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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 $X$? –  Ilya Mar 1 '12 at 8:01
    
@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... –  Ahoura Ghotbi Mar 1 '12 at 8:13
    
Would you fix it now in the original post? –  Ilya Mar 1 '12 at 8:14

1 Answer 1

You're right, this is just the formula for the number of ways of choosing $x$ out of $n$ tosses. To get the probabilities, you need to multiply this by the probabilities of the individual results. Thus, to get $x$ heads and $n-x$ tails out of $n$ tosses, you have $\binom nx$ ways of arranging the $x$ heads among the $n$ tosses, and for each of these arrangements the probability for it to occur contains a factor $p$ for each heads and a factor $q$ for each tails and is therefore $p^xq^{n-x}$. The probability for all $\binom nx$ of the events is therefore $\binom nx p^xq^{n-x}$. You can check this result by using the binomial expansion to find

$$\sum_{x=0}^n\binom nxp^xq^{n-x}=(p+q)^n=1^n=1\;,$$

so the total probability for any number of heads from $0$ to $n$ to occur is $1$, as it should be.

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