Question

If A, B, and C are mutually exclusive events with P(A) 0.2, P(B): 0.3, and P(C) 0.4, determine the following probabilities:

Answer 1

Answer:

a. 0.9

b. 0.024

c. 0.06

d. 0.2

e. 0.336

Step-by-step explanation:

Since A, B and C are mutually exclusive then P (A U B U C)=P(A)+P(B)+P(C) and P(A∩B∩C)=P(A)*P(B)*P(C).

a.

P (A U B U C)=P(A)+P(B)+P(C)

P (A U B U C)=0.2+0.3+0.4=0.9

b.

P(A∩B∩C)=P(A)*P(B)*P(C)

P(A∩B∩C)=0.2*0.3*0.4=0.024

c.

P(A∩B)=P(A)*P(B)

P(A∩B)=0.2*0.3=0.06

d.

P[(AUB)∩C]=P(AUB)*P(C)

P(AUB)=P(A)+P(B)=0.2+0.3=0.5

P[(AUB)∩C]=0.5*0.4=0.20

e.

P(A')=1-P(A)=1-0.2=0.8

P(B')=1-P(B)=1-0.3=0.7

P(C')=1-P(C)=1-0.4=0.6

P(A'∩B'∩ C')=P(A')*P(B')*P(C')

P(A'∩B'∩ C')=0.8*0.7*0.6=0.336