Answer to Question #140492 in Statistics and Probability for Bryar

Question

Suppose you just received a shipment of thirteen televisions. Two of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work The probability that at least one of the two televisions does not work?

Answer 1

Total number of televisions is 13

Number of defective televisions is 2

Hence number of working televisions is 11

P(first works) $=\frac{11}{13}$

P(second works given first work) $=\frac{11-1}{13-1}=\frac{10}{12}$

P(both work) = P(first works) * P(second works given first work) $=\frac{11}{13}\cdot\frac{10}{12}=\frac{55}{78}$

or:

Number of selecting 2 working televisions is combination:

$C(11,2)=\frac{11!}{2!9!}=55$

Number all possible outcomes selecting 2 televisions:

$C(13,2)=\frac{13!}{2!11!}=78$

P(both work) $=\frac{C(11,2)}{C(13,2)}=\frac{55}{78}$

P(at least 1 does not work) = 1 – P(both work) = $1-\frac{55}{78}=\frac{23}{78}$