Question

In one city, the probability that a person will pass his or her driving test on the first attempt is 0.68. Eleven people are selected at random from among those taking their driving test for the first time. What is the probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

Answer 1

Answer:

0.0308 = 3.08% probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they pass the test on their first attempt, or they do not. Each person is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In one city, the probability that a person will pass his or her driving test on the first attempt is 0.68.

This means that $p = 0.68$

Eleven people are selected at random from among those taking their driving test for the first time.

This means that $n = 11$

What is the probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

This is

$P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)$.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{11,2}.(0.68)^{2}.(0.32)^{9} = 0.0009$

$P(X = 3) = C_{11,3}.(0.68)^{3}.(0.32)^{8} = 0.0057$

$P(X = 4) = C_{11,4}.(0.68)^{4}.(0.32)^{7} = 0.0242$

$P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0057 + 0.0242 = 0.0308$.

0.0308 = 3.08% probability that among these 11 people, the number passing the test is between 2 and 4 inclusive