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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?

योग

उत्तर

Let p be the probability of success that people are right-handed

⇒ `p = 90/100 = 9/10`

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

∴ X has a binomial distribution with

`n = 10, p = 9/10, q = 1/10`

∴ P (X = r) = nCr (q)n-r pr

Required probability = P(at most 6 of 10 people are right-handed)

= P (X ≤ 6) = 1 - P (7 ≤ X ≤ 10)

`= 1 - sum_(r =17)^10 ""^10C_r (9/10)^r (1/10)^(10-r)`

`= 1-  sum_(r=7)^10 ""^10C_r (0.9)^r (0.1)^(10-r)`

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अध्याय 13: Probability - Exercise 13.6 [पृष्ठ ५८२]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 13 Probability
Exercise 13.6 | Q 4 | पृष्ठ ५८२

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