Question

A manufacturer uses high speed weighing and packing machines with accuracy adequate to ensure that only one bag in 20 is likely to be underweight. When the machine are set to weigh a given weight. A customer receives a consignment of 60 bags. What are the chances that four or more will be underweight?

Answer 1

Please note ^nC_r = \frac{n!}{(n-r)! * r!}

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Also sum of probabilities from 0 till n = 1, i.eP(0) + P(1) + P(2) +.......+P(n) = 1

Here n = 60, p = 1 / 20 = 0.05, q = 1 – p = 0.95

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P(X 4) = P(4) + P(5) + ...... + P(60) = 1 - [P(0) + P(1) + .P(2) + P(3)]

P(X = 0) = ⁶⁰C₀ * (0.05)⁰ * (0.95)^60-0 = 0.0461

P(X = 1) = ⁶⁰C₁ * (0.05)¹ * (0.95)^60-1 = 0.1455

P(X = 2) = ⁶⁰C₂ * (0.05)² * (0.95)^60-2 = 0.2259

P(X = 3) = ⁶⁰C₃ * (0.05)³ * (0.95)^60-3 = 0.2298

Therefore, P(X 4) = 1 - (0.0461 + 0.1455 + 0.2259 + 0.2298) = 0.3257