A and B be two events such that Q8

A and B toss a coin alternatively till one of them gets a head and wins the game. If A begins the game, the probability B wins the game is

Q9

Suppose X B(np) and P(X = 5). If p > 1/2, then

Q10

In a game called “odd man out man out”, m(m > 2) persons toss a coin to determine who will buy refreshments for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is