Question

An online test allows a maximum of 10 attempts. Abdulllah will attempt the test until he
passes it. In each attempt his chance of passing the test is 40%. Find the following:

a) What is the probability that he is able to pass the test?
b) What is the expected number of attempts to pass the test?
c) What is the variance of number of attempts?
d) Given that he passed the test, what is the probability that he passed in less than 5 attempts?
e) Given that he passed the test, what is the expected number of attempts?

Answer 1

Total number of attempts = 10

Probability of passing in each attempt = 0.40

(a) The probability that he is able to pass the test is the probability that he passes in one of the 10 attempts. Assumping that the probability of passing and failing is the same in each attempt and is independent of the previous attempts, this can be denoted using the geometric series as follows:

Passes in first attempt + Does not pass in first attempt but passes in second + ... + Does not pass in first 9 attempts but passes in the 10th attempt

Using the formula for sum of n terms of a geometric series gives us:

So, the probability that he is able to pass the test is 0.9940.

(b) The given situation can be modelled using a type 2 geometric distribution. The random variable X which denotes the number of attempts before he passes the test can be determined as a geometric distribution with parameters n=10 and p=0.40.

So, the expected number of attempts is the mean of the above distribution given by 1/p, i.e. 1/0.40 = 2.5.

(c) Similary, the variance of the number of attempts is given by (1-p)/p2 i.e. (1-0.4)/0.42 = 3.75

(d) The probability that he passed in less than 5 attempts, given that he passed the test is given as:

P(Passes and Passes in Less than 5 attempts | Pass)

Now its reasonable that P(Passes and Passes in Less than 5 attempts) = P(Passes in Less than 5 attempts), So we have:

P(Passes in Less than 5 attempts) \ P(Passes)

The numerator can be solved using geometrc series as before, so we have:

P(Passes in Less than 5 attempts) =

P(Passes in Less than 5 attempts) =

and, P(Passes) = 0.9940 from (a)

Thus, the probability that he passed in less than 5 attempts, given that he passed the test is 0.9222/0.9940 = 0.9278

(e) The new expected number of attempts given that he passed the test is given as:

2.5/0.9940 = 2.51