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Sagot :
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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