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# Whats the probability that in three rolls of a fair die, 6

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CEO
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Whats the probability that in three rolls of a fair die, 6 [#permalink]

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30 Sep 2003, 12:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Whats the probability that in three rolls of a fair die, 6 comes up at least twice.
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01 Oct 2003, 05:52
Using Binomial Distribution,

The probability that an event will happen EXACTLY k times out of the n tries = nck * p^k * (1-p)^(n-k)

Probability of getting a 6 atleast 2 times = Probability of getting 6 exactly 2 times + Probability of getting 6 exactly 3 times

Probability of getting a 6 with a fair die = 1/6

Probability of getting 6 exactly 2 times = 3C2 (1/6)^2 + (1/6)^1 = 3/216
Probability of getting 6 exactly 3 times = 3C3 (1/6)^3 + (1/6)^0 = 1/216

So, Probability of getting a 6 atleast 2 times = 3/216 + 1/216 = 1/54

Am, I right?
Senior Manager
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01 Oct 2003, 06:58
Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36
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02 Oct 2003, 06:48
Vicky wrote:
Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36

Vicki,

In your notes, 5/216 is the probability of getting a 6 on the first 2 tries only, right? Should we not take in to account the cases where we can have 6's on the 2nd and 3rd tries and also 1st and 3rd tries?
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02 Oct 2003, 08:46
yes amarsesh i made a mistake...
we should include getting 6 for 2nd, 3rd and 1st, 3rd time as well.
That will make it = 5*3/216 + 1/216 = 16/216 = 2/27
correct me if i am wrong.
thanks
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02 Oct 2003, 08:53
Vicky wrote:
yes amarsesh i made a mistake...
we should include getting 6 for 2nd, 3rd and 1st, 3rd time as well.
That will make it = 5*3/216 + 1/216 = 16/216 = 2/27
correct me if i am wrong.
thanks

Yes, you are right. I made a mistake in my calculations. I took 1/6 for the prob. that 6 is not obtained when I should have taken that as 5/6

It's obvious that I was not looking into my mistakes

Last edited by amarsesh on 02 Oct 2003, 12:06, edited 1 time in total.
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02 Oct 2003, 12:00
i did the same mistake vicks did...but i think binomial is a great way to solve problems where we have a fixed # of trials and the prob of success and failure is known.. i will try to use it in future..

Three ways we can get two six's.

6 # 6
# 6 6
6 6 #

Two six's = 3* (1/6)^2 * 5/6 =15/216

Three six's = 1/6^3 = 1/216

Probability = 16/216 = 2/27

thanks
praetorian
Re: Probability # 1   [#permalink] 02 Oct 2003, 12:00
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