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:

a. P (A U B U C)

b. P (A∩B∩C)

c. P (A∩B)

d. P[ (AUB) ∩C]

e. P (A'∩B'∩ C')

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