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In a city 40% are vegetarians. In a random sample of 6 persons, find the probability that (i) 3 are vegetarians (ii) at least one is a vegetarian. |
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Answer» Given : p = 40% = 0.4 q = 1 – p = 70.6 6n = 6 Let ‘x’ be the number of vegetarians is a Binomial variate with the parameters n = 6 p = 0.4 and q = 0.6, then, The p.m.f. is – p(x) = ncxpxqn – x ; x = 0,1,2, …… n = 6Cx (0.4)x (0.6)6 – 3 ; x = 0,1,2 ……… 6 (i) p(3 are vegetarians) = p(x = 3) = 6C3 (0.4)3 (0.6)6-3 = 20 x 0.064 x 0.216 = 0.2765 (ii) p(atleast one is a vegetarian) = p(x ≥ 1)= 1 – p(x < 1) = 1 – p(x = 0) = 1 – 6C0 (0.4)0 (0.6)6-0 = 1 – 1 x 1 x 0.046 = 0,9554 |
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