A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is

Solution :

Probability of getting success, p = \(1\over 6\)

and probability of failure, q = \(5\over 6\)

Now, we must get 2 sixes in seven throws, so probability is \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)

and the probability that 8th throw is \(1\over 6\).

\(\therefore\)   Required Probability = \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)\(1\over 6\)

= \(^7C_2\times 5^5\over {6^8}\)

Similar Questions

The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s second win occurs at the third test is

Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an even sum.

If A and B are two mutually exclusive events, then

A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is \(1\over 2\), \(1\over 3\) and \(1\over 4\). Probability that the problem is solved is

Leave a Comment

Your email address will not be published. Required fields are marked *