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Probability

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Source: — Problem Solving |

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by Brent@GMATPrepNow » Sun Jun 23, 2013 1:33 pm
arvindsekar wrote:What is the probability of getting exactly 2 sixes on three rolls of a fair six-sided die?
There are 3 different ways to get exactly two 6's.
#1) 6 on 1st roll, 6 on 2nd roll, non-6 on 3rd roll
#2) 6 on 1st roll, non-6 on 2nd roll, 6 on 3rd roll
#3) non-6 on 1st roll, 6 on 2nd roll, 6 on 3rd roll

So, P(exactly two 6's) = P(6 on 1st, 6 on 2nd, non-6 on 3rd OR 6 on 1st, non-6 on 2nd, 6 on 3rd OR non-6 on 1st, 6 on 2nd, 6 on 3rd)
= P(6 on 1st, 6 on 2nd, non-6 on 3rd) + P(6 on 1st, non-6 on 2nd, 6 on 3rd) + P(non-6 on 1st, 6 on 2nd, 6 on 3rd)
= (1/6)(1/6)(5/6) + (1/6)(5/6)(1/6) + (5/6)(1/6)(1/6)
= 5/216 + 5/216 + 5/216
= 15/216
= [spoiler]5/72[/spoiler]



Aside: notice that P(6 on 1st, 6 on 2nd, non-6 on 3rd) = P(6 on 1st, non-6 on 2nd, 6 on 3rd) = P(non-6 on 1st, 6 on 2nd, 6 on 3rd)
So, once you find one of these probabilities, then you automatically know the other two.

Cheers,
Brent
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by mgm » Mon Jun 24, 2013 4:36 am
Brent,

Thanks for the explanation , do you happen to have a list of good probability questions posted on btg handy ? For example in this question if the events were not independent then solution will be different. I really found it helpful to solve from your list of overlapping sets questions ...
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by Brent@GMATPrepNow » Mon Jun 24, 2013 6:07 am
If you want to focus on one topic at a time, you can use BTG's tagging feature to isolate one concept. For example, here are all of the questions tagged as probability questions: https://www.beatthegmat.com/forums/tags/ ... robability

See the left side of that linked page for more tag options.

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by Yaj » Wed Jul 10, 2013 11:59 am
Hi Brent,

Your explanation was really helpful!

I have a doubt (since I am just getting started, please excuse the v basic Qs I shoot) -

Why are we looking at OR (+) & not AND (*) here = 5/216 + 5/216 + 5/216 ? Since we are looking at all conditions working out (i.e 1/6*1/6*5/6) THREE times in a row, would not AND come in here ?

Thanks in advance
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by Brent@GMATPrepNow » Wed Jul 10, 2013 3:23 pm
Yaj wrote:Hi Brent,

Your explanation was really helpful!

I have a doubt (since I am just getting started, please excuse the v basic Qs I shoot) -

Why are we looking at OR (+) & not AND (*) here = 5/216 + 5/216 + 5/216 ? Since we are looking at all conditions working out (i.e 1/6*1/6*5/6) THREE times in a row, would not AND come in here ?

Thanks in advance
The solution requires ANDs and ORs. I highlighted the ORs because this is the part that students often miss.

As you suggested, parts such as P(6 on 1st, 6 on 2nd, non-6 on 3rd) are actually AND probabilities.
So, P(6 on 1st, 6 on 2nd, non-6 on 3rd) = P(6 on 1st AND 6 on 2nd AND non-6 on 3rd)
= P(6 on 1st) x P(6 on 2nd) x P(non-6 on 3rd)
= 1/6 x 1/6 x 6/6
= 5/216

I hope that helps.

Cheers,
Brent
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by send2dar » Sat Jul 13, 2013 6:25 am
Brent, I liked your explanation...
Thanks
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by Matt@VeritasPrep » Sat Jul 13, 2013 9:56 am
Just to add one more point to this discussion:

Once you figure out the probability of getting exactly two sixes in some order - say a 6, a 6, then a non-6 - all you have to do is figure out how the number of possible arrangements, then multiply the probability of one arrangement by the number of arrangements.

For example, the odds of going

Six, Six, Non-Six

are 1/6 * 1/6 * 5/6

There are three ways to arrange the letters SSN, so there are three possible arrangements.

So the probability is 1/6 * 1/6 * 5/6 * 3.

The more complex these sort of problems get, the better it is to think of the arrangements combinatorially.
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by vipulgoyal » Sun Jul 14, 2013 11:03 pm
experts I was trying to get it by (1 - odds = desired) approach, tried multiple times, please shed some light, 1 - {p(getthing one 6) + P(getting 3 6)}
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by Brent@GMATPrepNow » Mon Jul 15, 2013 6:21 am
vipulgoyal wrote:experts I was trying to get it by (1 - odds = desired) approach, tried multiple times, please shed some light, 1 - {p(getting one 6) + P(getting 3 6)}
Yeesh - that's a lot of work. But let's try it out to show that both ways do, indeed, work:

P(exactly two 6's) = 1 - P(not two 6's)
= 1 - P(zero 6's OR one 6 OR three 6's)
Aside: You forgot to mention the option of zero 6's in your post

Let's calculate P(zero 6's OR one 6 OR three 6's)
P(zero 6's OR one 6 OR three 6's) = P(zero 6's) + P(one 6) + P(three 6's)
= (5/6)(5/6)(5/6) + {(1/6)(5/6)(5/6) + (5/6)(1/6)(5/6) + (5/6)(5/6)(1/6)} + (1/6)(1/6)(1/6)
= 125/216 + {75/216} + 1/6
= 201/216

So, P(exactly two 6's) = 1 - 201/216
= 15/216
= [spoiler]5/72[/spoiler]

Cheers,
Brent
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