Questions: 65% of the people in Missouri pass the driver's test on the first attempt. A group of 7 people took the test. What is the probability that at most 3 in the group pass their driver's tests in their first attempt? Round your answer to three decimal places. Remember: 65% = 0.65. Select the correct answer below: 0.944 0.144 0.468 0.200

65% of the people in Missouri pass the driver's test on the first attempt. A group of 7 people took the test. What is the probability that at most 3 in the group pass their driver's tests in their first attempt? Round your answer to three decimal places.

Remember: 65% = 0.65.

Select the correct answer below:
0.944
0.144
0.468
0.200

Transcript text: 65\% of the people in Missouri pass the driver's test on the first attempt. A group of 7 people took the test. What is the probability that at most 3 in the group pass their driver's tests in their first attempt? Round your answer to three decimal places. Remember: $65 \%=0.65$. Select the correct answer below: 0.944 0.144 0.468 0.200

Solution

Solution Steps

Step 1: Define the Problem

We are given that $65\%$ of the people in Missouri pass the driver's test on the first attempt, which translates to a probability of success $p = 0.65$. A group of $n = 7$ people took the test, and we need to find the probability that at most $3$ people pass their driver's tests on their first attempt.

Step 2: Calculate Individual Probabilities

We will calculate the probabilities of exactly $0$, $1$, $2$, and $3$ successes using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where $q = 1 - p = 0.35$.

For $x = 0$: \[ P(X = 0) = \binom{7}{0} \cdot (0.65)^0 \cdot (0.35)^7 = 0.001 \]
For $x = 1$: \[ P(X = 1) = \binom{7}{1} \cdot (0.65)^1 \cdot (0.35)^6 = 0.008 \]
For $x = 2$: \[ P(X = 2) = \binom{7}{2} \cdot (0.65)^2 \cdot (0.35)^5 = 0.047 \]
For $x = 3$: \[ P(X = 3) = \binom{7}{3} \cdot (0.65)^3 \cdot (0.35)^4 = 0.144 \]

Step 3: Calculate Cumulative Probability

To find the probability that at most $3$ people pass, we sum the probabilities of $0$, $1$, $2$, and $3$ successes:

\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]

Calculating this gives:

\[ P(X \leq 3) = 0.001 + 0.008 + 0.047 + 0.144 = 0.200 \]