6=2(y+2) I. Don’t get it
2 answers:
You might be interested in
Answer:
(a) The joint PMF of W, L and T is:
(b) The marginal PMF of W is:
Step-by-step explanation:
Let <em>X</em> = number of soccer games played.
The outcome of the random variable <em>X</em> are:
<em>W</em> = if a game won
<em>L</em> = if a game is lost
<em>T</em> = if there is a tie
The probability of winning a game is, P (<em>W</em>) = 0.60.
The probability of losing a game is, P (<em>L</em>) = 0.30.
The probability of a tie is, P (<em>T</em>) = 0.10.
The sum of the probabilities of the outcomes of <em>X</em> 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 <em>n</em> 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 <em>X</em> follows a Multinomial distribution.
The joint probability mass function of <em>W</em>, <em>L</em> and <em>T</em> is:
The soccer tournament consists of <em>n</em> = 5 games.
Then the joint PMF of W, L and T is:
(b)
The random variable <em>W</em> is defined as the number games won in the soccer tournament.
The probability of winning a game is, P (W) = <em>p</em> = 0.60.
Total number of games in the tournament is, <em>n</em> = 5.
A game is won independently of the others.
The random variable <em>W</em> follows a Binomial distribution.
The probability mass function of <em>W</em> is:
Thus, the marginal PMF of W is: