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Yasin has a bag containing five red balls and two white balls. Yasin is going to pick a ball at random. How many more white balls should Yasin put in the bag so that it is just as likely that he will pick a red ball as a white ball? ​

User Skeept
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If Yasin has five red balls and two white balls, then the probability of picking a red ball is 5/7 and the probability of picking a white ball is 2/7. To make it just as likely that he will pick a red ball as a white ball, the probability of picking a red ball or a white ball must be 1/2.

Let x be the number of white balls Yasin needs to add to the bag. Then, the total number of balls in the bag will be 5 + x red balls and 2 + x white balls.

The probability of picking a red ball will be (5)/(7 + x) and the probability of picking a white ball will be (2 + x)/(7 + x).

We can set up an equation:

5 / (7 + x) = (2 + x) / (7 + x)

Solving for x, we get:

5(7 + x) = (2 + x)(7 + x)
35 + 5x = 14 + 9x + x^2
x^2 + 4x - 21 = 0
(x + 7)(x - 3) = 0

The solution x = 3 makes sense since if Yasin adds 3 more white balls, then the bag will have 5 red balls and 5 white balls, making it equally likely to pick a red ball or a white ball.
User The Black Horse
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