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Why can’t I use binomial distribution here? I am getting 5/16

Hello,

I have a fundamental doubt. When we choose an event like getting the same number on 2 die in a single throw, why isn’t each trial counted twice (i.e a 5,5 on both die counted twice) are we not excluding one trial by considering it as a single trial ?

I tried that problem several times but I came to the answer which was 5/16, which isn’t any of those choices. I used the binomial distribution to solve this problem. 6C3 times (1/2)^(3) times (1/2)^(3).

(1/2)^3 is for the head that is being tossed three times.

(1-1/2)^(6-3) for the probability of remainder that should not contain the head.

I am lost where I go wrong.

your solution is for getting atleast 3 heads so it includes 6c4 6C5 6c6..remove possibilities of 6c4,6c5,6c6..ans will be 3/16

no he applied the formula for exactly 3 heads.

But how do you do this problem without having to think out all the possible combinations?

Its B, because there are 12 possible “three heads in a row”

The answer is A. There are 2^6 = 64 possible combinations of heads and tails. Only four of them meet the requirement of exactly three heads in a row: HHHTTT, THHHTT, TTHHHT, TTTHHH. Thus the probability is 4/64 or 1/16.

Oxana,

There is a little twist to the problem. You have listed some of the possibilities that fit the requirement, but there are actually a few more ways of arranging all heads.

I see. So there are 8 more combinations like HHHTHT and THHHTH. Altogether 12/64 or 3/16. B