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Probability

by NarendraSure » Sat Jun 27, 2015 8:50 pm
Molly needs to roll a 1 in a die and she has 3 tries. What is the probability that she will succeed?


can some body answer this with an explanation.

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Naren
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by NarendraSure » Sat Jun 27, 2015 8:55 pm
Hi All,

Thanks for passing by, found the answer. If anyone is interested below is how.

Probability to roll a 1 at the first try: 1/6
Probability to roll a 1 at the second try: 5/6 * 1/6 = 5/36
Probability to roll a 1 at the third try: 5/6 * 5/6 * 1/6 = 25/216
Since she can succeed with one OR two OR three tries, we sum it: 91/216

P(2nd roll) is the probability that you DON'T roll a 1 on the 1st roll, because the only way to get to the 2nd roll is to avoid a 1 on the 1st roll. this is why we need to multiply the 1/6 by 5/6.



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Naren
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by Brent@GMATPrepNow » Sun Jun 28, 2015 9:23 am
NarendraSure wrote:Molly needs to roll a 1 in a die and she has 3 tries. What is the probability that she will succeed?


can some body answer this with an explanation.

Regards
Naren
We want P(at least one 1 in three rolls).

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least one 1 in three rolls) = 1 - P(not getting at least one 1 in three rolls)
What does it mean to not get at least one 1? It means getting zero 1's.
So, we can write: P(getting at least one 1) = 1 - P(getting zero 1's in three rolls)

P(get zero 1's in three rolls)
P(get zero 1's in three rolls) = P(no 1 on 1st roll AND no 1 on 2nd roll AND no 1 on 3rd roll )
= P(no 1 on 1st roll) x P(no 1 on 2nd roll) x P(no 1 on 3rd roll)
= 5/6 x 5/6 x 5/6
= 125/216

So, P(getting at least one 1) = 1 - 125/216
= [spoiler]91/216[/spoiler]

Cheers,
Brent
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by NarendraSure » Mon Jun 29, 2015 4:22 am
Thanks for the reply Brent.

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by Matt@VeritasPrep » Mon Jun 29, 2015 3:26 pm
Just to illustrate why Brent's approach is so good, consider the positive approach, in which we list all the cases.

To get exactly one 1, we would need to roll 1, not-1, not-1 in some order. That gives us

1XX = (1/6) * (5/6) * (5/6) = 25/216

But this could happen in any of THREE orders: 1XX, X1X, XX1. So we have 3 * (25/216) = 75/216.

Now we consider exactly two 1s, which is basically the same idea:

11X = (1/6) * (1/6) * (5/6) = 5/216

This could happen in three orders as well (11X, 1X1, X11), so 3 * (5/216) = 15/216.

We also have exactly three ones, which is 111 = (1/6)*(1/6)*(1/6) = 1/216.

Adding these up, we have 75/216 + 15/216 + 1/216 = 91/216. Yuck!

So doing P(At least one) = 1 - P(No ones) is much more efficient.
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by NarendraSure » Wed Jul 01, 2015 5:50 am
Yeah agree matt this is so direct
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by talaangoshtari » Wed Jul 01, 2015 6:12 am
In probability, we have a distribution called Geometric Distribution.
P(X = x) = p.(q)^(x - 1)
p = probability of success
q = probability of failure

Here we have,
P(1《 X 《 3) = P(X = 1) + P(X = 2) + P(X = 3)
= p + pq + pq^2 = (1/6) + (1/6)(5/6) + (1/6)(5/6)^2 = 91/216
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by NarendraSure » Wed Jul 01, 2015 1:49 pm
Hi talaangoshtari,

how do we need to determine the usage of this neat little formula.

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by talaangoshtari » Wed Jul 01, 2015 9:55 pm
Hi NarendraSure,

Geometric distribution computes the probabability of the first success after k - 1 failures. In this problem there are 3 scenarios:

1. We roll a 1 in a die in the first try, so we have one success.
2. We roll a 1 in a die in the second try, so we have one failure and one success.
3. We roll a 1 in a die in the third try, so we have two failures and one success.
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by nikhilgmat31 » Fri Jul 03, 2015 12:33 am
Brent's way to solve is simple calculation. Great.

But it is difficult to strike. But it is very good to solve such questions

P(A) = 1- P(Not A)
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