Question

If P(A) = 1/6 , P(B) = 5/12, and P(A\B) + P(B\A) = 7/10 Find P(AUB)

Answer 1

Probability is the likelihood or chance that an event will occur. The probability of P(AUB) is 1/2

Conditional probability

Probability is the likelihood or chance that an event will occur. Given the following parameters

If P(A) = 1/6

P(B) = 5/12

P(A\B) + P(B\A) = 7/10

Required

p(AUB)

Recall that:

P(A|B)=P(AnB)/P(B)

P(B|A) = P(BnA)/P(A)

P(AnB)/P(B) + P(BnA)/P(A) = 7/10
12/5P(AnB) + 6P(BnA) = 7/10

42/5P(BnA) = 7/10
6/5P(BnA) = 1/10
6P(BnA) = 1/2
P(BnA) = 1/12

Determine P(AUB)

P(AUB) = P(A) + P(B) - P(AnB)
P(AUB) = 1/6 + 5/12 - 1/12
P(AUB) = 1/6 + 4/12
P(AUB) = 2+4/12
P(AUB) = 1/2

Hence the probability of P(AUB) is 1/2

Learn more on probability here: brainly.com/question/24756209