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I have to solve this problem:

Three urns each contain ten gems:

  • Urn 1 contains 6 rubies and 4 emeralds.
  • Urn 2r contains 8 rubies and 2 emeralds.
  • Urn 2e contains 4 rubies and 6 emeralds.

The following procedure is used to select two gems. First, one gem is drawn at random from urn 1. If this first gem is a ruby, then a second gem is drawn at random from urn 2r; however, if the first gem is an emerald, then the second gem is drawn at random from urn 2e.

Suppose that this procedure is independently replicated three times. What is the probability that a ruby is obtained on the second draw exactly once?

My solution was to calculate the probability of obtaining a ruby on the second draw, where I got $0.64$ which is also correct according to the solutions.

Then to solve the actual problem I wanted to use R, writing

pbinom(1,3,0.64) - pbinom(0,3,0.64)

, getting $0.248832$ as an answer.

Yet the correct solution would be $0.3549$.

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  • $\begingroup$ Are the gems drawn with replacement? $\endgroup$ – T. Linnell Aug 1 '17 at 15:12
  • $\begingroup$ Yes (charminimum) $\endgroup$ – ToTom Aug 1 '17 at 15:14
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    $\begingroup$ I get the exact same answer as you in that case, $(0.64)^{1}*(0.36)^{2}*\left(\array{3\\1}\right)=0.248832$. Is the answer of $0.3549$ explained in any way? $\endgroup$ – T. Linnell Aug 1 '17 at 15:16
  • $\begingroup$ No, we got the sheet with all the answers which our TA (teaching assistant) calculated. There was nothing wrong in the last 50 exercises, but maybe he made a mistake this time. $\endgroup$ – ToTom Aug 1 '17 at 15:18
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    $\begingroup$ I also match your answer...haven't been able to think of any variant that yields $.3549$. $\endgroup$ – lulu Aug 1 '17 at 15:27
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If I get this right, we replicate procedure three times and want to know probability of obtaining precisely one ruby during the second draw.

We can avoid a ruby on the second draw with probability $$0.6\cdot 0.2 + 0.4\cdot 0.6 = 0.36$$ We can obtain a ruby on second draw with probability (following computation really unnecessary)$$0.6\cdot 0.8 + 0.4\cdot 0.4 = 0.64 $$ There are three ways this can happen so your solution must be correct.

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  • $\begingroup$ Hmm thank you. I guess the answer of the solutions is simply wrong. $\endgroup$ – ToTom Aug 1 '17 at 19:15

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