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 1 What is the probability of getting at least one six in a single throw of three unbiased dice ? A. 31/216 B. 91/216 C. 125/216 D. 81/216
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Explanation

Three cases arise:

Case 1: When only one dice shows up a six

This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability

Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)

=(1/6)*(5/6)*(5/6)

=25/216

similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216

And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216

So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216

Case 2:When two dice show up a six

Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3

Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs

Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216

So, total probability = 3*(5/216) = 15/216

Case 3: When all dice show up a six

In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216

So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216

kishore 3/15/2014 1:03:33 PM

instead of considering 3 cases by subtracting the odd case is best on

max probabilty is 1

1- (no dies getting 6)=at least one die getting 6

so 1-(5*5*5/216)=1-(125/216)

=91/216

sankar 9/7/2014 8:34:09 AM

Thanks Kishore for even more short cut..

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