#### About Chris Lele

Chris Lele is the GRE and SAT Curriculum Manager (and vocabulary wizard) at Magoosh Online Test Prep. In his time at Magoosh, he has inspired countless students across the globe, turning what is otherwise a daunting experience into an opportunity for learning, growth, and fun. Some of his students have even gone on to get near perfect scores. Chris is also very popular on the internet. His GRE channel on YouTube has over 10 million views.You can read Chris's awesome blog posts on the Magoosh

GRE blog and

High School blog!You can follow him on

Twitter and

Facebook!

If you don’t want to particularly write all possibilities then I got another method- its (1/2)^4 +(1/2)^5 +(1/2)^5+(1/2)^4=3/16….that’s all.simply multiply probability of particular outcome and add the cases. For eg. HHHT(HT)-The terms outside the braces are fixed so they have probability (1/2)^4 and terms inside the braces can be permuted viz. They could be HT TH TT HH so they have probability of 1 similarly for rest three cases

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