Question

When trevor and brian play ping pong, the probability that trevor will win is 0.6. if trevor and brian play two games of ping pong, what is the probability that trevor wins the first game and brian wins the second game? assume games are independent?

Answer 1

The probability that Trevor wins is $p=0.6$ and the probability that Trevor loses is $q=1-p=1-0.6=0.4.$

The probability that Brian  wins is $0.4$ (Trevor loses) and the probability that Brian  loses is $0.6$ (Trevor wins).

There are four different possibilities for 2 ping pong games:

1. 1st and 2nd wins Trevor: $Pr=0.6\cdot 0.6=0.36;$
2. 1st wins Trevor, 2nd wins Brian: $Pr=0.6\cdot 0.4=0.24;$
3. 1st wins Brian, 2nd wins Trevor: $Pr=0.4\cdot 0.6=0.24;$
4. 1st and 2nd wins Brian: $Pr=0.4\cdot 0.4=0.16.$

Check that in total $0.36+0.24+0.24+0.16=1.$

Answer: the probability that Trevor wins the first game and Brian wins the second game is 0.24.