Question

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 and put into the latter and then 9 coins are taken out from the latter and put into the former. Then what is the Probablity that the "rupee" is still in the first purse:-

Answer 1

The probability that the rupee coin is still in the first purse after transferring coins between two purses is 1/10 or 10%.

Step-by-step explanation:

The question is about calculating the probability that a rupee coin is still in the first purse after a series of coin transfers between two purses.

Initially, the first purse has one rupee coin and nine coins of five paise, and the second purse has ten coins of five paise. If we transfer nine coins from the first purse to the second and then transfer nine coins back, we want to determine the probability that the rupee coin remains in the first purse.

When we take out the nine coins from the first purse, we have ten possible sets of coins that can be drawn, each including the unique rupee coin and eight five paise coins. Whether or not the rupee coin is drawn in the first draw, each of these sets of nine coins will then go into the second purse.

Since the second purse originally had only five paise coins, when we draw nine coins again, that set will contain, at most, one rupee coin. If the rupee coin was transferred to the second purse in the first draw, there’s a chance we could draw it back in the second. However, if it wasn't transferred in the first place, it cannot possibly be drawn in the second draw, ensuring it stays in the first purse.

The probability that the rupee coin is included in the first draw of nine coins from the first purse is 9/10 (since nine out of the ten coins being drawn could be the rupee). Consequently, the probability that the rupee coin is not included and remains in the first purse is 1 - 9/10 = 1/10.