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katrin [286]
4 years ago
6

You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first

Mathematics
1 answer:
Tomtit [17]4 years ago
4 0

Answer:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=3, p=0.1)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

Y \sim Binom(n=5, p=0.05)

The probability mass function for the Binomial distribution is given as:

P(Y)=(nCy)(p)^y (1-p)^{n-y}

Where (nCx) means combinatory and it's given by this formula:

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

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alison, beatrice and chole each had some books. alison gave beatrice and chole some books that doubled the number of books they
slamgirl [31]

alison, beatrice and chole each of them have at first 56, 8, 32 books.

alison, beatrice and chole each had some books. alison gave beatrice and chole some books that doubled the number of books they had, lastly chole gave alison and beatrice some books that doubled the number of books they had. each of them had 32 books at the end.

<h3>What is Working backward?</h3>

Working backward is process in which calculation done in reverse direction.

let alison, beatrice and chole has x,y and z books initially,

alison gave beatrice and chole some books that doubled the number of books they had
alison, beatrice and chole has (96-2y-2z), (2y) , (2z)
chole gave alison and beatrice some books that doubled the number of books they had. each of them had 32 books at the end.

alison, beatrice and chole has 2(96-2y-2z), 2(2y) , (96-2(96-2y-2z)-4y)

at the end,
beatrice = 32

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y = 8

chole =32

96-2(96-2y-2z)-4y)=32
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alison = 96-y-z

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Thus the required alison, beatrice and chole each of them have at first 56, 8, 32 books.

Learn more about working backward here:

brainly.com/question/14636831

#SPJ1



8 0
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