Probability Help!!!

Brent@GMATPrepNow wrote:

manihar.sidharth wrote:
P(exactly one 6)
Consider one case where we get exactly one 6: six on first roll, non-six on second roll, non-six on third roll, and non-six on forth roll.
To make this solution less cumbersome, we'll denote this outcome as {6, ~6, ~6, ~6)
So, P(6, ~6, ~6, ~6) = (1/6)(5/6)(5/6)(5/6)

We also need to consider the outcome {~6, 6, ~6, ~6)
So, P(~6, 6, ~6, ~6) = (5/6)(1/6)(5/6)(5/6)

Notice that we can place the 6 in 4 different places to get exactly one 6, so...
P(exactly one 6) = (4)(1/6)(5/6)(5/6)(5/6) = 500/1296

P(exactly two 6's)
Let's examine one possible outcome: {6, 6, ~6, ~6}
P(6, 6, ~6, ~6) = (1/6)(1/6)(5/6)(5/6)
Of course, {6, 6, ~6, ~6} is just one possible way to get exactly two 6's.
In how many different ways can we get exactly two 6's? Well, in how many different ways can we select 2 places for the 6's?
There are 4 different rolls, and we need to select 2 of them to be 6's. This can be accomplished in 4C2 ways (6 ways).
So, P(exactly two 6's) = (6)(1/6)(1/6)(5/6)(5/6) = 150/1296

P(exactly three 6's)
Let's examine one possible outcome: {6, 6, 6, ~6}
P(6, 6, 6, ~6) = (1/6)(1/6)(1/6)(5/6)
In how many different ways can we select 3 places for the 6's?
There are 4 different rolls, and we need to select 3 of them to be 6's. This can be accomplished in 4C3 ways (4 ways).
So, P(exactly three 6's) = (4)(1/6)(1/6)(1/6)(5/6) = 20/1296

P(exactly four 6's)
There's only one way to accomplish this.
P(6, 6, 6, 6) = (1/6)(1/6)(1/6)(1/6) = 1/1296

So... P(at least one 6) = (500/1296) + (150/1296) + (20/1296) + (1/1296)
= 671/1296
= D

Cheers,
Brent

tisrar02 wrote:In my opinion, when you are dealing with a probability question that states "at least one" of anything, that means you can have 1 or 2 or 3 or whatever. So therefore you want to figure out what the probability of having NONE of the desired outcomes. Because the greatest probability of anything is 1, you would then subtract 1 from the number you received to get the answer you desire.

1-none= at least one

Trying to figure out EVERY case is a lot of work and not recommended on the GMAT.

Hope this helps

Best of luck Sid

Brent@GMATPrepNow wrote:

tisrar02 wrote:In my opinion, when you are dealing with a probability question that states "at least one" of anything, that means you can have 1 or 2 or 3 or whatever. So therefore you want to figure out what the probability of having NONE of the desired outcomes. Because the greatest probability of anything is 1, you would then subtract 1 from the number you received to get the answer you desire.

1-none= at least one

Trying to figure out EVERY case is a lot of work and not recommended on the GMAT.

Hope this helps

Best of luck Sid

Agreed!
That said, if someone failed to use the complement for this question, they would need to solve it using a different approach.

Alternatively, the question could have asked for something like P(exactly two 6's), in which case, we'd need to use the method described above.

Cheers,
Brent