Please solve with explanation

Need help ppl answer needed asap

Nady [450]

Answer:

Step-by-step explanation:

4 0

3 years ago

You play in a soccer tournament, that consists of 5 games. Each game you win with probability .6, lose with probability .3, and

nasty-shy [4]

Answer:

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

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

(b) The marginal PMF of W is:

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

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:

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

The soccer tournament consists of n = 5 games.

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

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

(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:

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

Thus, the marginal PMF of W is:

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

3 0

3 years ago

Marvin counts six $5 bills. write each value that he counts. what pattern do you see in the one digits of the values he counts?

spin [16.1K]

Answer: If the number of bills counted is odd, the ones digit is 5; if the number of bills is even, the ones digit is 0.

Explanation

1) Simulate the counting of the bills:

number of bills counted          value            ones digit

1                                             5                         5

2                                             5 + 5 = 10       0

3                                             10 + 5 = 15         5

4                                             15 + 5 = 20         0

5                                              20 + 5 = 25        5

6                                              25 + 5 = 30        0

2) Pattern: you can see that the ones digit alreternate: 5, 0, 5, 0, 5, 0, ...

If the number of bills counted is odd the ones digit is 5, if the number of bills is even the ones digit is 0.

4 0

4 years ago

Two computers working together can finish a search in 30 seconds. One of these computers can finish in 50 seconds. How long woul

vivado [14]

It would take the second computer 75 seconds because:

$\frac{1}{30} - \frac{1}{50} = \frac{1}{75}$

6 0

3 years ago

Meg rowed her boat upstream a distance of 45 mi and then rowed back to the starting point. The total time of the trip was 14 hou

Montano1993 [528]

Recall your d = rt.

so hmm is about the same as the other one... so say the boat has a speed rate of "b", well, going downstream the boat goes say " b + 2 ", because the current is adding to it, and going upstream is " b - 2 " because the current is eroding from it.

So if the whole trip took 14 hours, say it took "t" hours going upstream, then going downstream it took the slack or " 14 - t ".

Also notice, the trip forth and back is the same 45 miles.

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Upstream&45&b-2&t\\
Downstream&45&b+2&14-t
\end{array}
\\\\\\
\begin{cases}
45=t(b-2)\implies \frac{45}{b-2}=\boxed{t}\\\\
45=(b+2)(14-t)\\
----------\\
45=(b+2)\left( 14- \boxed{\frac{45}{b-2}}\right)
\end{cases}
\\\\\\
\cfrac{45}{b+2}=14- {\cfrac{45}{b-2}}\implies \cfrac{45}{b+2}=\cfrac{14(b-2)~-~45}{b-2}$

$\bf \cfrac{45}{b+2}=\cfrac{14b-28-45}{b-2}\implies \cfrac{45}{b+2}=\cfrac{14b-73}{b-2}
\\\\\\
45(b-2)=(b+2)(14b-73)\implies 45b-90=14b^2-45b-146
\\\\\\
0=14b^2-90b-56\implies 0=7b^2-45b-28
\\\\\\
0=(7b+4)(b-7)\implies b=
\begin{cases}
-\frac{4}{7}\\\\
\boxed{7}
\end{cases}$

6 0

3 years ago