Probability
Probability
Allen Star Batch
Grade 12
Question:
There are two purses. The first contains 9 fifty paise coins and a one-rupee coin, while the second purse has 10 fifty-paise coins. Nine coins are transferred from the first purse to the second randomly. Then nine coins are transferred from the second purse to the first randomly. The probability of finding a one rupee coin in the first purse after these transfers is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - p = \ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: Use law of total probability with two cases: (Eā) one-rupee coin transferred in first step with probability 1/10, and (Eā) one-rupee coin not transferred with probability 9/10. Then calculate conditional probability that the rupee coin returns in the second transfer, accounting for the different compositions of the second purse in each case.
The problem involves transferring 9 coins from the first purse to the second, and we need to find the probability that a rupee coin is in the first purse after the transfer. Using the law of total probability with events $E_1$ (one rupee transferred) and $E_2$ (one rupee not transferred), we get $P(A) = P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right) = \frac{9}{19} \cdot \frac{9}{10} + 1 \cdot \frac{10}{19} = \frac{81 + 190}{190} = \frac{10}{19}$.
Correct Answer: 9