Finding the Minimum Number of White Balls in a Container with 27 Balls

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SUMMARY

The discussion focuses on calculating the minimum number of white balls required in a container of 27 balls to ensure that the probability of drawing two black balls without replacement is less than 23/30. The probability of drawing two white balls is represented by the formula [n(n-1)]/702, where n is the number of white balls. The inequality (n^2 - n)/702 < 23/30 must be solved to determine the minimum value of n. The correction to the original problem statement clarified the drawing process, confirming it involves drawing without replacement.

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ghostfirefox
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We have 27 balls in the container, some of which are white and some black. How many white balls in the container must be at least, so that the probability that two black balls were drawn at random without a return was less than 23/30?
 
Last edited:
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ghostfirefox said:
We have 27 balls in the container, some of which are white and some black. How many white balls in the container must be at least, so that the probability that two balls were drawn at random without a draw was less than 23/30?

What?? "Two balls are drawn at random without a draw"?? What does that mean? Did you mean "two white balls are drawn without a black ball being drawn"? Is this drawing without replacement? If so let "n" be the number of white balls in the container. The probability the first ball drawn is white is n/27. If that happens there are 26 balls left in the container, n-1 of them white. The probability the second ball drawn is also white is (n-1)/26. The probability both balls are white is [n(n-1)]/702. You want to solve the inequality (n^2- n)/702< 23/30.
 
You're right a I missed a word. I corrected the exercise. Thank for your response.
 
Last edited:

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