Question

Suppose that we have a sample space S = {E 1, E 2, E 3, E 4, E 5, E 6, E 7}, where E 1, E 2, ..., E 7 denote the sample points. The following probability assignments apply: P(E 1) = 0.1, P(E 2) = 0.15, P(E 3) = 0.15, P(E 4) = 0.2, P(E 5) = 0.1, P(E 6) = 0.05, and P(E 7) = 0.25.
A= {E1, E4, E6}
B= {E2, E4, E7}
C= {E2, E3, E5, E7}
(a) Find P(A), P(B), and P(C). (b) What is P(A ∩ B)? (c) What is P(A ∪ B)? (d) Are events A and C mutually exclusive?

Answer 1

Answer:

(a) The probability of P(A), P(B), and P(C) are 0.35, 0.6 and 0.65 respectively.

(b) The probability of P(A ∩ B) is 0.2.

(c) The probability of P(A ∪ B) is 0.75.

(d) Events A and C mutually exclusive because the intersection of set A and C is null set or ∅.

Step-by-step explanation:

The given sample space is

$S=\{E_1,E_2,E_3,E_4,E_5,E_6,E_7\}$

$P(E_1)=0.1, P(E_2)=0.15,P(E_3)=0.15,P(E_4)=0.2,P(E_5)=0.1,P(E_6)=0.05, P(E_7)=0.25$

It is given that

$A=\{E_1,E_4,E_6\}$

$B=\{E_2,E_4,E_7\}$

$C=\{E_2,E_3,E_5,E_7\}$

(a)

$P(A)=P(E_1)+P(E_4)+P(E_6)=0.1+0.2+0.05=0.35$

$P(B)=P(E_2)+P(E_4)+P(E_7)=0.15+0.2+0.25=0.6$

$P(C)=P(E_2)+P(E_3)+P(E_5)+P(E_7)=0.15+0.15+0.1+0.25=0.65$

Therefore the probability of P(A), P(B), and P(C) are 0.35, 0.6 and 0.65 respectively.

(b)

A ∩ B represent the common elements of set A and set B.

$A\cap B=\{E_4\}$

$P(A\cap B)=P(E_4)=0.2$

The probability of P(A ∩ B) is 0.2.

(c)

A ∪ B represent all the elements of set A and set B.

$A\cup B=\{E_1,E_2,E_4,E_6,E_7\}$

$P(A\cup B)=P(E_1)+P(E_2)+P(E_4)+P(E_6)+P(E_7)$

$P(A\cup B)=0.1+0.15+0.2+0.05+0.25=0.75$

The probability of P(A ∪ B) is 0.75.

(d)

Set A and C has no common element. So, the intersection of set A and C is empty set.

Yes, events A and C mutually exclusive because the intersection of set A and C is null set or ∅.