Answer:
Explanation:
Given
Probability of clearing exam in 1 st attempt=0.5
Probability of clearing exam in 2 nd attempt=0.7
Probability of clearing exam in 3 rd attempt=0.8
Probability that she passes the exam
![=P(1 st\ attempt)+P(1\ fail)\cdot P(2\ pass)+P(1\ fail)\cdot P(2\ fail)\cdot P(3 rd pass)](https://img.qammunity.org/2020/formulas/mathematics/college/6eqir1acr8n5qzf0t7pnpu951lrgrwomcq.png)
![P=0.5+0.5* 0.7+0.5* 0.3* 0.8=0.97](https://img.qammunity.org/2020/formulas/mathematics/college/1zd8hqyueni7et32acndsogag8gaizyib0.png)
(b)P(Pass qualification on 2nd try|passes qualification)
P(Pass qualification on 2nd try|passes qualification)
![=(P(fail\ on\ 1 )\cdot P(pass\ in\ 2nd))/(P(passes))](https://img.qammunity.org/2020/formulas/mathematics/college/c38emnpgxiq4uwm9921y4ymwiq8anaa7cd.png)
P(Pass qualification on 2nd try|passes qualification)
![=(0.5* 0.7)/(0.97)=0.3608](https://img.qammunity.org/2020/formulas/mathematics/college/694k3q2k8ejzjw846jqxtav4tb2fgltemz.png)