Saved Bookmarks
| 1. |
Bag A contains 2 white and 3 red balls, and bag B contains 4 white and 5 red balls. One ball is drawn at randow from one of the bags and it is found to be red. Find the probability that it was drawn from bag B. |
|
Answer» SOLUTION :Let `E_1`= ENVENT of CHOOSING bag A, `E_2`= event of choosing bag B, and E= event of drawing a red ball. Then, `P(E-1)=1/2 and P(E-2)=1/2` Also, `P(E//E_1)`= event of drawing a red ball from bag `A=3/5`, and `P(E//E-2)` = event of drawing a red ball from bag `B=5/9`. Probability of drawing a ball from B, it being given that it is red `=P(E_2//E)` `=(P(E//E_2).P(E_2))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2))`[by Bayes's theorem] `=((5/9xx1/2))/((3/5xx1/2)+(5/9xx1/2))=25/52`. Hence, the REQUIRED probability is`25/52`. |
|