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The equation for binomial distribution is as follows,

$$P(x) = \binom{n}{x}\cdot p^x \cdot (1 - p)^{n-x}$$

My question is, why does it multiply the odds of an event firing with the odds of an event not firing? I do not see the logic behind this, for example,

if a pitcher has a 20% chance of hitting a home run, and six balls were thrown at him, what are the odds that he will get at least two home runs?

Logically, it makes sense that the answer to this question will be $1$ - the chance of missing 4 balls

So logically, we should be able to say the answer is, $1 - \dbinom{6}{4} \cdot \left(\dfrac{4}{5}\right)^{4}$ but of course this gives an answer greater than one and is thus impossible. So my question is, where is the logic between multiplying the odds of the event not firing by the odds of the event firing?

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  • $\begingroup$ $1 - \binom{6}{4} * (\frac{4}{5})^{4}$ is not greater than $1$. $\endgroup$ – 5xum Mar 24 '14 at 13:53
  • $\begingroup$ Do 'pitchers' hit homeruns? Is $p$ here the chance to 'miss'? Confusing. $\endgroup$ – drhab Mar 24 '14 at 13:58
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As you are doing n tries, you have to succeed x times and fail (n-x) times. hence $p^x$ and $(1-p)^{n-x}$. Then, for each value of x, you have to determine how many patterns will realize exactly x wins and (n-x) fails. So you choose x among n wins in the series of n tries. Hence $\binom{n}{x}$.

In your example, "at least 2 home runs" is 2, 3, 4, 5 or 6 homeruns. You spare yourself some work if you view it as "the complementary of 0 or 1 homeruns".

So

$$1- \binom{6}{0} \cdot p^0 \cdot (1-p)^6-\binom{6}{1} \cdot p^1 \cdot (1-p)^5$$.

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