Question

ch 12 a fair die is rolled and the sample space is given s={1,2,3,4,5,6}. let a={1,2} and b={3,4}. which of the following statements is true? a and b are mutually exclusive (no elements in common). p(a or b)=4/6. p(a or b) = p(a)+p(b). p(a or b) = p(a)+p(b)-p(a and b). question 14 ch 12 let x be the number of girls in a family of three children. does the table below show a valid assignment of probabilities to values of x? x 0 1 2 3 probability 0.125 0.375 0.375 0.125 yes, because all possible values of x are included. no, because it is not possible for x to be equal to 0. yes, because all probabilities are between 0 and 1 and they sum to 1. no, because x could take on other values.

Explanation:

Step1: Check mutual - exclusivity of A and B

Set A = {1, 2} and set B = {3, 4}. Since A ∩ B=∅, A and B are mutually exclusive.

Step2: Calculate P(A) and P(B)

The sample space S = {1, 2, 3, 4, 5, 6}, n(S)=6. For A = {1, 2}, n(A)=2, so P(A)=$\frac{n(A)}{n(S)}=\frac{2}{6}$. For B = {3, 4}, n(B)=2, so P(B)=$\frac{n(B)}{n(S)}=\frac{2}{6}$.

Step3: Calculate P(A or B)

By the addition rule for mutually - exclusive events, P(A or B)=P(A)+P(B). Since A and B are mutually exclusive, P(A and B) = 0. P(A or B)=$\frac{2}{6}+\frac{2}{6}=\frac{4}{6}$. The general addition rule is P(A or B)=P(A)+P(B)-P(A and B), and when A and B are mutually exclusive, P(A and B) = 0, so P(A or B)=P(A)+P(B).

Step4: Check the probability assignment for X

The possible values of X (number of girls in a family of 3 children) are 0, 1, 2, 3. The probabilities are P(X = 0)=0.125, P(X = 1)=0.375, P(X = 2)=0.375, P(X = 3)=0.125. The sum of the probabilities is 0.125 + 0.375+0.375 + 0.125=1, and each probability is between 0 and 1.

Answer:

A. A and B are mutually exclusive (no elements in common).
B. P(A or B)=4/6.
C. P(A or B)=P(A)+P(B).
D. P(A or B)=P(A)+P(B)-P(A and B).
E. Yes, because all probabilities are between 0 and 1 and they sum to 1.