Question

Consider this scenario: A traffic signal for eastbound traffic is red for 15 seconds, yellow for 5 seconds, and green for 30 seconds. The traffic policemen want to know the probability that out of the next x number of eastbound cars that arrive at that signal(where x>3), 3 of them would be stopped by a red light. What is the probability that out of the next 8 eastbound cars that arrive at the signal, exactly 3 will be stopped by a red light? 0.254 0.552 0.806 0.448

Answer 1

P(3 reds out of 8) 
= 8C3 * [15 / (15 + 5 + 30)]^3 * [(5 + 30) / (15 + 5 + 30)]^(8 - 3) 
= 8C3 * 0.3^3 * 0.7^5 
= 0.25412184 
= 25.4% 
=~ 0.3

Answer 2

Answer:

0.254

Step-by-step explanation:

Given : A traffic signal for eastbound traffic is red for 15 seconds, yellow for 5 seconds, and green for 30 seconds.

To Find: What is the probability that out of the next 8 eastbound cars that arrive at the signal, exactly 3 will be stopped by a red light?

Solution:

A traffic signal for eastbound traffic is red for 15 seconds, yellow for 5 seconds, and green for 30 seconds.

Total time = 15+5+30=50

So, probability of occurring red light = $\frac{15}{50}=0.3$

So, Probability of not occurring red light = $1-0.3=0.7$

Now we are supposed to find  the probability that out of the next 8 eastbound cars that arrive at the signal, exactly 3 will be stopped by a red light

So, we will use binomial

$P(X=r)=^nC_r p^rq^{n-r}$

Substitute n = 8

r = 3

p is the probability of success that is probability of occurring red light = 0.3

q is the probability of failure that is probability of not occurring red light=0.7

So, $P(X=3)=^{8}C_{3} (0.3)^3 (0.7)^{5}$

$P(X=3)=\frac{8!}{3!(8-3)!}(0.3)^3 (0.7)^{5}$

$P(X=3)=0.254$

Thus the probability that out of the next 8 eastbound cars that arrive at the signal, exactly 3 will be stopped by a red light is 0.254

Hence Option A is true.