Question #150817
Suppose you just received a shipment of 7 televisions. 4 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?

1. The probability that both televisions work is ______
1
Expert's answer
2020-12-15T02:41:50-0500
P(both work)=(32)(40)(72)P(both \ work)=\dfrac{\dbinom{3}{2}\dbinom{4}{0}}{\dbinom{7}{2}}

=3!2!(32)!17!2!(72)!=321=17=\dfrac{\dfrac{3!}{2!(3-2)!}\cdot1}{\dfrac{7!}{2!(7-2)!}}=\dfrac{3}{21}=\dfrac{1}{7}


P(at least 1 does not work)P(at\ least\ 1 \ does\ not\ work)

=1P(both work)=117=67=1-P(both \ work)=1-\dfrac{1}{7}=\dfrac{6}{7}



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