Question

A basketball team is to play two games in a tournament. The probability of winning the first game is .10.1 the first game is won, the probability of winning the second game is 15. If the first game is lost, the probability of winning the second game is 25. What is the probability the first game was won if the second game is lost? Express the answer with FOUR decimal points.

Answer 1

`P(W_1/W_2')=0.1110`

Explanation:

Probability of winning the first game be considering the given factors be,
`W_1=0.1`

Probability of winning the second game be considering the given factors be,
`W_2`= probability of winning the second game when the first game is won + probability of winning the second game when the first game is lost:

`P(W_2)=P(W_1).P(W_2/W_1)+P(W_1').P(W_2/W_1')`

`P(W_2)=0.1* 0.15+0.9* 0.25`

`P(W_2)=0.24`

Hence the probability of losing the second game:

`P(W_2')=1-P(W_2)`

`P(W_2')=0.76`

Probability of winning the first game when the second game is won:

`P(W_1/W_2)=(P(W_2/W_1).P(W_1))/(P(W_2))`

`P(W_1/W_2)=(0.15* 0.1)/(0.24)`

`P(W_1/W_2)=0.0625`

Probability of winning the first game be considering the given factors,
`W_1`= probability of winning the first game when the second game is won + probability of winning the first game when the second game is lost:

`P(W_1)=P(W_2).P(W_1/W_2)+P(W_2').P(W_1/W_2')`

`0.1=0.24*0.0625+0.76* P(W_1/W_2')`

`P(W_1/W_2')=0.1110`

Answer 2

`P((A)/(B'))`=0.111

Explanation:

Given:

The probability of winning the first game is 10.1

The first game is won

The probability of winning the second game is 15

If the first is lost, the probability of winning the second game is 25

Solution:

`P(B)=P(A)P((B)/(A))+P(A')P((B)/(A'))\\ =0.1(0.15)+(0.3)*0.25)\\P(B)=0.24 ------(1)\\P((A)/(B))=(P((B)/(A))P(A) )/(P(B))\\ =(0.15(0.1))/(0.24)\\ =0.0625 ------(2)\\P(B')=1-P(B)=0.76 ------(3)\\P(A)=P(B)P((A)/(B))+P(B')P((A)/(B'))\\0.1=0.24(0.0625)+0.76(p((A)/(B') ))\\P((A)/(B'))=0.111`