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If 10 coins were flipped, what would be the probability of 9 of them landing on heads? Would it be 1/512 or would the probability be different considering that 1 coin failed at landing on heads? Thank you

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  • $\begingroup$ $\frac{1}{512}=\frac{1}{2^9}$ is the probability that in nine flips, all nine are heads, or equivalently that in ten flips, the first nine are heads and the tenth is either. $\endgroup$ – JMoravitz Sep 6 '16 at 18:26
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There are $2^{10}$ (equally probable) outcomes of $10$ coin flips. There are $10$ different possible outcomes where exactly 9 coins land on heads: $HHHHHHHHHT$, $HHHHHHHHTH$, ... $THHHHHHHHH$. So the probability is $\frac{10}{2^{10}}$.

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Define a coin ending up being heads a success, with probability $p$. Then, by the binomial distribution, we have

$$\binom{10}{9}p^{9}(1-p)^{1}.$$

If the coin is fair, $p=\frac{1}{2}$ and the probability becomes $$\binom{10}{9}\left(\frac{1}{2}\right)^{10}=10\left(\frac{1}{2}\right)^{10}\approx 1\%.$$

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  • $\begingroup$ So would that make a 1/102.4 chance then? $\endgroup$ – user329957 Sep 6 '16 at 18:39
  • $\begingroup$ @user329957 Yes! $\endgroup$ – Bobson Dugnutt Sep 6 '16 at 18:47

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