A) \[\frac{5}{7}\]
B) \[\frac{2}{7}\]
C) \[\frac{3}{7}\]
D) \[\frac{4}{7}\]
Correct Answer: C
Solution :
| Let \[{{\operatorname{E}}_{1}}\]and \[{{\operatorname{E}}_{2}}\]be the events of the boy watching DOORDARSHAN and TEN SPORTS, respectively. It is given that |
| \[\operatorname{P}\left( {{E}_{1}} \right)=\frac{1}{5}\] and \[\operatorname{P}\left( {{E}_{2}} \right)=\frac{4}{5}\] |
| Let \[\operatorname{E}\]be the boy falls asleep. Again by hypothesis \[\operatorname{P}\left( E/{{E}_{1}} \right)=\frac{3}{4}and\,P\left( E/{{E}_{2}} \right)=\frac{1}{4}\] |
| Now \[\operatorname{E}=E\cap \left( {{E}_{1}}\cup {{E}_{2}} \right)=\left( {{E}_{1}}\cap E \right)\cup \left( {{E}_{2}}\cap E \right)\] |
| So that, \[\operatorname{P}\left( E \right)=P\left( {{E}_{1}} \right)P\left( E/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( E/{{E}_{2}} \right)\] |
| By Bayes? theorem \[\operatorname{P}\left( {{E}_{1}}/E \right)=\frac{P\left( {{E}_{1}} \right)P\left( E/{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)P\left( E/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( E/{{E}_{2}} \right)}\] |
| \[=\frac{\left( 1/5 \right)\times \left( 3/4 \right)}{\left( 1/5 \right)\times \left( 3/4 \right)+\left( 4/5 \right)\times \left( 1/4 \right)}=\frac{3}{7}\] |
You need to login to perform this action.
You will be redirected in
3 sec
