Final answer:
The probability of not getting a 0 in K random digits is (0.9)^K; the probability that neither 0 nor 1 appears is (0.8)^K; and the probability that at least one of 0 and 1 does not appear is 1 - (0.01)^K.
Step-by-step explanation:
The question asks about finding the probability of certain events when K random digits are selected. Specifically, it inquires about the probability that in K random digits, (a) 0 doesn't appear, (b) neither 0 nor 1 appears, and (c) at least one of 0 and 1 does not appear. We can use the given probability addition formula P(A U B) = P(A) + P(B) - P(A AND B) to solve different probability scenarios.
For case (a), assuming there are 10 possible digits (0 to 9), the probability of not getting a 0 each time is 9 out of 10 or 0.9. Since we are looking at K independent events, the probability that 0 never appears is (0.9)^K.
For case (b), not getting a 0 or a 1 on a single random selection would have a probability of 8 out of 10 or 0.8. Thus, the probability that neither 0 nor 1 appears over K selections is (0.8)^K.
For case (c) we can utilize the formula P(A U B) = P(A) + P(B) - P(A AND B), where A is the event '0 does not appear' and B is the event '1 does not appear'. The probability that at least one of the two digits 0 and 1 does not appear is 1 - P(both 0 and 1 appear), which can be calculated as 1 - (P(A') * P(B')), where P(A') is the probability that 0 appears and P(B') is the probability that 1 appears. This amounts to 1 - (0.1 * 0.1) = 1 - 0.01
= 0.99 for one selection, and for K selections, it would be 1 - (0.01)^K.