For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
a) (1/6)^4
b) 2(1/6)^3 +(1/6)^4
c) 3((1/6)^3)*(5/6)+(1/6)^4
d) 4((1/6)^3)*(5/6)+(1/6)^4
e) 6((1/6)^3)*(5/6)+(1/6)^4
D i want some easy way to do this question, because such type of questions will consume more than a minute to do in real gmat test
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Is their any easy way to do this question?
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Jim@StratusPrep
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This just requires a good understanding of probability to do in the appropriate amount of time.
You have 2 scenarios: 1) all four are 2's and 2) There are three 2's
1) Here you can have a 2, 2, 2 and X (standing for a number other that 2.
If you did it in this order the probability would be 1/6 * 1/6 * 1/6 * 5/6. However, because the "X" can be in any of the four spaces, the solution must be multiplied by 4:
2, 2, 2, X
2, 2, X, 2
2, X, 2, 2
X, 2, 2, 2
Or..... 4*(1/6)^4*(5/6)
2) This is the easy one -- (1/6)^4
Your answer is the sum: 4*(1/6)^4*(5/6) + (1/6)^4
You have 2 scenarios: 1) all four are 2's and 2) There are three 2's
1) Here you can have a 2, 2, 2 and X (standing for a number other that 2.
If you did it in this order the probability would be 1/6 * 1/6 * 1/6 * 5/6. However, because the "X" can be in any of the four spaces, the solution must be multiplied by 4:
2, 2, 2, X
2, 2, X, 2
2, X, 2, 2
X, 2, 2, 2
Or..... 4*(1/6)^4*(5/6)
2) This is the easy one -- (1/6)^4
Your answer is the sum: 4*(1/6)^4*(5/6) + (1/6)^4
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Brent@GMATPrepNow
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P(roll 2 at least 3 times) = P(roll 2 three times OR roll 2 four times)sana.noor wrote:For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
a) (1/6)^4
b) 2(1/6)^3 +(1/6)^4
c) 3((1/6)^3)*(5/6)+(1/6)^4
d) 4((1/6)^3)*(5/6)+(1/6)^4
e) 6((1/6)^3)*(5/6)+(1/6)^4
= P(roll 2 three times) + P(roll 2 four times)
P(roll 2 three times)
There are 4 different ways we can roll a 2 three times:
- 2, 2, 2, non-2
- 2, 2, non-2, 2
- 2, non-2, 2, 2
- non-2, 2, 2, 2
The fortunate thing is that the probability of each occurrence is equal. Here's what I mean:
P(2, 2, 2, non-2) = (1/6)(1/6)(1/6)(5/6) = [(1/6)^3](5/6)
Similarly, P(non-2, 2, 2, 2) = (5/6)(1/6)(1/6)(1/6) = [(1/6)^3](5/6)
So, P((roll 2 three times) equals the sum of the four probabilities = 4[(1/6)^3](5/6)
P(roll 2 four times)
P(roll 2 four times) = P(2, 2, 2, 2)
= (1/6)(1/6)(1/6)(1/6)
= (1/6)^4
So, P(roll 2 at least 3 times) = 4[(1/6)^3](5/6) + (1/6)^4
= D
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Brent
Brent Hanneson - Creator of GMATPrepNow.com
