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April 19, 2026, 09:41:29 pm
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Author Topic: Quick Question 2002 Exam  (Read 594 times)  Share 

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davyp3

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Quick Question 2002 Exam
« on: October 26, 2010, 08:37:35 pm »
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A bag contains twenty casino chips which are used to represent cash. Five of the chips have value $10 each and
the other fifteen chips have value $5 each. If four casino chips are drawn from the bag at random without
replacement, what is the probability that there will be at least one chip of value $10?

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|ll|lll|

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Re: Quick Question 2002 Exam
« Reply #1 on: October 27, 2010, 06:47:09 pm »
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Is the answer A?

You take the total probability - Pr(No chips of value $10)
Only in this case is it possible for ALL chips to be $5 - so that none of them is worth $10 itself.

C and D is definitely wrong because it is done with the chips being replaced and well, it's still wrong nevertheless :P
B only speaks for one circumstance - that is if only ONE chip of value $10 is chosen. Pr(AT LEAST one chip of $10) includes 5C2 x 15C2, as well as 5C3 x 15C1 and 5C4 x 15C0

E uses binomal distribution (I think) :P
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TyErd

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Re: Quick Question 2002 Exam
« Reply #2 on: October 28, 2010, 03:29:48 pm »
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Oh I hate these questions! How do you actually go about doing them without a tree diagram?
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dopplereffect

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Re: Quick Question 2002 Exam
« Reply #3 on: October 28, 2010, 05:43:58 pm »
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1- Pr(not getting any $10)

= 1 - 15/20 * 14/19 * 13/18 * 12/17 = 232/323

By evaluating each answer, A = 232/323

Therefore A.
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