InterviewSolution
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InterviewSolution
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There are three bags B_1,B_2 and B_3. The bag B_1 contains 5 red and 5 green balls, B_2 contains 3 red 5 green balls, and B_3 contains 5 red and 3 green balls. Bags B_1, B_2 and B_3 have probabilities (3)/(10),(3)/(10) and (4)/(10) respectively of being chosen. A bag is selected at random and a ball is chosen at random form the bag. then which of the following options is/are correct ? |
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Answer» Probability that the CHOSEN ball is green, GIVEN that the selected bag is `B_3` equals `(3)/(8)`. It is given that there are three bags, `B_1,B_2` and `B_3` and probabilities of being chosen, `B_1,B_2 and B_3` are RESPECTIVELY. `THEREFORE P(B_1)=(3)/(10),P(B_2)=(3)/(10) and P(B_3)=(4)/(10)` Now, probability that the chosen ball is green, given that selected bag is `B_3P((G)/(B_3))=(3)/(8)` Now, probability that the chosen selected bag is `B_3`, given that the chosen ball is green `=P((B_3)/(G))` `=(P((G)/(B_3))P(B_3))/(P((G)/((B_1))P(B_1))+P((G)/(B_2))P(B_2)+P((G)/(B_3))P(B_2))["by Baye's theorem"]` `=(((3)/(8)xx(4)/(10)))/(((5)/(10)xx(3)/(10))+((5)/(8)xx(3)/(10))+((3)/(8)xx(4)/(10)))=((1)/(2))/((1)/(2)+(5)/(8)+(1)/(2))=(4)/(13)` Now, probability that the chosen ball is green `=P(G)=P(B_1)P((G)/(B_1))+P(B_2)P((G)/(B_2))+P(B_3)P((G)/(B_3))` [By using theorem of total probability] `=((3)/(10)xx(5)/(10))+((3)/(10)xx(5)/(8))+((4)/(10)xx(3)/(8))` `=(3)/(20)+(3)/(16)+(3)/(20)=(12+15+12)/(80)=(39)/(80)` Now, probability that the selected bag is `B_3` and the chosen ball is green `=P(B_3)xxP((G)/(B_3))=(4)/(10)xx(3)/(8)=(3)/(20)` Hence, OPTIONS (a) and (c) is are correct. |
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