Question

At a certain intersection, the light for eastbound traffic is red for 15 seconds, yellow for 5 seconds, and green for 30 seconds. What is the probability that the light is red? Using the binomial distribution, what is the probability that out of the next eight eastbound cars that arrive randomly at the light, exactly three will be stopped by a red light?
a)P(red)=.15,P(exactly3getstopped)=.08

b)P(red)=.15,P(exactly3getstopped)=.97

c)P(red)=.3,P(exactly3getstopped)=.8

d)P(red)=.3,P(exactly3getstopped)=.25

Answer 1

The Probability of red light is 0.3

P(red) = 0.3

The probability that out of the next eight eastbound cars that arrive randomly at the light, exactly three will be stopped by a red light be 0.25.

P(exactly three get stopped by red light)=0.25

The correct option is (d) P(red)=.3,P(exactly 3 get stopped)=.25

What is Binomial distribution?

A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times.

The occurrence,

• Red light= 15 sec,
• Yellow light= 5 sec,
• Green light= 15 sec

p= probability of a red light

 =$\frac{15}{15+5+30}$

 = $\frac{15}{50}$

 = 0.3

Hence, Probability of getting red light be,

P (red)= 0.3

Now, the Binomial distribution formula,

Probability=( $_r^nC$)$\;p^{r} \;q^{n-r}$

where, r = number of times for a specific outcome within n trials

$_r^nC$   = number of combinations

p = probability of success on a single trial

q = probability of failure on a single trial

n = number of trial

According to question,

• r= 3 (Exactly three)
• n=8 (eight eastbound)
• p(red) =0.3
• q= 1 -p

          = 1- 0.3

          = 0.7

Now using the formula, ( $_r^nC$)$\;p^{r} \;q^{n-r}$

P( exactly 3  get stopped) = ( $_3^8C$)$\;(0.3)^{3} \;(0.7)^{8-3}$

                                           

                                            = $\frac{8!}{3!(8-3)!}$ $(0.027) \; (0.7)^{5}$

                                           =  $\frac{8!}{3!(5)!}$$(0.027) \; (0.16807)$

                                           = $\frac{8 * 7 * 6 }{3 * 2}$ x 0.027 x 0.16807

                                           = 56 x 0.00453789

                                           = 0.25412184

                                          ≈ 0.25

Hence, Probability of next eight eastbound so that exactly three will stopped by a red light be 0.25.

Learn more about Binomial distribution here:

brainly.com/question/13634543

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