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I am trying to solve the following problem :

A biased coin which has P(heads) = p = .7 and p(tails) = q = .3 is tossed 3 times. the coin is tossed in such a way that the outcomes on each toss are independent. we obtain the probability of 0,1,2 and 3 heads using the formula : $$ C = n! / (x!(n-x)!) $$

i) P(X=0) = ii) P(X=1) =

now I solved the above question using the formula however I got 1 as my answer and the formula that was mentioned is to find the combinations possible, not the probability. Could anyone explain how I would need to use the above formula to get the answer?

Variable Hint : - x = X - X = number of heads you want to achieve - N = total number of tries (3)

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    Your notation is inconsistent: what are $n,x$ and $X$?2012-03-01
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    @Ilya sorry about that, n is the number of tries (3) x is the number of heads you want to achieve and X = x technically...2012-03-01
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    Would you fix it now in the original post?2012-03-01

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