Chapter 3, Problem 3.1TI

The sample space S Is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).

a. St _____________________

Let event A = the sum Is even and event B = the first number Is prime.

b. A= ______________

c. P(A)= _____________

d. AANDB= .AORB _____ ___

e. P(A AND B) ______, P(A OR B) = _________

f. B’ ___________ ___________

g. P(A) + P(A’) = ___________

h. P(AB) = __________, PBA) = : are the probabilities equal?

To determine

(a)

The sample space.

Answer to Problem 3.1TI

Thus, the sample space is, $S=\left\{\begin{array}{l}\left(1,1\right)\left(1,2\right)(1,3)(1,4)\\ (2,1)(2,2)(2,3)(2,4)\\ (3,1)(3,2)(3,3)(3,4)\end{array}\right\}$.

Explanation of Solution

Given:

S is the ordered pair of two whole numbers.

Calculation:

The sample space S is,

$S=\left\{\begin{array}{l}\left(1,1\right)\left(1,2\right)(1,3)(1,4)\\ (2,1)(2,2)(2,3)(2,4)\\ (3,1)(3,2)(3,3)(3,4)\end{array}\right\}$

Conclusion:

Thus, the sample space is, $S=\left\{\begin{array}{l}\left(1,1\right)\left(1,2\right)(1,3)(1,4)\\ (2,1)(2,2)(2,3)(2,4)\\ (3,1)(3,2)(3,3)(3,4)\end{array}\right\}$.

To determine

(b)

P(A) and P(B).

Answer to Problem 3.1TI

Thus, $A=\left\{(1,1)(1,3)(2,2)(2,4)(3,1)(3,3)\right\}$ and $B=\left\{(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}$.

Explanation of Solution

Given:

Event A is equal to - the sum is even and event B is equal to - the first number is prime.

Calculation:

$A=\left\{(1,1)(1,3)(2,2)(2,4)(3,1)(3,3)\right\}$

$B=\left\{(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}$

Conclusion:

Thus, $A=\left\{(1,1)(1,3)(2,2)(2,4)(3,1)(3,3)\right\}$ and $B=\left\{(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}$.

To determine

(c)

P(A) and P(B).

Answer to Problem 3.1TI

Thus, $P(A)=$

$\frac{1}{2}$ and $P(B)=$

$\frac{2}{3}$.

Explanation of Solution

Calculation:

$\begin{array}{l}P(A)=\frac{n(A)}{n(S)}\\ n(A)=6\text{ and }n(S)=12\\ P(A)=\frac{6}{12}=\frac{1}{2}\\ P(B)=\frac{n(B)}{n(S)}\\ n(B)=8\text{ and }n(S)=12\\ P(B)=\frac{8}{12}=\frac{2}{3}\end{array}$

Conclusion:

Thus, $P(A)=$

$\frac{1}{2}$ and $P(B)=$

$\frac{2}{3}$.

To determine

(d)

A and B and A or B.

Answer to Problem 3.1TI

Thus,

$A\text{ and }B=\left\{(2,2)(2,4)(3,1)(3,3)\right\}$ ; $A\text{ or }B=\left\{(1,1)(1,3)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}$.

Explanation of Solution

Calculation:

A and B is determined by the common terms between A and B

$A\text{ and }B=\left\{(2,2)(2,4)(3,1)(3,3)\right\}$

A or B is determined by all the terms in A and B and in A or B

$A\text{ or }B=\left\{(1,1)(1,3)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}$

Conclusion:

Thus,

$A\text{ and }B=\left\{(2,2)(2,4)(3,1)(3,3)\right\}$ ; $A\text{ or }B=\left\{(1,1)(1,3)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}$.

To determine

(e)

P(A and B) and P(A or B).

Answer to Problem 3.1TI

Thus, $P(A\text{ and }B)=$

$\frac{1}{3}$ ; $P(A\text{ or }B))=$

$\frac{5}{6}$.

Explanation of Solution

Calculation:

$\begin{array}{l}P(AandB)=\frac{n(AandB)}{n(S)}\\ n(AandB)=4\text{ and }n(S)=12\\ P(AandB)=\frac{4}{12}=\frac{1}{3}\\ P(AorB)=\frac{n(AorB)}{n(S)}\\ n(AorB)=10\text{ and }n(S)=12\\ P(AorB))=\frac{10}{12}=\frac{5}{6}\end{array}$

Conclusion:

Thus, $P(A\text{ and }B)=$

$\frac{1}{3}$ ; $P(A\text{ or }B))=$

$\frac{5}{6}$.

To determine

(f)

B^’ and P (B^’).

Answer to Problem 3.1TI

Thus, $B^{’}=\left\{(1,1)(1,2)(1,3)(1,4)\right\}$ ; $P(B^{\text{'}})=$

$\frac{1}{3}$.

Explanation of Solution

Calculation:

B^’ is determined by the terms which are not in set B compared with S.

$\begin{array}{l}S=\left\{\begin{array}{l}\left(1,1\right)\left(1,2\right)(1,3)(1,4)\\ (2,1)(2,2)(2,3)(2,4)\\ (3,1)(3,2)(3,3)(3,4)\end{array}\right\}\\ B=\left\{(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(3,4)\right\}\\ B^{’}=\left\{(1,1)(1,2)(1,3)(1,4)\right\}\end{array}$

$\begin{array}{l}P(B^{\text{'}})=\frac{n(B^{\text{'}})}{n(S)}\\ n(B^{\text{'}})=4\text{ and }n(S)=12\\ P(B^{\text{'}})=\frac{4}{12}=\frac{1}{3}\end{array}$

Conclusion:

Thus, $B^{’}=\left\{(1,1)(1,2)(1,3)(1,4)\right\}$ ; $P(B^{\text{'}})=$

$\frac{1}{3}$.

To determine

(g)

B^’ and P (B^’).

Answer to Problem 3.1TI

Thus, $P(A)+P(A^{\text{'}})=1$.

Explanation of Solution

Calculation:

$\begin{array}{l}P(A)=1-P(A^{\text{'}})\\ \therefore P(A)+P(A^{\text{'}})=1\end{array}$

Conclusion:

Thus, $P(A)+P(A^{\text{'}})=1$.

To determine

(h)

Whether P(A|B) and P(B|A) are equal or not.

Answer to Problem 3.1TI

Thus, P (A|B) and P (B|A) are not equal.

Explanation of Solution

Calculation:

$\begin{array}{l}P\left(A|B\right)=\frac{P\left(AandB\right)}{P\left(B\right)}\\ P\left(AandB\right)=\frac{1}{3}P\left(B\right)=\frac{2}{3}\\ P\left(A|B\right)=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{3}\times \frac{3}{2}\\ P\left(A|B\right)=\frac{1}{2}\end{array}$

$\begin{array}{l}P\left(B|A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}\\ P\left(AandB\right)=\frac{1}{3}P\left(A\right)=\frac{1}{2}\\ P\left(B|A\right)=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}\times \frac{2}{1}\\ P\left(B|A\right)=\frac{2}{3}\end{array}$

Conclusion:

Thus, P(A|B) and P(B|A) are not equal.