Question

An urn contains ten marbles, of which 4 are green, 3 are blue, and 3 are...

An urn contains ten marbles, of which 4 are green, 3 are blue, and 3 are red. Three marbles are to be drawn from the urn, one at a time without replacement.

(a) Let ?? be the event that the ?th marble drawn is green. Find ?(?1 ∩ ?2 ∩ ?3), which is the probability that all three marbles drawn are green. Hint: Use the Multiplicative Law of Probability

(b) Now let ?? be the random variable defined to be the number of green marbles drawn. Explain why ?? must be hypergeometric, give the values of the parameters (?, ?, ?), and use the hypergeometric pmf to find ?(? = 3), which should agree with the probability found in (a).

(c) Give the expected value and variance of ?.

(a)

Total marbles =10

Green marbles =4

?(?1 ∩ ?2 ∩ ?3) =(4/10)(3/9)(2/8) =24/720 =0.0333

(b)

The random variable 'Y=y' is a Hypergeometric Distribution because - A sample of size of "n" is randomly selected without replacement from a population of size "N" in which "r" items are treated as successes and "N - r" items are treated as failures and y =number of successes in the sample of size "n".

Here, Success =Green marble

The parameters are: N =Total number of marbles =10; n =number of marbles drawn =3; r =number of Green marbles =4

Formula: P(Y =y) =C(r, y)*C(N-r, n-y)/C(N, n)

P(Y =3) =C(4, 3)*C(10-4, 3-3)/C(10, 3) =4(1)/120 =0.0333

Thus, the probabilities found in (a) and (b) are same, i.e., agreed with each other.

(c)

Expected value of Y = n.r/N =3(4)/10 =12/10 =1.2

Variance of Y =(n.r/N)[(N-r)/N][(N-n)/(N-1)] =1.2[6/10][7/9] =50.4/90 =0.56