Veronica says that the expression

Calculate the (modeled) probability P(E) using the given information, assuming that all outcomes are equally likely.

Luba_88 [7]

The probability P(E) is 13/15.

According to the statement

we have given that the S = {1, 3, 5, 7, 9}, E = {1, 5, 7}

And we have to find the probability P(E).

So, For this purpose

Recall the formula for the probability of an event E in case when all outcomes are equally likely:

P(E)= n(E) /n(S)

in which S is the sample space.

But we have the S = {1, 3, 5, 7, 9},

so, n(S) = Sum of all outcomes / number of outcomes

n(S) = 1+3+5+7+9 /5

n(S) = 25 /5

n(S) = 5 and

For n(E) = Sum of all outcomes / number of outcomes

n(E) = 1+5+7 /3

n(E) = 13 /3

Substitute these values in the above written formula then

P(E)= n(E) /n(S)

P(E)= (13/3)/ 5

P(E)= 13 /15

So, The probability P(E) is 13/15.

Learn more about the PROBABILITY here brainly.com/question/25870256

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5 0

1 year ago

A variable is a placeholder for an unknown value

frosja888 [35]

Answer:

Im going to guess that this is a true or false question so..

True :)

5 0

2 years ago

WILL MARK YOU BRAINLIEST HURRY PLZ IM ON A TIME LIMIT

alexdok [17]

Answer:

9

Step-by-step explanation:

according to the question

27×⅓

9

7 0

2 years ago

7^2+7^7 answer in exponential form

Helen [10]

It could also be 8.23592 x 10 ^ 5

5 0

2 years ago

Assume the random variable x has a binomial distribution with the given probability of obtaining a success. Find the following.

borishaifa [10]

Answer:

P(x > 10) = 0.6981.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question:

$n = 14, p = 0.8$

P(x>10)

$P(x > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 11) = C_{14,11}.(0.8)^{11}.(0.2)^{3} = 0.2501$

$P(X = 12) = C_{14,12}.(0.8)^{12}.(0.2)^{2} = 0.2501$

$P(X = 13) = C_{14,13}.(0.8)^{13}.(0.2)^{1} = 0.1539$

$P(X = 14) = C_{14,14}.(0.8)^{14}.(0.2)^{0} = 0.0440$

$P(x > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.2501 + 0.2501 + 0.1539 + 0.0440 = 0.6981$

So P(x > 10) = 0.6981.

8 0

3 years ago