Sue either travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4 . The probability that she catches the bus from the shop is 0.7 . Show the probability that sue walks at least one way is 0.72

P(A∪B)=0.72. The answer is provided below.

Step-by-step explanation:

Sue travels by bus or walks whenever she goes to the shops.

Probability of (catching the bus to shop ), P(A)= 0.4

Probability of (catching the bus from the shop ), P(B) = 0.7

Both of the events A and B are independent events.

Therefore, P(A∩B) = 0.4 × 0.7

 = 0.28

Probability that Sue walks one way will be = 1 - P(A∩B)

= 1 - 0.28

= 0.72

Hence, the probability that Sue will walk at one way is 0.72

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)