Question

65% of the people in Missouri pass the driver’s test on the first attempt. A group of 5 people took the test. What is the probability that less than 3 in the group pass their driver's tests in their first attempt?

Round your answer to three decimal places.

Remember: 65% = 0.65.

Answer 1

0.235

Explanation:

The parameters of this binomial experiment are:

n=5 trials

p=0.65

x= less than 3 successes

This means that x takes on more than one value. Less than 3 means that x can be 0,1, or 2 successes.

Now we have that

P(X=0)=5C0⋅0.650⋅0.355=5!0!5!⋅0.650⋅0.355≈1⋅1⋅0.0053≈0.005

P(X=1)=5C1⋅0.651⋅0.354=5!1!4!⋅0.651⋅0.354≈5⋅0.65⋅0.015≈0.049

and

P(X=2)=5C2⋅0.652⋅0.353=5!2!3!⋅0.652⋅0.353≈10⋅0.4225⋅0.0429≈0.181

So P(X<3)=0.005+0.049+0.181=0.235

Answer 2

Explanation:

This problem involves a binomial distribution where the probability of success (passing on the first attempt) is 0.65, and the probability of failure (not passing on the first attempt) is 0.35. We want to find the probability that less than 3 out of 5 people pass on their first attempt.

We can use the binomial probability formula to calculate this:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Where X is the number of people who pass on their first attempt.

P(X = 0) = (5 choose 0) * 0.65^0 * 0.35^5 = 0.0024

P(X = 1) = (5 choose 1) * 0.65^1 * 0.35^4 = 0.0289

P(X = 2) = (5 choose 2) * 0.65^2 * 0.35^3 = 0.1323

So,

P(X < 3) = 0.0024 + 0.0289 + 0.1323 = 0.1636

Therefore, the probability that less than 3 out of 5 people pass their driver's tests on the first attempt is approximately 0.164, rounded to three decimal places.