PROBABILITY
An insurance salesman sells policies to 5 men, all of identical age in good health. according to the actuarial tables, the probability that a man of this particular age will be alive 30years hence is 2/3, find the probability that in 30 years, a) all men will be alive b) at least 3 men will be alive c) only 2 men will be alive d) at least 1 man will be alive,
Answers
n = 5p = 2/3q = 1/3P(x) = C(n,r) * p^x * q^(n-x)a) P(x=5) = C(5,5) * (2/3)^5 = 32/243b) P(x>=3) = C(5,3) * (2/3)^3 * (1/3)^2 + C(5,4) * (2/3)^4 * (1/3)^1 + C(5,5) * (2/3)^5 = 192/243c) P(x=2) = C(5,2) * (2/3)^2 * (1/3)^3 = 40/243d) P(x>=1) = 1 - P(x=0) = 1 - C(5,0) * (1/3)^5 = 242/243
20 March 2016
n = 5p = 2/3q = 1/3P(x) = C(n,r) * p^x * q^(n-x)a) P(x=5) = C(5,5) * (2/3)^5 = 32/243b) P(x>=3) = C(5,3) * (2/3)^3 * (1/3)^2 + C(5,4) * (2/3)^4 * (1/3)^1 + C(5,5) * (2/3)^5 = 192/243c) P(x=2) = C(5,2) * (2/3)^2 * (1/3)^3 = 40/243d) P(x>=1) = 1 - P(x=0) = 1 - C(5,0) * (1/3)^5 = 242/243
06 April 2016
18 eggs = 3%1% = 18 divided by 318/3 = 6Therefore 1% = 6We need to find out how many eggs are in the whole batch.Whole tells us that we need to find 100%6 x 100 (%) = 600.There are 600 in the whole batch.Or100% = 18/3 x 100
06 April 2016
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