Question

A probability experiment is conducted in which the sample space of the experiment is S 1,2,3,4, 5,6,7,8, 9, 10, 11, 12]. Let event E 12, 3, 4, 5, 6,7), event F 15,6, 7, 8,9j, event G [9,10, 11, 12], and event H2, 3, 4). Assume that each outcome is equally likely.
5. List the outcomes in E and F. Are E and F mutually exclusive?
6. List the outcomes in Fand G.Are Fand G mutually exclusive?
7. List the outcomes in For G. Now find P(F or G) by counting the number of outcomes in For G. Determine P(F or G) using the General Addition Rule. NV

Answer 1

Answer:

$E\ n\ F = \{1, 5, 6,7\}$

$F\ n\ G =\{9\}$

$F\ U\ G =\{1,5,6, 7, 8,9,10, 11, 12\}$

$P(F\ U\ G) = \frac{3}{4}$

Step-by-step explanation:

Given

$S = \{1,2,3,4, 5,6,7,8, 9, 10, 11, 12\}$

$E = \{1,2, 3, 4, 5, 6,7\}$

$F =\{1,5,6, 7, 8,9\}$

$G = \{9,10, 11, 12\}$

$H = \{2, 3, 4\}$

Solving (5): E and F

Sets of E and F = E n F

$E\ n\ F = \{1,2, 3, 4, 5, 6,7\}\ n\ \{1,5,6, 7, 8,9\}$

List out common elements

$E\ n\ F = \{1, 5, 6,7\}$

They are not mutually exclusive because $n(E\ n\ F) \neq 0$

Solving (6): F and G

Sets of F and G = F n G

$F\ n\ G =\{1,5,6, 7, 8,9\}\ n\ \{9,10, 11, 12\}$

List out common elements

$F\ n\ G =\{9\}$

They are not mutually exclusive because $n(F\ n\ G) \neq 0$

Solving (7): F or G and P(F or G)

Sets of F or G = F U G

$F\ U\ G =\{1,5,6, 7, 8,9\}\ U\ \{9,10, 11, 12\}$

List all elements without repetition

$F\ U\ G =\{1,5,6, 7, 8,9,10, 11, 12\}$

Solving P(F U G)

$P(F\ U\ G) = \frac{n(F\ U\ G)}{n(S)}$

$n(F\ U\ G) = 9$

$n(S) = 12$

Hence;

$P(F\ U\ G) = \frac{9}{12}$

Divide by 3

$P(F\ U\ G) = \frac{3}{4}$