Question

You play in a soccer tournament, that consists of 5 games. Each game you win with probability .6, lose with probability .3, and tie with probability .1, independently. Let W be the number of games you win, L be the number of games you lose, and T be the number of games you tie. a. What is the joint pmf of W, L,T? b. What is the marginal pmf of W?

Answer 1

(a) The joint PMF of W, L and T is:

`P(W,\ L,\ T)={5\choose (n_(W)!* n_(L)!* n_(T)!)}* [0.60]^{n_(W)}* [0.30]^{n_(L)}* [0.10]^{n_(T)}`

(b) The marginal PMF of W is:

`P(W=w)={5\choose n_(W)!}* 0.60^{n_(W)}* (1-0.60)^{n-n_(W)}`

Explanation:

Let X = number of soccer games played.

The outcome of the random variable X are:

W = if a game won

L = if a game is lost

T = if there is a tie

The probability of winning a game is, P (W) = 0.60.

The probability of losing a game is, P (L) = 0.30.

The probability of a tie is, P (T) = 0.10.

The sum of the probabilities of the outcomes of X are:

P (W) + P (L) + P (T) = 0.60 + 0.30 + 0.10 = 1.00

Thus, the distribution of W, L and T is a appropriate probability distribution.

(a)

Now, the outcomes W, L and T are one experiment.

The distribution of n independent and repeated trials, each having a discrete number of outcomes, each outcome occurring with a distinct constant probability is known as a Multinomial distribution.

The outcomes of X follows a Multinomial distribution.

The joint probability mass function of W, L and T is:

`P(W,\ L,\ T)={n\choose (n_(W)!* n_(L)!* n_(T)!)}* [P(W)]^{n_(W)}* [P(L)]^{n_(L)}* [P(T)]^{n_(T)}`

The soccer tournament consists of n = 5 games.

Then the joint PMF of W, L and T is:

`P(W,\ L,\ T)={5\choose (n_(W)!* n_(L)!* n_(T)!)}* [0.60]^{n_(W)}* [0.30]^{n_(L)}* [0.10]^{n_(T)}`

(b)

The random variable W is defined as the number games won in the soccer tournament.

The probability of winning a game is, P (W) = p = 0.60.

Total number of games in the tournament is, n = 5.

A game is won independently of the others.

The random variable W follows a Binomial distribution.

The probability mass function of W is:

`P(W=w)={5\choose n_(W)!}* 0.60^{n_(W)}* (1-0.60)^{n-n_(W)}`

Thus, the marginal PMF of W is:

`P(W=w)={5\choose n_(W)!}* 0.60^{n_(W)}* (1-0.60)^{n-n_(W)}`