Answer to Question #263054 in Statistics and Probability for Nathan

Question

A small online shop has only three drivers (Sandra, Peter, and Josh) to deliver orders in a certain area in Johannesburg South. Sandra delivers twice as much orders as Peter delivers and Josh delivers 10% of the orders. Sandra is late 10% of her deliveries, Peter is late 20% of his deliveries and Josh is late 50% of his deliveries. If a delivery is on time (not late), what is the probability that Sandra delivered this order?

Answer 1

Let S, P, and J be the event that order deliver by Sandra, Peter and Josh. Sandra delivers twice as much orders as Peter

S=2P

Josh deliver 10 % of orders

$S + P + 10 = 100 \\ 2P + P +10 = 100 \\ P = 30 \; \% \\ S = 60 \; \% \\ P(S) = 0.60 \\ P(P) = 0.30 \\ P(J) = 0.10$

Let D be the event that order deliver on time and $\bar{D}$ be the event that order deliver not on time.

$P(\bar{D} |S) = 0.10 \\ P(\bar{D}|P) = 0.20 \\ P(\bar{D}|J) = 0.50$

Probability that order not delivered

$P(\bar{D}) = P(\bar{D} |S) \times P(S) + P(\bar{D}|P) \times P(D) + P(\bar{D}|J) \times P(J) \\ = 0.10 \times 0.60 + 0.20 \times 0.30 + 0.50 \times 0.10 \\ = 0.06 + 0.06 +0.05 \\ = 0.17 \\ P(D) = 1 -0.17 = 0.83$

Probability that Sandra deliver order on times

$P(S|D) = \frac{P(D|S)P(S)}{P(D)} \\ = \frac{(1- P(\bar{D} |S) )P(S)}{P(D)} \\ = \frac{(1- 0.10) \times 0.60}{0.83} \\ = \frac{0.54}{0.83} \\ = 0.6506$