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A bag contains five white and three blue marbles. Successive marbles are drawn at random from the bag. What is the probability that the first blue marble is the third marble drawn.

Answer in back of book: 5/28

My answer: 1/8

My thinking:

P(First ball not being the first blue marble) = 7/8

P(Second ball not being the first blue marble) = 6/7

P(Third ball being the first blue marble) = 1/6

P(final) = 7/8 * 6/7 * 1/6 = 1/8

Where am I going wrong?

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    $\begingroup$ The probability that the first ball is not the first blue one is $\frac 58$. $\endgroup$ – lulu Jan 4 '16 at 13:20
  • $\begingroup$ Ah , I understand now. Actually I misunderstood the question. I thought each of the blue marbles were sort of assigned a first, second and third. So for the first selection one could select any white marble or the second or third blue marble but not the first blue marble. Thanks $\endgroup$ – Kantura Jan 4 '16 at 13:25
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Saying P(First ball not being the first blue marble) = 7/8 is wrong.

Took me a while to see why you'd think this. I suspect you have a picture in your head, where one of the marbles is "the first blue marble", and the other seven are not the first blue marble. No - if any one of the three blue marbles is drawn first then it becomes "the first blue marble".

Instead you want to consider P(the first marble is not blue). Which is of course 5/8.

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  • $\begingroup$ Yes David, you suspected correctly. That is how I interpreted the question. But I understood my misunderstanding from lulu's comment above. Cheers $\endgroup$ – Kantura Jan 4 '16 at 13:29

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