Balls are chosen at random

  • #1
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Main Question or Discussion Point

This problem appeared in a problem set which I encountered on the internet

In a game, balls are labeled by integer numbers. One chooses three different integer numbers between 1 and 10. Two balls are picked at the same time, at random from a box. If they are part of the three earlier chosen numbers, the player wins. What's the probability that the player will win?

The given answer is 1/15. But I found 1/90. The probability that the first ball is labeled by one of the chosen numbers is 1/10 and the second is 1/9. And I considered that picking two balls at the same time is equivalent to picking them in sequence. So (1/10) (1/9) = 1/90.
 

Answers and Replies

  • #2
mjc123
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The probability that the first ball is labeled by one of the chosen numbers is 1/10
That's the probability that it matches a specific one of the chosen numbers. But there were 3 chosen numbers, so the probability of it matching any of them is 3/10. Likewise, the probability of the second ball matching either of the remaining numbers is 2/9. So the overall probability is 6/90 = 1/15.
 
  • #3
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That's the probability that it matches a specific one of the chosen numbers. But there were 3 chosen numbers, so the probability of it matching any of them is 3/10. Likewise, the probability of the second ball matching either of the remaining numbers is 2/9. So the overall probability is 6/90 = 1/15.
Yea, I quickly realised that after my post. (3/10)(2/9) = 1/15.
 

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