Question

How to calculate questions (c) and (d)? Please give me detailed steps, thanks!

(a) There is a probability of 0.2 that a toy punched by a trainee operator will contain an error. Find the probability that a random sample of 4 toys punched by a trainee operator will contain just 1 error-free toy.

(b) There is a probability of 0.05 that a toy punched by an experienced operator will contain an error. Find the probability that in a random sample of 20 toys punched by an experienced operator there will be no more than 3 toys containing errors.

(c) Of the total number of toy punched in a day by a pool of operators, 4% are punched by trainee operators and 96% by experienced operators. Show that the probability that a toy chosen at random from a daily output in this pool contains an error is equal to 0.056.

(d) Given that 2 toys chosen at random from daily output contain error, calculate the probability that one toy was punched by a trainee operator and the other was punched by an experienced operator.

Answer 1

(c)

Given,

P(error | trainee) = 0.2

P(error | experienced) = 0.05

P(trainee) = 4% = 0.04

P(experienced) = 96% = 0.96

By law of total probability,

P(error) = P(trainee) P(error | trainee) + P(experienced) P(error | experienced)

= 0.04 * 0.2 + 0.96 * 0.05

= 0.056

(d)

By Bayes theorem,

P(trainee | error) = P(error | trainee) * P(trainee) / P(error)

= 0.2 * 0.04 / 0.056

= 0.1428571

P(experienced | error) = 1 - P(trainee | error) = 1 - 0.1428571 = 0.8571429

Probability that one toy was punched by a trainee operator and the other was punched by an experienced operator

= Probability that first toy containing error was punched by a trainee operator and second toy was punched by an experienced operator + Probability that first toy containing error was punched by a experienced operator and second toy was punched by trainee operator

= 0.1428571 * 0.8571429 + 0.8571429 * 0.1428571

= 0.2448979