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Ineedhelp
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P(A)+P(NOT A) = 1

by Ineedhelp Tue Dec 11, 2012 11:15 am

Molly is rolling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 4, she will stop rolling the cube. What is the probability that Molly will roll the die less than 3 times before stopping?

Please solve, and explain in detail. I'm having problems in this area. Thanks.
hiteshwd
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Re: P(A)+P(NOT A) = 1

by hiteshwd Tue Dec 11, 2012 3:32 pm

Sure

For Molly to stop maximum after three throws, 4 will have to appear either in 1st, 2nd or 3rd throw

So we will have 3 cases i.e. getting a 4 in first throw, in second throw and in third throw

1) probability of getting a 4 in first throw: 1/6
2) probability of getting a 4 in second throw: 5/6 * 1/6
3) probability of getting a 4 in third throw: 5/6 * 5/6 * 1/6

Hence, answer = 1/6 + 5/36 + 25/216= 91/216

Let me know in case you need any further help
jlucero
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Re: P(A)+P(NOT A) = 1

by jlucero Wed Dec 12, 2012 5:00 pm

Nice explanation hiteshwd, but be careful as the question asks for the probability of rolling less (this should say fewer!) than 3 times- getting a 4 on the first or second roll.

1/6 + (5/6 * 1/6) = 1/6 + 5/36 = 11/36
Joe Lucero
Manhattan GMAT Instructor