Respuesta :
Answer:
(a) The joint PMF of W, L and T is:
[tex]P(W,\ L,\ T)={5\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [0.60]^{n_{W}}\times [0.30]^{n_{L}}\times [0.10]^{n_{T}}[/tex]
(b) The marginal PMF of W is:
[tex]P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}[/tex]
Step-by-step 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:
[tex]P(W,\ L,\ T)={n\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [P(W)]^{n_{W}}\times [P(L)]^{n_{L}}\times [P(T)]^{n_{T}}[/tex]
The soccer tournament consists of n = 5 games.
Then the joint PMF of W, L and T is:
[tex]P(W,\ L,\ T)={5\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [0.60]^{n_{W}}\times [0.30]^{n_{L}}\times [0.10]^{n_{T}}[/tex]
(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:
[tex]P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}[/tex]
Thus, the marginal PMF of W is:
[tex]P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}[/tex]
