Question 1

Numbered disks are placed in a box and one disk is selected at random. There are 6 red disks numbered 1through 6, and 7 yellow disks numbered 7 through 13. In an experiment a disk is selected, the number andcolor noted, replaced, and then a second disk is selected. Is this an example of independence?

Accepted Answer

The answer of Numbered disks are placed in a box...

Question 2

A single die is rolled twice. Find the probability of getting a 2 the first time and a 2 the second time. Express theprobability as a simplified fraction.

Accepted Answer

The probability of rolling a 2 on a single die is $\frac{1}{6}$. Since the rolls are independent, the probability of rolling a 2 twice in a row is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.

Question 3

A bag contains 10 white, 12 blue, 13 red, 7 yellow, and 8 green wooded balls. A ball is selected from the bag, itscolor noted, then replaced. You then draw a second ball, note its color and then replace the ball. What is theprobability of selecting one white ball and one blue ball? Round to the nearest ten-thousandth.

Accepted Answer

The probability of selecting one white ball and one blue ball, with replacement, can be calculated by considering the two possible sequences: white then blue, and blue then white. The total number of balls is 10 + 12 + 13 + 7 + 8 = 50.- Probability of drawing a white ball first and then a blue ball: (10/50) * (12/50) = 120/2500- Probability of drawing a blue ball first and then a white ball: (12/50) * (10/50) = 120/2500Adding these probabilities gives: 240/2500 = 0.096. However, since the question asks for the answer rounded to the nearest ten-thousandth, the correct rounding or calculation approach based on the given options seems to have been misinterpreted in the explanation. The correct calculation for the scenario described should indeed result in a probability that matches one of the provided options more closely. Given the options, the closest correct approach to calculating the probability of drawing one white and one blue ball in any order, with replacement, would typically involve the principles outlined but seems to have been inaccurately represented in the final step of calculation or option selection. Re-evaluating the calculation accurately:The probability of selecting one white and one blue in any order, considering the correct calculation method and rounding to the nearest ten-thousandth, should be evaluated correctly based on the principles of probability, taking into account the total number of balls and the specific counts of white and blue balls. The initial approach correctly identifies the need to calculate the probability for both sequences (white then blue, and blue then white) and to sum these probabilities. However, the final probability should be accurately calculated and matched against the provided options, ensuring it reflects the correct application of probability principles for this scenario with replacement.

Question 4

Suppose a basketball player is an excellent free throw shooter and makes 91% of his free throws (i.e., he has a91% chance of making a single free throw). Assume that free throw shots are independent of one another.Suppose this player gets to shoot three free throws. Find the probability that he misses all three consecutive freethrows. Round to the nearest ten-thousandth.

Accepted Answer

The probability of missing a single free throw is 1 - 0.91 = 0.09. Since the shots are independent, the probability of missing three consecutive free throws is $0.09^3 = 0.000729$, which rounds to 0.0007.

Question 5

True or False: Mutually exclusive events are always independent.

Accepted Answer

Mutually exclusive events cannot happen at the same time, and if one event happens, it affects the probability of the other, making them dependent, not independent.

Question 6

A human gene carries a certain disease from the mother to the child with a probability rate of 39%. That is,there is a 39% chance that the child becomes infected with the disease. Suppose a female carrier of the gene hasfour children. Assume that the infections of the four children are independent of one another. Find theprobability that all four of the children get the disease from their mother. Round to the nearest thousandth.

Accepted Answer

The probability that all four children get the disease is $0.39^4 = 0.023$, rounded to the nearest thousandth.

Question 7

A bag contains 10 white, 12 blue, 13 red, 7 yellow, and 8 green wooded balls. A ball is selected from the bag, itscolor noted, then replaced. You then draw a second ball, note its color and then replace the ball. What is theprobability of selecting 2 red balls? Round to the nearest ten-thousandth.

Accepted Answer

The probability of selecting a red ball on the first draw is 13/50. The probability of selecting a red ball on the second draw is also 13/50. The probability of selecting 2 red balls is (13/50) * (13/50) = 0.0676.

Question 8

Two events are __________________ if the occurrence if the occurrence of event E in a probability experimentdoes not affect the probability of event F in the same experiment.

Accepted Answer

Independent events are defined by the fact that the occurrence of one event does not affect the probability of the occurrence of another event in the same experiment.

Question 9

You are dealt one card from a 52 card deck. Then the card is replaced in the deck, the deck is shuffled, and youdraw again. Find the probability of getting a picture card the first time and a club the second time. Express theprobability as a simplified fraction.

Accepted Answer

The answer of You are dealt one card from a...

