For one roll of a certain die, the probability of rolling...

[rasa.petrauskaite]

Source:
This problem is from Manhattan GMAT Word Translations Strategy Guide (4th edition), p. 94, #14.

Problem:
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

Answer: d

What I don't understand:
Why is it that we multiply the term "((1/6)^3)*(5/6)" by 4, and we don't multiply "(1/6)^4" by 4?

Thank you for your time.

-Rasa

[mschwrtz]

((1/6)^3)*(5/6) represents the probability of YYYN (where Y means rolling a two and N means not rolling a two). YYYN is just one of four EQUALLY LIKELY ways to arrive at three twos:

NYYY
YNYY
YYNY
YYYN

Since each of these outcomes is equally likely, rather than calculate them separately and add the probabilities, we can just multiply the probability of YYYN by 4.

On the other hand, there is only one way to get YYYY, so there's no need to multiply (1/6)^4 by 4.

[rasa.petrauskaite]

Thank you. What you explain makes sense. But I'm confused as to why this reasoning works to solve the problem.

How do we know that we should multiply by the number of possible combinations? Alternatively, why should we add the probabilities of the different ways that we can achieve a two on 1 roll out of 4?

I know that your method gives the right answer. But I don't understand the reasoning behind it.

- Rasa

[gregoryssmith]

If you have a chance (desired/possible) of 1/6 of getting a two, then you want to roll another two then you are increasing the number of possible outcomes.
getting a two: one out of 6
1, 2*, 3, 4, 5, 6

getting 2 twos: one out of 36
{1, 1} {1, 2} {1, 3} {1, 4} {1, 5} {1, 6}
{2, 1} {2, 2}* {2, 3} {2, 4} {2, 5} {2, 6}
{3, 1} {3, 2} {3, 3} {3, 4} {3, 5} {3, 6}
{4, 1} {4, 2} {4, 3} {4, 4} {4, 5} {4, 6}
{5, 1} {5, 2} {5, 3} {5, 4} {5, 5} {5, 6}
{6, 1} {6, 2} {6, 3} {6, 4} {6, 5} {6, 6}

generally you take every possibility of the first dice roll and think of every possibility of the 2nd dice roll, or use the shortcut by multiplying them together (namely 1/6 *1/6) like in the book with the menu example.

[mschwrtz]

Thanks Gregory,

Does that answer your concern Rasa?

Gregory's answer is mostly directed at this part of your question: How do we know that we should multiply by the number of possible combinations?

About the other part, Alternatively, why should we add the probabilities of the different ways that we can achieve a two on 1 roll out of 4?, because each is a distinct way to get the same result. We can't calculate probability if we don't take very favorable or unfavorable outcome into account.

[rasa.petrauskaite]

Yes, it makes sense. Thank you, both.

[tim]

You're welcome..

[lirann]

as a followup, if the question had asked for at least 3 out of 5 attempts would you have to find 3times, 4 times and 5 times and then add them up?

can you please tell me if my answer is correct?

5times = (1/6)^5
3times = 10(1/6)^3 * (5/6)^3
4times = 5(1/6)^4 * (5/6)^2

[tim]

your exponents on the 5/6's are each one too high, but other than that you've got the right idea..