Asandwich shop delivers lunches by bicycle to nearby office buildings. unfortunately, sometimes the delivery is made later than promised. a delay can occur either because food preparation takes too long, or because the bicycle rider gets lost. last month the food preparation took longer than expected or the rider got lost of the time. during the same month, the food preparation took longer than expected times and the bicycle rider got lost times. there were deliveries made during the month. for a randomly selected delivery last month, what is the probability that both the food preparation took too long and the rider got lost?

1/200

Step-by-step explanation:

Some missing piece of information:

The food preparation took too long or the rider got lost 7% of the time.

The food preparation took longer than expected 11 times and the bicycle rider got lost 4 times.

There were 200 deliveries made during the month.

Therefore:

The probability that food preparation took longer, P(long) = 11/200

probability that the rider got lost, P(lost) = 4/200

the probability that rider got lost or food preparation took long, P(long or lost) = 7/100

From probability addition rule,

P(A or B) = P(A) + P(B) - P(A and B)

hence, P(A and B) = P(A) + P(B) - P(A or B)

In this case, we have

P(long and lost) = P(long) +P(lost) - P(long or lost)

= 11/200 + 4/200 - 7/100

= 1/200

step-by-step explanation:

sry but any answer choices