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10.Suppose that you have bought 3 large glass-fronted cabinets. You have three workers, A, B and C, to help you with transporting the cabinets from the furniture store to your home, and you want them to finish it in (at most) one trip each. From past experience, it is known that the probability that A breaks something in a given trip is 10%, the probability that B breaks something in a given trip is 15%, and the probability that C breaks something in a given trip is 20%. The events that A, B, or C breaks something in a given trip are independent of each other. You come up with three strategies: (a) A transports all the furniture, (b) A transports two cabinets and B transports one cabinet, or (c)

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Answer:

Step-by-step explanation:

To find the expected no of broken cabinets for each strategy.

If A transport all the furniture

Suppose X₁ = number of furniture A broken

X₁ follows a binomial with n = 3, p = 0.10

E(X₁) = np

E(X₁) = 3 × 0.1

E(X₁) = 0.3

If A transport two (2) cabinets & B transport one (1) cabinet

expected no. of the broken cabinet by A out of 2 is

= 2 × 0.1

= 0.2

Suppose X₂ denotes the no of broken furniture by B

E(X₂) = np

where n= 1 and p = 0.15

E(X₂) = np

E(X₂) = 1 × 0.15

E(X₂) = 0.15

Thus. the expected number of broken cabinets = 0.2 + 0.15 = 0.35

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