Question

Show the derivation of the formulaP((A ∩ BC) ∪ (AC ∩ B)) = P(A) + P(B) − 2P(A ∩ B)Which gives the probability that exactly one of the events A and B will occur? [Compare with the formula P(A ∪ B) = P(A) + P(B) − P(A ∩ B), which gives the probability that at least one of the events A and B will occur.]

Answer 1

Answer:

We are to prove and show the derivation of a particular formula which is clearly shown with detailed explanation below.

Step-by-step explanation:

P[(A∩BC)∪(AC∩B)] = P(A) +P(B) - 2P(A∩B)

We are to prove that what we have on the left part is equal to what is shown on the right part.

The event of the left part indicates that;

either A but not B or either B but not A will occur.   (since A intersect B complement and A complement intersect B respectively.

Now; the event definitely occurs as a union of two disjoint set.

This is so because:

(A∩B')∩(A'∩B)

= ∅∩∅

= ∅

The two set A and B in both cases are said to be disjoint, since they do not have any common elements. As such, the intersection of the two sets is said to be an empty set and the cardinality of the intersection of the two disjoint set is zero.

However, The probabilities of the union of these two disjoint sets can now be written as the sum of their probabilities.

Given that:

P[(A∩B')∪(A'∩B)]

Since;

P(A) = P(A∩B') + P(A∩B)

P(A∩B') = P(A) -  P(A∩B)           --------- equation (1)

Also, Since;

P(B) =  P(A'∩B) + P(A∩B)

P(A'∩B) = P(B) - P(A∩B)           ----------- equation (2)

Equating the union of equation (1) and equation (2) together, we have:

P(A∩B') ∪ P(A'∩B) = P(A) -  P(A∩B) + P(B) - P(A∩B)

P[ (A∩B') ∪ (A'∩B) ] = P(A) + P(B) -  P(A∩B) - P(A∩B)

∴   P[ (A∩B') ∪ (A'∩B) ] = P(A) + P(B) - 2P (A∩B)