Step-by-step explanation:
since he was already on time on the first day, the question is reduced to the probability of being on time on exactly one of the next 2 days.
the probability to be on time on exactly one day means to be on time on one day and to be late on the other.
the probability to be late is the opposite of the probability to be on time, so
1 - 0.7 = 0.3
so, the probability for this combined event for one of the day combinations is
0.7 × 0.3 = 0.21
how many day combinations (1 day on time, the other late) can we have ? 2 (either on time on the first and late on the second, or late on the first and on time on the second).
so, we need to multiply this with the basic probability and get
0.21 × 2 = 0.42
so, b. is the right answer.
