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I am doing the following question.

If i have a box of $20$ soccer balls and the independent chance of a soccer ball of being flat is $0.1$. What is the probability of having at least $4$ flat soccer balls.

Now given the question I have decided to calculate $$ P(X=0 \vee 1 \vee 2 \vee 3) $$ and then take this from one.

Starting with $0$ I get an answer of approx. $0.12$ using the binomial distribution equation. But for $P(X=1)$ I am getting $2.4\ldots$ something by going $$ \binom{20}{1} \times 0.9^1 (1-0.9)^{19} $$

I am confused, why am I not getting $.270$ which is the answer I want.

Sorry for tags I have no idea how to put them in.

So my question, why am i getting the wrong answer for my equation?

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    $\begingroup$ Doesn't that mean that on average only $20 \cdot 0.1 = 2$ flat soccer balls are in the box? $\endgroup$ – mvw Dec 18 '14 at 13:27
  • $\begingroup$ "What is the probability of having at least 4 flat soccer balls." $\endgroup$ – Sim Dec 18 '14 at 13:37
  • $\begingroup$ Probably :) related to probability-of-3-heads-in-10-coin-flips $\endgroup$ – mvw Dec 18 '14 at 13:44
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First,

$$\binom{20}1(0.9)^1(1-0.9)^{19}$$

isn’t what you want: you got the exponents back to front. You want the probability that exactly one ball is flat, which is

$$\binom{20}1(0.9)^{19}(1-0.9)^1\approx0.27017\;.$$

Secondly, your arithmetic went astray somewhere, because

$$\binom{20}1(0.9)^1(1-0.9)^{19}=1.8\times10^{-18}$$

exactly.

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