The probability that my bus is late on any day is 0.2. The probability that it rains tomorrow is 0.4. If the weather and the bus are independent, what is the probability that it rains AND my bus is late?

The probability that it rains and the bus is late is 1/7

Step-by-step explanation:

Practically, we can apply the Bayes’ theorem to solve this.

Mathematically, we use the Bayes’ problem as follows;

P( rain| late) = P(rain ^ late)/P(late) = P( late|rain) • P(rain)/[P(late|rain)P(rain) + P(not late|no rain)P(no rain)]

Where P(no rain) = 1-P(rain) = 1-0.4 = 0.6

P(on time) = 1-P(late) = 1-0.2 = 0.8

Kindly recall that P of raining = 0.4 and the probability that the bus is late is 0.2

Substituting these values into the Bayes’ equation above, we have;

P( rain| late) = (0.2)(0.4)/(0.2)(0.4) + (0.8)(0.6)

= 0.08/(0.08 + 0.48) = 0.08/0.56 = 1/7

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