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Dima020 [189]
3 years ago
12

A basketball scout has developed a screening tool to identify future NBA all-stars. The screening tool is fairly reliable; 95% o

f all future NBA all-stars who are evaluated are identified as future NBA all-stars, and 95% of basketball players who are not future NBA all-stars and are evaluated are identified as not future NCAA all-stars. A basketball guru tells the scout that exactly one player of the 100 at a basketball camp is a future NBA all-star (and the scout believes the guru). The scout offers a recordbreaking contract on the spot to the first basketball player his screening tool identifies as a future NBA all-star. What is the probability this player truly is a future NBA all-star
Mathematics
1 answer:
Over [174]3 years ago
3 0

Answer:

The value is P(F) =0.0095

Step-by-step explanation:

From the question we are told that

    The  probability that a player is  being identified as an NBA all-stars is 95% = 0.95

     The number of player in the camp is  n =  100

Generally the probability that a player becomes the first person to be evaluated is mathematically represented as

         p(1) = \frac{1}{n}

=>      p(1) = \frac{1}{100}

=>     p(1) = 0.01

Generally the probability that the first basketball player his screening tool identifies as a future NBA all-star is truly a future NBA all-star

      P(F) =  P(1) *  0.95

=>    P(F) =0.01 * 0.95

=>    P(F) =0.0095

             

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Fred has a bag of coin with only nickels and quarters. He got $20 by
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Answer:

75 quarters 25 nickels

Step-by-step explanation:

q=quarters n=nickels

q+n=100   n=100-q

0.25q +0.05(100-q)=20

From now I will multiply  by 100 to avoid decimals

25q + 5(100-q)=2000

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2 years ago
Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre
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Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

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PROBLEM 2:

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So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

3 0
3 years ago
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