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julsineya [31]
3 years ago
6

The probability of event A is x, and the probability of event B is y. If the two events are independent, which of these conditio

ns must be true?
a. P(B|A) = y
b. P(A|B) = y
c. P(B|A) = x
d. P(A and B) = x + y

e. P(A and B) = x/y

P(A)
Mathematics
2 answers:
zzz [600]3 years ago
5 0

If two events are independent, then

Pr(A\cap B)=Pr(A)\cdot Pr(B).

Use formulas for conditional probabilities:

Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)},\\ \\ Pr(B|A)=\dfrac{Pr(A\cap B)}{Pr(A)}.

For independent events these formulas will be:

Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}=\dfrac{Pr(A)\cdot Pr(B)}{Pr(B)}=Pr(A),\\ \\ Pr(B|A)=\dfrac{Pr(A\cap B)}{Pr(A)}=\dfrac{Pr(A)\cdot Pr(B)}{Pr(A)}=Pr(B).

Now in your case Pr(A)=x,\ Pr(B)=y and Pr(A|B)=x,\ Pr(B|A)=y, Pr(A\cap B)=x\cdot y.

This shows that the only correct choice is A.

Novay_Z [31]3 years ago
5 0
The right answer for the question that is being asked and shown above is that: "d. P(A and B) = x + y." The probability of event A is x, and the probability of event B is y. If the two events are independent, the condition must be true is this d. P(A and B) = x + y

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Mr. Ortiz has 3/4 pound of oatmeal.
ki77a [65]

Answer:

\frac{1}{4}\ pound

Step-by-step explanation:

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5 0
3 years ago
Three numbers are in the ratio 3:9:10. If 10 is added to the last number, then the three numbers form an arithmetic progression.
MrRa [10]

Answer:

<em> The numbers are 6, 18, and 30 </em>

Step-by-step explanation:

If the three numbers are in the ratio of 3:9:10,

let the numbers be 3x, 9x and 10x.

<em>If 10 is added to the last number to form an arithmetic progression</em>

<em>Then, 3x 9x (10x+10) are the progression</em>

The common difference of an arithmetic progression (d) = T₂ - T₁ = T₃ - T₂

T₂-T₁ = T₃ - T₂ .............. Equation 1

Where T₁ = first term of the progression, T₂ = Second term of the progression, T₃ = third term of the progression

<em>Given: T₁ = 3x, T₂ = 9x, T₃ = 10x +10</em>

<em>Substituting these values into equation 1</em>

<em>9x-3x = (10x+10)-9x</em>

<em>Solving the equation above,</em>

<em>3x = 10+x</em>

<em>3x-x = 10</em>

<em>2x = 10</em>

<em>x = 10/2</em>

<em>x = 2.</em>

<em>Therefore the numbers are 6, 18, and 30 </em>

<em />

7 0
3 years ago
Please help (question is in the attachment) :(
Kitty [74]

Answer:

The answer is D

Step-by-step explanation:

Because both when we multiply we will get the same answer

8 0
2 years ago
Read 2 more answers
-2 + (-3) x (-8 + 4)
Ymorist [56]

Answer:

10

Step-by-step explanation:

-2+(-3)*(-8+4)

-2+(-3)*(-4)

-2+12

10

4 0
3 years ago
Read 2 more answers
The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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!}  

Solution to the problem

Part a

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

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
3 years ago
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