Answer:
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
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:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
The probability is P(X = 3), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 3) = h(3,15,5,8) = \frac{C_{8,3}C_{7,2}}{C_{15,5}} = 0.3916[/tex]
For which the equivalent fraction is option c.
A similar problem is given at https://brainly.com/question/24826394
