Question

A road is constructed so that the right-hand turn lane at an intersection has a capacity of three (3) cars. Suppose that 30% of cars approaching the intersection want to turn right. If a string of 15 cars approaches the intersection, what is the probability that the lane will be insufficiently large to hold all the cars waiting to turn right?

Answer 1

Let $X=$ the number of  cars waiting to turn right: $X\sim Bin(n, p).$

Given $n=15, p=0.3, 1-p=1-0.3=0.7.$

$$P(X\leq3)=P(X=0)+P(X=1)+$$

$$+P(X=2)+P(X=3)$$

$$=\dbinom{15}{0}(0.3)^0(0.7)^{15-0}+\dbinom{15}{1}(0.3)^1(0.7)^{15-1}$$

$$+\dbinom{15}{2}(0.3)^2(0.7)^{15-2}+\dbinom{15}{3}(0.3)^3(0.7)^{15-3}$$

$$\approx0.1727$$

Then

$$P(X>3)=1-P(X\leq3)$$

$$\approx1-0.1727=0.8273$$

The probability that the lane will be insufficiently large to hold all the cars waiting to turn right is 0.8273.