Question 10

A human gene carries a certain disease from the mother to the child with a probability rate of 51%. That is,there is a 51% chance that the child becomes infected with the disease. Suppose a female carrier of the gene hasfive children. Assume that the infections of the five children are independent of one another. Find theprobability that at least one of the children get the disease from their mother. Round to the nearest thousandth.

Accepted Answer

The probability that at least one child gets the disease is 1 minus the probability that none of the children get the disease. The probability that a child does not get the disease is 49% (100% - 51%). For five children, this is $0.49^5$. Thus, the probability at least one gets the disease is $1 - 0.49^5 \approx 0.972$.

Question 11

After completing an inventory of three warehouses, a golf club shaft manufacturer described its stock of 12, 246shafts with percentages given in the table. Is the event of selecting a shaft independent of the warehouse?

Accepted Answer

The event of selecting a shaft is not independent of the warehouse because the probability of selecting a specific type of shaft varies depending on which warehouse you are choosing from. Independence would require that the choice of shaft is not affected by the warehouse, which is not the case here as the distribution of shafts is different across warehouses.

Question 12

A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for themachine to work properly. Assume the probability of one part working does not depend on the functionality ofany of the other parts. Also assume that the probabilities of the individual parts working are P(A) = P(B) = 0.93,P(C) = 0.98, and P(D) = 0.96. Find the probability that at least one of the four parts will work. Round to sixdecimal places.

Accepted Answer

The probability that at least one of the four parts will work is calculated by subtracting the probability that none of the parts work from 1. The probability that a part fails is 1 minus the probability that it works. Therefore, the probability that all parts fail is (1 - 0.93) * (1 - 0.93) * (1 - 0.98) * (1 - 0.96) = 0.07 * 0.07 * 0.02 * 0.04. Subtracting this product from 1 gives the probability that at least one part works, which is very close to 1, indicating a very high probability that at least one component will function.

Question 13

There are 30 chocolates in a box, all identically shaped. There are 11 filled with nuts, 10 filled with caramel,and 9 are solid chocolate. You randomly select one piece, eat it, and then select a second piece. Is this anexample of independence?

Accepted Answer

The selection of the second chocolate is dependent on the outcome of the first selection because the first selection changes the total number of chocolates and the composition of chocolates remaining in the box.

Question 14

Two events are ________________ if the occurrence of event E in a probability experiment changes theprobability of event F in the same experiment.

Accepted Answer

Two events are considered dependent if the occurrence of one event affects the probability of occurrence of the other event. This means that knowing something about whether one event has occurred changes the likelihood of the other event occurring.

Question 15

A game has three outcomes. The probability of a win is 0.4, the probability of tie is 0.5, and the probability of aloss is 0.1. What is the probability of not winning in a single play of the game.

Accepted Answer

The probability of not winning is the sum of the probabilities of a tie and a loss, which is 0.5 + 0.1 = 0.6.

Question 16

A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for themachine to work properly. Assume the probability of one part working does not depend on the functionality ofany of the other parts. Also assume that the probabilities of the individual parts working are P(A) = P(B) = 0.99,P(C) = 0.91, and P(D) = 0.93. Find the probability that the machine works properly. Round to the nearestten-thousandth.

Accepted Answer

The probability that the machine works properly is the product of the probabilities of each component working: 0.99 * 0.99 * 0.91 * 0.93 = 0.8295.

Question 17

Suppose that events E and F are independent, P(E) = 0.8 and P(F ) = 0.9. What is the P(E and F )?

Accepted Answer

For independent events E and F, P(E and F) = P(E) * P(F) = 0.8 * 0.9 = 0.72.

Question 18

If you toss a fair coin 12 times, what is the probability of getting all heads? Express the probability as asimplified fraction.

Accepted Answer

The probability of getting heads on a single toss of a fair coin is $\frac{1}{2}$. For 12 independent tosses, the probability of getting all heads is $(\frac{1}{2})^{12} = \frac{1}{4096}$.

Question 19

The complement of 4 heads in the toss of 4 coins is

Accepted Answer

The complement of getting 4 heads in 4 coin tosses is not getting 4 heads in any of the tosses, which means getting at least one tail in any of the tosses.

Question 20

What is the probability that in three consecutive rolls of two fair dice, a person gets a total of 7, followed by atotal of 11, followed by a total of 7? Round to the nearest ten-thousandth.

Accepted Answer

The probability of getting a total of 7 on two fair dice is 6/36 = 1/6. The probability of getting a total of 11 on two fair dice is 2/36 = 1/18. So, the probability of getting a total of 7, followed by a total of 11, followed by a total of 7 is (1/6) * (1/18) * (1/6) = 1/324 = 0.0031. Rounded to the nearest ten-thousandth, this is 0.0031.

Quiz 5: Probability