at a high school, the probability that a student is a senior is 0.25. the probability that a student plays a sport is 0.20. the probability that a student is a senior and plays a sport is 0.08. what is the probability that a randomly selected student plays a sport, given that the student is a senior?

Answers

Answer 1

Answer:

Answer:

0.32

Step-by-step explanation:

We have been given that at a high school, the probability that a student is a senior is 0.25. The probability that a student plays a sport is 0.20. The probability that a student is a senior and plays a sport is 0.08.

We will use conditional probability formula to solve our given problem. $P(B|A)=\frac{P(\text{A and B)}}{P(A)}$ , where,

$P(B|A)$ = The probability of event B given event A.

$P(\text{A and B)}$ = The probability of event A and event B.

$P(A)$ =Probability of event A.

Let A be that the student is senior and B be the student plays a sport.

P(A and B) = Probability that student is a senior and plays a sport.

$P(B|A)=\frac{\text{Probability that a student is a senior and plays a sport}}{\text{Probability that a student is senior}}$

Upon substituting our given values we will get,

$P(B|A)=(0.08)/(0.25)$

$P(B|A)=0.32$

Therefore, the probability that a randomly selected student plays a sport, given that the student is a senior will be 0.32.

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Step-by-step explanation:

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the answer is 8 because if you divide 52 by 4 it gives 13, do the same to the top and you get 8

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A charity is selling raffle tickets at $10 apiece. Ten third-place winners will be drawn, each winning $20. Five second-place winners will be drawn, each winning $50. Finally, one grand prize of $200 will be awarded. If 100 tickets are sold, what is the expected value of one ticket?

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In a different plan for area codes, the first digit could be any number from 4 through 8, the second digit was either 3, 4, 5, or 6, and the third digit could be any number exceptnbsp 2 or 5. with this plan, how many different area codes are possible?

Answers

The possible area codes for different plan are $160$ .

Step-by-step explanation:

Given: In a different plan for area codes, the first digit could be any number from $4$ through $8$ , the second digit was either $3,\;4,\;5,\;\rm{or}\;6$ and the third digit could be any number except $2\;\rm{or}\;5$ .

As mentioned in question:

Area codes are created by digits as:

first digit could be any number from $4$ through $8$ are $4,\;5,\;6,\;7,\;8$ .

Second digit are $3,\;4,\;5,\;\rm{or}\;6$

And third digit couldbe any number except $2\;\rm{or}\;5$ are $1,3,4,6,7,8,9,0$ .

Total number of area codes are $5*4*8=160$ .

Hence, the possible area codes are $160.$

Learn about possible codes here:

brainly.com/question/23225459?referrer=searchResults

Final answer:

In a different plan for area codes, there are 140 possible codes.

Explanation:

In a different plan for area codes, the first digit could be any number from 4 through 8, the second digit was either 3, 4, 5, or 6, and the third digit could be any number except 2 or 5. To calculate the total number of possible area codes, we need to multiply the number of choices for each digit.

For the first digit, there are 5 choices (4, 5, 6, 7, 8).

For the second digit, there are 4 choices (3, 4, 5, 6).

For the third digit, there are 7 choices (0, 1, 3, 4, 6, 7, 8).

To find the total number of possible area codes, we multiply these choices together: 5 x 4 x 7 = 140.

Learn more about Area codes here:

brainly.com/question/33856218

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