Question

Activity Two: Discrete Probability Distributions in Action (Example #2)

Suppose a health insurance company can resolve 60% of claims using a computerised system, the remaining needing work by humans. On a particular day, 10 claims arrived, assuming claims are independent, what is the probability that

Q2.1) Either 3 or 4 (inclusive) claims require work by a human?

Q2.2) No more than 7 claims require work by a human?

Answer 1

P(work by human) = 1 - 0.6 = 0.4

n = 10

it is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

1) P(3 < X < 4) = P(X = 3) + P(X = 4)

                         = 10C3 * (0.4)^3 * (0.6)^7 + 10C4 * (0.4)^4 * (0.6)^6

                         = 0.4658

2) P(X < 7) = 1 - P(X > 7)

                   = 1 - (P(X = 8) + P(X = 9) + P(X = 10))

                   = 1 - (10C8 * (0.4)^8 * (0.6)^2 + 10C9 * (0.4)^9 * (0.6)^1 + 10C10 * (0.4)^10 * (0.6)^0)

                   = 1 - 0.0123

                   = 0.9877