Question

Suppose that a sample space has five equally likely experimental outcomes: E1, E2, E3, E4, E5. Let A = {E1, E2} B = {E3, E4} C = {E2, E3, E5}

a. Find P(A), P(B), and P(C).
b. Find P(A U B). Are A and B mutually exclusive?
c. Find A^c, C^c, P(A^c), and P(C^c).
d. Find A U B^c and P(A U B^c).
e. Find P(BUC).

Answer 1

Answer with Step-by-step explanation:

We are given that a sample space=S={
`E_1,E_2,E_3,E_4,E_5`}

A={
`E_1,E_2`}

B={
`E_3,E_4`}

C={
`E_2,E_3,E_5`}

a.We have to find P(A),P(B) and P(C)

We know that probability=
`(Number\;of \;favorable\;events)/(total\;number\;of outcomes)`

Total number of outcomes=5

Number of outcomes favorable to event A=2

Number of outcomes favorable to event B=2

Number of outcomes favorable to event C=3

Therefore, P(A)=
`(2)/(5)`

`P(B)=(2)/(5)`

`P(C)=(3)/(5)`

b.
`A\cup B}`={
`E_1,E_2,E_3,E_4`}

`A\cap B=\phi`

`P(A\cup B)=(4)/(5)`

Yes, A and B are mutually exclusive because A and B are disjoint events.

c.
`A^c`={
`E_3,E_4,E_5`}

`C^c`={
`E_1,E_4`}

`P(A^c)=(3)/(5)`

`P(C^c}=(2)/(5)`

d.
`B^c`={
`E_1,E_2,E_5`}

`A\cup B^c`={
`E_1,E_2,E_5`}

`P(A\cup B^c)=(3)/(5)`

e.
`B\cup C`={
`E_2,E_3,E_4,E_5`}

`P(B\cup C)=(4)/(5)`

Answer 2

Given information: U = {E1, E2, E3, E4, E5}, A = {E1, E2} B = {E3, E4} C = {E2, E3, E5}

Total number of outcome = 5

From the given information, we get

`n(U)=5,n(A)=2, n(B)=2, n(C)=3`

Formula for probability:

`Probability=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}`

(a)

`P(A)=(n(A))/(n(U))=(2)/(5)`

`P(B)=(n(B))/(n(U))=(2)/(5)`

`P(C)=(n(C))/(n(U))=(3)/(5)`

(b)

We need to find P(A U B) if A and B are mutually exclusive.

`P(A\cap B)=0`

`P(A\cup B)=P(A)+P(B)-P(A\cap B)`

`P(A\cup B)=P(A)+P(B)`

`P(A\cup B)=(2)/(5)+(2)/(5)=(4)/(5)`

(c)

`A^c=U-A=\{E1, E2, E3, E4, E5\}-\{E1, E2\}=\{E3, E4, E5\}`

Number of elements in
`A^c` = 3

`n(A^c)=3`

`P(A^c)=(n(A^c))/(n(U))=(3)/(5)`

`C^c=U-C=\{E1, E2, E3, E4, E5\}-\{E2, E3, E5\}=\{E1, E4\}`

Number of elements in
`C^c` = 2

`n(C^c)=2`

`P(C^c)=(2)/(5)`

(d)

`A\cup B^c=A+B^c=\{E1, E2\}+\{E1, E2,E5\}=\{E1, E2,E5\}`

`P(A\cup B^c)=(n(C^c))/(n(U))=(3)/(5)`

(e)

`B\cup C=B+C=\{E3, E4\}+\{E2, E3, E5\}=\{E2, E3, E4, E5\}`

`n(B\cup C)=4`

So,

`P(B\cup C)=(n(B\cup C))/(n(U))=(4)/(5)`