Hello. I have another concern with the Probability course exam, specifically in question 13. The question clearly stated that you only get to pick one ball per draw. Therefore, by the time you get to pick the second ball, the sample space would only have 53 balls, since you kept the first one to yourself.

Hey Carl, Thank you for your question! Initially, we have 54 balls in the bowl - 42 black and 12 white. The number of ways in which we can draw 2 balls out of 54, with no condition about the order in which we draw them, is "54-choose-2". Those are all possible events, so the number goes as the denominator of the fraction. Now we ask "in how many ways can we choose 1 white ball out of 12"? There are 12 ways to do that, so it goes as one of the factors in the numerator. The second factor in the numerator is obtained by asking the question "in how many ways can we choose 1 black ball out of 42"? Indeed, there are 42 ways to do this. What you say is absolutely true - after drawing the first ball, we are left with 53 balls. However, our condition is that we draw a ball from one color and the second ball we draw is of the other color. It has now come to my attention that there is a typo in the denominator in the explanation. We will have this corrected. Hope this answers your question! Kind regards, 365 Hristina

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Carl Angelo Catchola

Last answered:

16 Jun 2022

Posted on:

16 Jun 2022

Resolved: Probability - course exam question 13

in Probability / Course Introduction

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Hristina Hristova

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Posted on:

16 Jun 2022

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Resolved: Probability - course exam question 13

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