Answer:
Step-by-step explanation:
Given
[tex]P_1=[/tex]Probability of clearing exam in 1 st attempt=0.5
[tex]P_2=[/tex]Probability of clearing exam in 2 nd attempt=0.7
[tex]P_3=[/tex]Probability of clearing exam in 3 rd attempt=0.8
Probability that she passes the exam[tex]=P(1 st\ attempt)+P(1\ fail)\cdot P(2\ pass)+P(1\ fail)\cdot P(2\ fail)\cdot P(3 rd pass)[/tex]
[tex]P=0.5+0.5\times 0.7+0.5\times 0.3\times 0.8=0.97[/tex]
(b)P(Pass qualification on 2nd try|passes qualification)
P(Pass qualification on 2nd try|passes qualification)[tex]=\frac{P(fail\ on\ 1 )\cdot P(pass\ in\ 2nd)}{P(passes)}[/tex]
P(Pass qualification on 2nd try|passes qualification)[tex]=\frac{0.5\times 0.7}{0.97}=0.3608[/tex]