Question

A commuter drives to work each morning. The route she takes each day includes 25

stoplights. Assume the probability each stoplight is red when gets to it, is 0.35 and that these

stoplights are independent. What is the distribution for X, the number of times she must stop for

a red light on her way to work? Evaluate 𝑃(𝑋 = 4) 𝑎𝑛𝑑 𝑃(𝑋 ≥ 5)

Answer 1

Since each stoplight has the same probability being red and all the stotlights are independent, then X has binomial distribution Bin(n, p), where n - number of spotlights, p - probability of being red. In the given case we have

X ~ Bin(25, 0.35)

As i understand, by "the number of times she must stop for a red light on her way to work" means "the number of times she most likely will stop for a red light on her way to work", which means we should evaluate the expected value of X

E(X) = np = 25*0.35 = 8.75

Since for any given day the number of stops must be integer, then we should find out whether it's 8 or 9 times

$P(X=8)={25 \choose 8}*0.35^8*0.65^{17}=0.1607$

$P(X=9)={25 \choose 9}*0.35^9*0.65^{16}=0.1635$

The answer is 9

$P(X=4)={25 \choose 4}*0.35^4*0.65^{21}=0.0224$

$P(X≥5)=1-P(X<5)=1-P(X=4)-P(X=3)-P(X=2)-P(X=1)-P(X=0)=1-0.0224-0.0076-0.0018-0.0003-0.0000=0.9679$