Theory of Expectation :: Problems on Profits, Business : Probability Distribution

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A building contractor is considering one of the two contracts for new down town buildings "A" and "B". It has been estimated that a profit of Rs. 2,00,000 would be made on building "A". Bidding costs for the contractor on building "A" would be Rs. 10,000. On building "B", the estimated profit is Rs. 5,00,000 and bidding costs would be Rs. 20,000. The probability of the contract being awarded is 2/5 for building "A" and 1/5 for building "B" (Assume bidding costs are incurred only in case the contract is not obtained).

i) What is the contractor's expectation for building A?
ii) What is the contractor's expectation for building B?
iii) For which Job should the contractor bid?

Net Answers :
[Expectation(A): +Rs.74,000 ; Expectation(B): +Rs.84,000 ;
The contractor can bid for both the contracts. However, if he can bid only for one of the contracts, he should prefer contract "B".]

Solution

Let "x", "y' indicate the profit made by the contractor on the contract "A" and contract "B" respectively.
For "Contract A"
The expected earnings/profit that can be made by the contractor would be

Rs. 2,00,000 if he is awarded the contract

− Rs. 10,000 if he is not awarded the contract

If he is not awarded the contract, he would make a loss equal to the bidding cost, since he is to bear the bidding cost if he is not awarded the contract.
⇒ The values carried by the variable ("x") would be either − 10,000 or + 2,00,000
⇒ "X" is a discrete random variable with range = {− 10,000, + 2,00,000}
"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"
Probability for the contract

to be awarded the Contract "A" = 2/5 i.e. 0.4
⇒ P(Awarded) = 0.4

not to be awarded the Contract "A" = 3/5 i.e. 0.6
⇒ P(Not Awarded) = 0.6

Considering the two events of "Awarded" or "Not Awarded" to be the only possibilities, they are exhaustive events
⇒ P("Awarded" ∪ "Not Awarded") = 1 → (1)
Since there can either be atleast one error or there cannot be any error, the two events of "Awarded" or "Not Awarded" are mutually exclusive
⇒ P("Awarded" ∩ "Not Awarded") = 0
(Or) P("Awarded" ∩ "Not Awarded") = P("Awarded") + P("Not Awarded") → (2)
From (1) and (2) we can write
P(""Awarded" ∪ "Not Awarded"") = P("Awarded") + P("Not Awarded") = 1
⇒ P("Awarded") + P("Not Awarded") = 1
⇒ 0.4 + P("Not Awarded") = 1
⇒ P("Not Awarded") = 1 − 0.4
⇒ P("Not Awarded") = 0.6
Probability that the profit earned by the contractor would be

− Rs. 10,000

⇒ P(X = − 10,000) = P (Not Awarded)

= 0.6

+ Rs. 2,00,000

⇒ P(X = + 2,00,000) = P (Awarded)

= 0.4

The probabilty distribution of "x" would be

x − 10,000 + 2,00,000

P(X = x) or P 0.6 0.4

Calculations for Mean

x P Px

− 10,000 0.6 − 6,000

+ 2,00,000 0.4 80,000

Total 1 74,000

The contractors expected profit from Contract "A"
⇒ Expectation of "x"

⇒ E (x) = Σ px

= + Rs. 74,000

For "Contract B"
The expected earnings/profit that can be made by the contractor would be

Rs. 5,00,000 if he is awarded the contract

− Rs. 20,000 if he is not awarded the contract

If he is not awarded the contract, he would make a loss equal to the bidding cost, since he is to bear the bidding cost if he is not awarded the contract.
⇒ The values carried by the variable ("y") would be either − 20,000 or + 5,00,000
⇒ "Y" is a discrete random variable with range = {− 20,000, + 5,00,000}
"Y" represents the random variable and P(Y = x) represents the probability that the value within the range of the random variable is a specified value of "y"
Probability for the contract

to be awarded the Contract "B" = 1/5 i.e. 0.2
⇒ P(Awarded) = 0.2

not to be awarded the Contract "B" = 4/5 i.e. 0.8
⇒ P(Not Awarded) = 0.8

Considering the two events of "Awarded" or "Not Awarded" to be the only possibilities, they are exhaustive events
⇒ P("Awarded" ∪ "Not Awarded") = 1 → (1)
Since there can either be atleast one error or there cannot be any error, the two events of "Awarded" or "Not Awarded" are mutually exclusive
⇒ P("Awarded" ∩ "Not Awarded") = 0
(Or) P("Awarded" ∩ "Not Awarded") = P("Awarded") + P("Not Awarded") → (2)
From (1) and (2) we can write
P(""Awarded" ∪ "Not Awarded"") = P("Awarded") + P("Not Awarded") = 1
⇒ P("Awarded") + P("Not Awarded") = 1
⇒ 0.2 + P("Not Awarded") = 1
⇒ P("Not Awarded") = 1 − 0.2
⇒ P("Not Awarded") = 0.8
Probability that the profit earned by the contractor would be

− Rs. 20,000

⇒ P(X = − 20,000) = P (Not Awarded)

= 0.8

+ Rs. 5,00,000

⇒ P(X = + 5,00,000) = P (Awarded)

= 0.2

The probabilty distribution of "y" would be

x − 20,000 + 5,00,000

P(Y = y) or P 0.8 0.2

Calculations for Mean

y P Py

− 20,000 0.8 − 16,000

+ 5,00,000 0.2 1,00,000

Total 1 84,000

The contractors expected profit from Contract "B"
⇒ Expectation of "y"

⇒ E (y) = Σ py

= + Rs. 84,000

The contractor can bid for both the contracts. However, if he can bid only for one of the contracts, he should prefer contract "B".

Credit : Vijayalakshmi Desu