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

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.

Cheers,
Brent

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

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