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Probability for exact number of outcomes

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Probability for exact number of outcomes

by arnabis2good » Sat May 07, 2011 9:56 pm
What is the probability of getting exactly two 2s when a dice is rolled 5 times?

I am curious to know the formula or trick to solve problems where it states 'exact' number of outcomes. For example - if a coin is tossed 3 times, I can figure out the probability of exact two heads is 3/8 (because the result set is small) but not able to figure out a way to put that mathematically
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by manpsingh87 » Sat May 07, 2011 11:07 pm
arnabis2good wrote:What is the probability of getting exactly two 2s when a dice is rolled 5 times?

I am curious to know the formula or trick to solve problems where it states 'exact' number of outcomes. For example - if a coin is tossed 3 times, I can figure out the probability of exact two heads is 3/8 (because the result set is small) but not able to figure out a way to put that mathematically
when six dices are rolled 5 times,, total no. of possible outcomes would be, 6^5;

now we want exactly two 2's; out of the five draws, any two draws where 2 will occur can be selected in 5C2 ways; and in remaining 3 draws; any of the remaining 5 numbers other than 2 can come in 5^3 ways,,
hence required probability is 5C2*5^3/6^5;
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by arnabis2good » Sat May 07, 2011 11:45 pm
Thank you so much, that helped a lot.

I am using Manhattan's set of guides for preparation. Probability problem solving approach is not so impressive. Any suggestion what I can follow and what level of question really appears on GMAT?
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by GMATGuruNY » Sun May 08, 2011 3:30 am
What is the probability of getting exactly two 2s when a dice is rolled 5 times?


P(exactly n times) = P(one way) * total possible ways.

Let T = two and N = not two.

One way to get exactly 2 T's is to get T on each of the first 2 rolls and N on the each of the last 3 rolls:
P(TTNNN) = 1/6 * 1/6 * 5/6 * 5/6 * 5/6 = 5^3/6^5.

Any arrangement of TTNNN will yield an outcome with exactly 2 T's. Thus, to count the total number of ways to get exactly 2 T's, we count the number of ways to arrange TTNNN:
Total possible ways = number of ways to arrange TTNNN = 5!/2!3! = 10.

P(exactly 2 T's) = 5^3/6^5 * 10.
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by arnabis2good » Sun May 08, 2011 8:57 am
Thanks guys. Both explanations are just perfect.
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