Five firms, F1, F2F5 each offer bids on three separate contracts, C1, C2 and C3. Any one firm will be awarded at most one contract. The contracts are quite different, so an assignment of C1 to F1, say, is to be distinguished from an assignment of C2 to F1.

a. How many sample points are there altogether in this experiment involving assignment or contracts to the firms? (No need to list them all.)
b. Under the assumption of equally likely sample points, find the probability that F3 is awarded under a contract.

a. 60

b. 3/5

Step-by-step explanation:

a. There are 5 firms and at a time only 3 firms can bid. The order of assignment is important since it is mentioned in the question that the assignments for all firms are different. So, the total number of sample points are:

Sample points = ⁵P₃ = 60

b. Any one firm will be awarded at most one contract so at one time, only 3 out of 5 firms will be awarded a contract. F3 can be awarded either one of the three contracts (3 possible ways) and the rest of the 2 contracts will be awarded to the other 4 firms. These 4 firms can be awarded the contracts in ⁴P₂ = 12 ways

F3 can be awarded either one of the 3 contracts so,

No. of ways F3 can be awarded a contract = 3 x 12 = 36

Total number of sample points = 60

Probability that F3 will be awarded a contract = 36/60 = 3/5

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step-by-step explanation:

step-by-step explanation:

milk=16 large

soda=280 kids

juice box=82 regular

63 soda +63 regular

124 large total