Question

The probability of the first operator typing a programme correctly is 0.4, and the probability of the second operator typing a programme correctly is 0.7. Each operator types two programmes. Find the probability that: a) all the programs are typed incorrectly; b) two contain errors; c) at least one of the programs is error-free.

Answer 1

Ответ:

a) Probability that all programs are typed incorrectly by both operators:

P(Operator 1 makes an error) = 1 - P(Operator 1 types correctly) = 1 - 0.4 = 0.6

P(Operator 2 makes an error) = 1 - P(Operator 2 types correctly) = 1 - 0.7 = 0.3

Since the events are independent:

P(Both operators make errors) = P(Operator 1 makes an error) * P(Operator 2 makes an error) = 0.6 * 0.3 = 0.18

b) Probability that two programs contain errors:

There are two ways this can happen: (1) Operator 1 makes an error and Operator 2 types correctly, or (2) Operator 1 types correctly and Operator 2 makes an error.

P(Operator 1 makes an error) * P(Operator 2 types correctly) + P(Operator 1 types correctly) * P(Operator 2 makes an error) = (0.6 * 0.7) + (0.4 * 0.3) = 0.42 + 0.12 = 0.54

c) Probability that at least one of the programs is error-free:

You can calculate the complementary probability, i.e., the probability that both programs contain errors and then subtract it from 1.

P(At least one error-free program) = 1 - P(Both programs contain errors) = 1 - (0.18) = 0.82

So, the probabilities are:

a) All programs are typed incorrectly: 0.18

b) Two programs contain errors: 0.54

c) At least one program is error-free: 0.82