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Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

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  • A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
Since $90\%$ of people are right handed, P (righthanded) $ = p = \large\frac{90}{100}=\frac{9}{10}$
Out of 10 selected persons, We want at most 6 right handed persons.
P (at most 6 right handed person) = $p(X\leq 6)$ =$1-P(X\geq 6)$ =$1-[p(X=7)+p(X=8)+p(X=10)]$
=$ 1 - \large\sum_{7}^{10} C^{10}_{r} $$\large\frac{9}{10}^{r} \frac{1}{10}^{10-r}$
=$1-[\large C^{10}_{7} p^{7}q^{3}+c^{10}_{8}p^{9}q^{1}+c^{10}_{10}p^{10}q_{10}]$
=$1-\large [C^{10}_{7}\left(\frac{9}{10}\right)^{7}\left(\frac{1}{10}\right)^{3}+C^{10}_{8}\left(\frac{9}{10}\right)^{8}\left(\frac{1}{10}\right)^{2}+C^{10}_{9}\left(\frac{9}{10}\right)^{9}\left(\frac{1}{10}\right)^{1}+C^{10}_{10}\left(\frac{9}{10}\right)^{10}\left(\frac{1}{10}\right)^{0}]$
=$1-\large[$$120 \large\frac{9^{7}}{10^{10}}$$+45\large \frac{9^{8}}{10^{10}}$$+10\large\frac{9^{9}}{10^{10}}$$+1\large\frac{9^{10}}{10^{10}}]$
=$1-\large\frac{9^{7}}{10^{10}}$$(120+45\times 9+10\times 9^{2}+9^{3})$
answered Jun 22, 2013 by balaji.thirumalai

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