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

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alok kumar singh | Contributor-Level 10

4 months ago

80. It is given that $90 %$ of people are right handed.

$p = P$ (right handed) $= 90 % = \frac{90}{100} = \frac{9}{10}$ and $q = P$ (left handed) $= 1 - \frac{9}{10} = \frac{1}{10}$

Using Binomial distribution,

Probability more than $6$ people are right handed is given by

$= \sum_{r = 7}^{10} {10c}_{r}$

$p^{r} q^{n - r}$

$= \sum_{r = 7}^{10} {10c}_{r}$

${(\frac{9}{10})}^{r} {(\frac{1}{10})}^{10 - r}$

Hence, probability of having at most $6$ right handed people:

$= 1 - P$ (more than $6$ people are right handed)

$= 1 - \sum_{r = 7}^{10} {10c}_{r}$

${(\frac{9}{10})}^{r} {(\frac{1}{10})}^{10 - r}$

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

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