Respuesta :
Answer:
[tex]P(\frac{A}{B'})[/tex]=0.111
Step-by-step 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:
[tex]P(B)=P(A)P(\frac{B}{A})+P(A')P(\frac{B}{A'})\\ =0.1(0.15)+(0.3)*0.25)\\P(B)=0.24 ------(1)\\P(\frac{A}{B})=\frac{P(\frac{B}{A})P(A) }{P(B)}\\ =\frac{0.15(0.1)}{0.24}\\ =0.0625 ------(2)\\P(B')=1-P(B)=0.76 ------(3)\\P(A)=P(B)P(\frac{A}{B})+P(B')P(\frac{A}{B'})\\0.1=0.24(0.0625)+0.76(p(\frac{A}{B'} ))\\P(\frac{A}{B'})=0.111[/tex]
Answer:
[tex]P(W_1/W_2')=0.1110[/tex]
Step-by-step explanation:
Probability of winning the first game be considering the given factors be, [tex]W_1=0.1[/tex]
Probability of winning the second game be considering the given factors be, [tex]W_2[/tex]= probability of winning the second game when the first game is won + probability of winning the second game when the first game is lost:
[tex]P(W_2)=P(W_1).P(W_2/W_1)+P(W_1').P(W_2/W_1')[/tex]
[tex]P(W_2)=0.1\times 0.15+0.9\times 0.25[/tex]
[tex]P(W_2)=0.24[/tex]
Hence the probability of losing the second game:
[tex]P(W_2')=1-P(W_2)[/tex]
[tex]P(W_2')=0.76[/tex]
Probability of winning the first game when the second game is won:
[tex]P(W_1/W_2)=\frac{P(W_2/W_1).P(W_1)}{P(W_2)}[/tex]
[tex]P(W_1/W_2)=\frac{0.15\times 0.1}{0.24}[/tex]
[tex]P(W_1/W_2)=0.0625[/tex]
Probability of winning the first game be considering the given factors, [tex]W_1[/tex]= probability of winning the first game when the second game is won + probability of winning the first game when the second game is lost:
[tex]P(W_1)=P(W_2).P(W_1/W_2)+P(W_2').P(W_1/W_2')[/tex]
[tex]0.1=0.24\times0.0625+0.76\times P(W_1/W_2')[/tex]
[tex]P(W_1/W_2')=0.1110[/tex]
