probabilities and describe.

Question

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Answer 1

Question

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

(b)

To determine

To find: probabilities and describe.

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

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Students have asked these similar questions

(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).

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= Consider the hypothesis test Ho: μ₁ = μ₂ against H₁ μ₁ μ2. Suppose that sample sizes are n₁ = 15 and n₂ = 15, that x1 = 4.7 and X2 = 7.8 and that s² = 4 and s² = 6.26. Assume that o and that the data are drawn from normal distributions. Use απ 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain ẞ = 0.05 if the true difference in means is - 2? Assume that α = 0.05. (a) The null hypothesis is 98.7654). rejected. The P-value is 0.0008 (b) The power is 0.94 . Round your answer to four decimal places (e.g. Round your answer to two decimal places (e.g. 98.76). (c) n₁ = n2 = 1 . Round your answer to the nearest integer.

Answer 2

Question

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

(b)

To determine

To find: probabilities and describe.

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

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Answer 3

Question

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

(b)

To determine

To find: probabilities and describe.

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

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Answer 4

Question

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

(b)

To determine

To find: probabilities and describe.

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

(b)

To determine

To find: probabilities and describe.

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Answer 6

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Answer 7

Chapter 10, Problem 10.13AYK

(a)

To determine

To find: probabilities and describe.

Answer 8

(a)

Expert Solution

Answer to Problem 10.13AYK

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Calculation:

Starting at the left of the tree diagram, a child with red hair falls into one of the four hair colors. The bottom leftmost branch gives the probability of a red hair child, so P(red hair) is 0.025.

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has red hair is 0.025.

The three segments going out from the hair branch point carry the conditional probabilities of eye color given hair color. For example, the top middle branch is P(blue eye|black hair) . Now use the general multiplication rule to find P(blue eye) . Recall that the general multiplication rule is:

P(A and B)=P(A)P(B|A)P(blue eye)={P(blue eye and black hair)+P(blue eye and brown hair)+P(blue eye and blonde hair)+P(blue eye and red hair)}={P(blue eye|black hair)P(black hair)+P(blue eye | brown hair)P(brown hair)+P(blue eye | blonde hair)P(blonde hair)+P(blue eye | red hair)P(red hair)}=0.030⋅0.061+0.259⋅0.461+0.562⋅0.453+0.473⋅0.025=0.388

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has blue eyes is 0.388.

The two segments going out from the eye color branch point carry the conditional probabilities of freckles given eye color when given hair color. For example, the top rightmost branch is P(freckles | blue eye | black hair) . Now use the general multiplication rule for three events to find P(freckles) . Recall that the general multiplication rule for three events is:

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Conclusion:

Therefore, if a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.253.

Answer 13

(b)

To determine

To find: probabilities and describe.

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Answer 14

(b)

To determine

To find: probabilities and describe.

Answer 15

(b)

Expert Solution

Answer to Problem 10.13AYK

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Use the general multiplication rule to find P(freckles and red hair) . Recall that the general multiplication rule for three events is:

P(A and B and C)=P(A)P(B|A)P(C|Aand B)⋅P(freckles and red hair)={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}={P(freckles and blue eye and red hair)+P(freckles and green eye and red hair)+P(freckles and brown eye and red hair)}=(0.025)(0.473)(0.857)+(0.025)(0.405)(0.767)+(0.025)(0.122)(0.778)=0.0203

If a child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles and red hair is 0.0203.

From part (a), P(red hair) is 0.025. The condition probability of freckles given red hair is

P(freckles | red hair)=P(freckles and red hair)P(red hair)=0.02030.025=0.812

If a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Conclusion:

Therefore,if a red hair child of Caucasian descent in Germany is randomly selected, the probability that the child has freckles is 0.812.

Answer 20

Ch. 10 - Prob. 10.1AYK Ch. 10 - Prob. 10.2AYK Ch. 10 - Prob. 10.3AYK Ch. 10 - Prob. 10.4AYK Ch. 10 - Prob. 10.5AYK Ch. 10 - Prob. 10.6AYK Ch. 10 - Prob. 10.7AYK Ch. 10 - Prob. 10.8AYK Ch. 10 - Prob. 10.9AYK Ch. 10 - Prob. 10.10AYK

probabilities and describe.

To find: probabilities and describe.

Answer to Problem 10.13AYK

Explanation of Solution

To find: probabilities and describe.

Answer to Problem 10.13AYK

Explanation of Solution

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Chapter 10 Solutions