Multiple choice

An urn contains 52 balls, numbered from 1 to 52. Let the balls be drawn one at a time and divided among four players $A, B, C, D$. The players $A $ gets the ball drawn on the first, fifth, ninth draws, ...etc., player $B$ gets the balls drawn on the second, sixth, tenth, ....draws and so on.Suppose that the balls numbered 1, 11, 31 and 41 are considered "Lucky". What is the probability that each player will have a 'Lucky' ball?

  1. $\displaystyle \dfrac{4!(13)^{4}}{2.49.50.51.52}$
  2. $\displaystyle \dfrac{2!(13)^{4}}{49.50.51.52}$
  3. $\displaystyle \dfrac{(13)^{4}}{49.50.51.52}$
  4. $\displaystyle \dfrac{4!(13)^{4}}{49.50.51.52}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

There are 52 balls and 4 players. Each player receives 13 balls. The total ways to distribute 52 balls into 4 groups of 13 is 52! / (13!)^4. The number of ways to distribute the 4 lucky balls such that each player gets one is 4! * (ways to distribute the remaining 48 balls). This leads to the probability calculation involving combinations, which simplifies to the provided answer D.

AI explanation

The 52 balls are distributed cyclically among the four players, meaning each player receives exactly 13 balls. The total number of ways to distribute the 4 lucky balls among the 4 players so that each gets exactly one is 4 factorial. For each player, there are 13 possible positions in their specific sequence of draws to receive their lucky ball, yielding 13 to the power of 4 combinations. The denominator is the total ways to place the 4 lucky balls among the remaining 48 non-lucky balls plus themselves, calculated as 52 multiplied by 51 multiplied by 50 multiplied by 49. This gives the final probability of 4!(13)^4 divided by 52*51*50*49.

Keep practicing — related questions