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)