Question

a television game show has 11 doors, of which the contestant must pick 2. behind 3 of the doors are expensive cars, and behind the other 8 doors are consolation prizes. the contestant gets to keep the items behind the 2 doors she selects. determine the probability that the contestant wins at least one car
the probability is (type an integer or a simplified fraction.)

Explanation:

Step1: Calculate total number of ways to pick 2 doors

Use combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 11\), \(r=2\).
\(C(11,2)=\frac{11!}{2!(11 - 2)!}=\frac{11\times10}{2\times1}=55\)

Step2: Calculate number of ways to pick no - car (i.e., 2 consolation prizes)

Here \(n = 8\) (number of consolation - prize doors), \(r = 2\)
\(C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28\)

Step3: Calculate probability of picking no - car

\(P(\text{no car})=\frac{C(8,2)}{C(11,2)}=\frac{28}{55}\)

Step4: Calculate probability of picking at least one car

Use the formula \(P(\text{at least one car})=1 - P(\text{no car})\)
\(P(\text{at least one car})=1-\frac{28}{55}=\frac{55 - 28}{55}=\frac{27}{55}\)

Answer:

\(\frac{27}{55}\)