Question

There are three modes of transporting material from Ontario to Florida, namely, by land, sea, or air. Also land transportation may be by rail or highway. About half of the materials are transported by land, 30% by sea, and the rest by air. Also, 40% of all land transportation is by highway and the rest by rail shipments. The percentages of damaged cargo are, respectively, 10% by highway, 5% by rail, 6% by sea, and 2% by air. (1) What percentage of all cargoes may be expected to be damaged? (2) If a damaged cargo is received, what is the probability that it was shipped by land? By sea? By air?

Answer 1

0.057

0.6140

0.3158

0.0701

Explanation:

Given that:

Let :

P(L) = Number transported by land = half = 50% = 0.5

P(S) = number transported by sea = 30% = 0.3

P(A) = Number transported by air = (100 - (50 + 30))% = 20% = 0.2

P(H) = highway transport = 40% of land transport = 0.4 * 0.5 = 0.2

P(R) = Rail shipment =(100- 40)% = 60% of land transport = 0.6 * 0.5 = 0.3

Percentage of damaged cargo :

Let probability of damage = P(d)

P(d | H) = 0.1

P(d | R) = 0.05

P(d | S) = 0.06

P(d | A) = 0.02

1) What percentage of all cargoes may be expected to be damaged

[P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

(0.1*0.2) + (0.05*0.3) + (0.06*0.3) + (0.02*0.2) = 0.057

(2) If a damaged cargo is received, what is the probability that it was shipped by ;

land?

([P(d | H)*p(H)] + [P(d | R)*p(R)]) / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

((0.1*0.2) + (0.05*0.3)) / (0.1*0.2) + (0.05*0.3) + (0.06*0.3) + (0.02*0.2)

= 0.035 / 0.057

= 0.6140

By sea?

[P(d | S)*p(S)] / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

(0.06 * 0.3) / 0.057

= 0.3158

By air?

[P(d | A)*p(A)] / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

(0.02 * 0.2) / 0.057

= 0.0701