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

AI Thread Summary
The discussion revolves around determining the minimum number of white balls in a container of 27 balls, ensuring the probability of drawing two black balls without replacement is less than 23/30. Clarification was sought regarding the phrasing of the drawing process, specifically whether it referred to drawing two white balls. The probability calculations involve using the number of white balls, denoted as "n," and setting up an inequality to solve for n. The correct approach involves calculating the probability of drawing two white balls and comparing it to the specified threshold. The exercise was corrected for clarity based on feedback received.
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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?
 
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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.
 
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