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jeka94
3 years ago
7

There were n balloons at the beginning of the party. By the end of the party, 5 of them had popped. Using n, write an expression

for the number of balloons that were left
Mathematics
2 answers:
frozen [14]3 years ago
8 0

Answer:

The expression for the number of balloons that were left is n-5.

Step-by-step explanation:

There were n balloons at the beginning of the party.

By the end of the party, 5 of them had popped.

Let the remaining balloons be represented by x

So, x = total balloons minus 5

This gives: x=n-5

Therefore, the expression for the number of balloons that were left is n-5.

NeX [460]3 years ago
6 0
I think the answer is n -5 

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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 <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:

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 <em>n</em> = 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 <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:

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
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