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:

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)

$P(A)=\frac{n(A)}{n(U)}=\frac{2}{5}$

$P(B)=\frac{n(B)}{n(U)}=\frac{2}{5}$

$P(C)=\frac{n(C)}{n(U)}=\frac{3}{5}$

(b)

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

$P(A\cap B)=0$

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

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

$P(A\cup B)=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}$

(c)

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

Number of elements in $A^c$ = 3

$n(A^c)=3$

$P(A^c)=\frac{n(A^c)}{n(U)}=\frac{3}{5}$

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

Number of elements in $C^c$ = 2

$n(C^c)=2$

$P(C^c)=\frac{2}{5}$

(d)

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

$P(A\cup B^c)=\frac{n(C^c)}{n(U)}=\frac{3}{5}$

(e)

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

$n(B\cup C)=4$

So,

$P(B\cup C)=\frac{n(B\cup C)}{n(U)}=\frac{4}{5}$

Answer 2

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=$\frac{Number\;of \;favorable\;events}{total\;number\;of outcomes}$

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)=$\frac{2}{5}$

$P(B)=\frac{2}{5}$

$P(C)=\frac{3}{5}$

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

$A\cap B=\phi$

$P(A\cup B)=\frac{4}{5}$

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

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

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

$P(A^c)=\frac{3}{5}$

$P(C^c}=\frac{2}{5}$

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

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

$P(A\cup B^c)=\frac{3}{5}$

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

$P(B\cup C)=\frac{4}{5}$