# Probability of weighted coin

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!

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.)

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.