ANHL player's labor negotiating committee is to be selected from 8 player representatives from the Eastern Conference and 7 player representatives from the Western Conference. Find the probability of selecting 3 Eastern Conference representatives and 2 Western Conference representatives. a. 1/ 3003
b. 1/3
c. 56/143
d. 140/429

Using the hypergeometric distribution, it is found that the probability is:

c. 56/143

The players are selected from the sample without replacement, hence, the hypergeometric distribution is used.

Hypergeometric distribution:

The parameters are:

In this problem:

The probability is P(X = 3), hence:

For which the equivalent fraction is option c.

A similar problem is given at link

21 / 143

Step-by-step explanation:

Given that:

Number of Eastern conference reps = 8

Number of western conference rep = 7

Probability of selecting 3 from Eastern reps and 2 from western reps

Probability = required outcome / Total possible outcomes

Total possible outcomes:

selection to be made = 3+ 2 = 5

Total Number of players = 8 +7 = 15

Total possible outcomes

Using combination formula :

nCr = n! / (n-r)!r!

15C5 = 15! / 10!5! = (15 * 14 * 13 * 12 * 11) / (5*4'3*2*1) = 360360 / 120 = 3003

Total possible outcomes = 3003

Required outcome :

8C3 * 7C2

8C3 = 56 ; 7C2 = 21

8C3 * 7C2 = 56 * 21 = 1176

required outcome / Total possible outcomes

= 1176 / 3003

= 21 / 143

7/2

(decimal: 3.5)

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step-by-step explanation: