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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? - Mathematics

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प्रश्न

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?

बेरीज

उत्तर

Let X be the number of people that are right-handed in the sample of 10 people.
X follows a binomial distribution with n = 10,

\[p = 90 % = 0 . 9 and q = 1 - p = 0 . 1\]

\[P(X = r) = ^{10}{}{C}_r (0 . 9 )^r (0 . 1 )^{10 - r} \]

\[\text{ Probability that at most 6 are right - handed } = P(X \leq 6)\]

\[ = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)\]

\[ = 1 - {P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}\]

\[ = 1 - \sum^{10}_{r = 7}{^{10}{}{C}_r} (0 . 9 )^r (0 . 1 )^{10 - r}\]

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पाठ 33: Binomial Distribution - Exercise 33.1 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 33 Binomial Distribution
Exercise 33.1 | Q 18 | पृष्ठ १३

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