Prove using only the axioms of probability that if A and B are events, then P(A...

Prove using only the axioms of probability that if A and B are events, then P(A ∪ B) ≤ P(A) + P(B)

Solutions

Expert Solution

From LHS

From RHS we have seen that we have to equate deduct P(A B) so P(A)+P(B) is greater than P(A U B)

If A and B are disjoint then the value of P(A B) is zero so the P(A)+P(B) is equal to P(A U B)

Here we see the

If A and B are independent then

P(A∪B)=P(A)+P(B)−P(A∩B)

this should be obvious as if we simply add the two probabilities together we are counting the overlap where both things happen twice.

where ∩ represents intersection or P(A and B)

the probability that both A and B happen.P(A and B)

the probability that both A and B happen and U represents union or P(A or B) the probability that at one or both of A and B happens.

From this it follows that P(A∩B)=P(A)+P(B)−P(A∪B)

P(A∩B)=P(A)+P(B)−P(A∪B)

It should also be obvious that

P(A∩B)=P(A)−P(A∩)

=P(B)−P(B∩)

Where represents not B i.e. the probability of B not happening.

P(AUB) = n(AUB) n(A) + n(B) - n(ANB) n(S) n(ANB) = P(A) + P(B) - P(ANB) n(S) n(A) n(B) n(s) * n(s) n(S)

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orchestra answered 1 year ago

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