98% of drivers stop at every red light. Some percentage of those that stop at every red light also stop completely at every stop sign. There is a 77% probability that a driver stops at a red light and stops completely at stop signs. Given that a driver stops at every red light, what is the probability that they also stop completely at every stop sign, rounded to the nearest whole percent?

79%
76%
28%
131%

1 answer

Let's denote the events as follows:

  • \( A \): The event that a driver stops at every red light.
  • \( B \): The event that a driver stops completely at every stop sign.

From the information provided in the problem:

  • \( P(A) = 0.98 \) (98% of drivers stop at every red light).
  • \( P(A \cap B) = 0.77 \), which means the probability that a driver both stops at every red light and stops completely at every stop sign is 77% of drivers.

Now we want to find the conditional probability \( P(B | A) \), which is the probability that a driver stops completely at every stop sign given that they stop at every red light. We can use the definition of conditional probability:

\[ P(B | A) = \frac{P(A \cap B)}{P(A)} \]

Substituting the known values into the formula, we find:

\[ P(B | A) = \frac{0.77}{0.98} \]

Now, let's perform the calculation:

\[ P(B | A) = \frac{0.77}{0.98} \approx 0.7857142857 \]

To convert this to a percentage, we multiply by 100:

\[ P(B | A) \approx 78.57% \]

Rounding to the nearest whole percent, we get:

\[ P(B | A) \approx 79% \]

Thus, the probability that a driver stops completely at every stop sign, given that they stop at every red light, is approximately \( 79% \).