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I'm having trouble with the following question. A weighted coin lands heads 2/3 of the time whereas it lands tails 1/3 of the time. If the coin is tossed 10 times what is the probability that it will land exactly 4 heads?

I would solve the problem doing (10 choose 4)(2/3)^4(1/3)^6 Kind of like a binomial distribution with probabilities of landing heads = 2/3 and n = 10. Is this correct? Similarly if I wanted to do the P(x<=4) where x is the probability of landing heads I would just sum up the probabilities from 10 choose 0 to 10 choose 4? Thanks guys!

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Yes.

(I would have left the answer at that, but the site requires a minimum of 30 characters in an answer. The OP has the right idea in both the "exactly four" and the "less than or equal to four" case.)

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That looks good to me.

To be clear,

(10 choose 4)(2/3)^4(1/3)^6 is the probability of getting exactly 4 heads.

More generally,

(10 choose n)(2/3)^n(1/3)^(10-n) is the probability of getting exactly n heads.

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