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?

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

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?

Solution:

Given that, $90 \%$ of the people are right handed

Let $p$ represent the probability of right-handed people and $q$ represent the probability of left-handed people.

$p=9 / 10$ and $q=1-9 / 10=1 / 10$

Using the binomial distribution, the likelihood of having more than 6 right-handed people is:

$\sum_{r=7}^{10} 10 c_{r} p^{r} q^{n-r}=\sum_{r=7}^{10} 10 C_{r}\left(\frac{9}{10}\right)^{r}\left(\frac{1}{10}\right)^{10^{-r}}$

Hence, the probability of having more than 6 right handed people:

$=1-\mathrm{P}$ (More than 6 people are right handed)

$=1-\sum_{r=7}^{10} 10 C_{r}(0.9)^{r}(0.1)^{10^{-r}}$

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