Multiple choice

In a purse there are 10 coins, all 5 paise except one which is a rupee. In another purse, there are 10 coins, all 5 paise 9 coins are taken out from the former purse & put into the latter. Then 9 coins are taken out from the latter & put into the former. Then the chance that the rupee is still in the first purse is

  1. $\dfrac{9}{19}$
  2. $\dfrac{10}{19}$
  3. $\dfrac{4}{9}$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a classic probability problem. After the transfers, the probability that the rupee coin remains in the first purse is 10/19.

AI explanation

When 9 coins are moved from the first purse to the second, the second purse ends up with 19 coins, consisting of 9 rupees and 10 5-paise coins. The probability that the 9 coins drawn from the second purse include the rupee is calculated as choosing 1 rupee and 8 5-paise coins out of the total ways to draw 9 coins. The number of favorable ways is 9C1 times 10C8, and the total number of ways is 19C9. This gives a probability of 9/19 that the rupee leaves the first purse, meaning the probability that the rupee is still in the first purse is 1 - 9/19 = 10/19.

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