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Challenging Task
a. A manufacturer knows that the condensers he makes contain on the average 1% of defectives. He packs them in boxes of 100. What is the probability that a box picked at random will contain 3 or more faulty condensers. b. The probability that an individual suffers from a bad reaction from a certain injection is 0.001, determine the probability that out of 2000 individuals. i. exactly 3 ii. more than 2 individuals will suffer from a bad reaction
a. P(X ≥ 3) = 1 – P(X < 3)
= 1 – P(X = 0) – P(X = 1) – P(X = 2)
= 1 – (100 choose 0) (0.01)^0 (0.99)^100 – (100 choose 1) (0.01)^1 (0.99)^99 – (100 choose 2) (0.01)^2 (0.99)^98
≈ 0.0047
Therefore, the probability that a box picked at random will contain 3 or more faulty condensers is approximately 0.0047.
b. Let Y be the number of individuals suffering from a bad
reaction out of 2000. Then Y follows a binomial distribution with n = 2000 and
p = 0.001 (since the probability of an individual suffering from a bad reaction
is 0.001).
(i) The probability that exactly 3 individuals will suffer from a bad reaction can be calculated as follows:
P(Y = 3) = (2000 choose 3) (0.001)^3 (0.999)^1997
≈ 0.180
Therefore, the probability that exactly 3 individuals will suffer from a bad reaction out of 2000 is approximately 0.180.
(ii) The probability that more than 2 individuals will suffer from a bad reaction can be calculated as follows:
P(Y > 2) = 1 – P(Y ≤ 2)
= 1 – P(Y = 0) – P(Y = 1) – P(Y = 2)
= 1 – (2000 choose 0) (0.001)^0 (0.999)^2000 – (2000 choose 1) (0.001)^1 (0.999)^1999 – (2000 choose 2) (0.001)^2 (0.999)^1998
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