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December 21, 2014

December 21, 2014

Posted by **christy** on Wednesday, February 15, 2012 at 8:53pm.

A= {getting a heart}, B= {getting an even number}

A= {getting a club}, B= {getting a red card}

A= {getting an ace}, B= getting a black card}

A= {getting a prime}, B= {getting a diamond}

Can you please show me how to work this

- Probability -
**MathMate**, Friday, February 17, 2012 at 6:58pmIn all the cases, the probabilities of P(A) are not mutually exclusive from P(B). For example, we can get a heart AND an even number, which is P(A∩B).

The event of getting a heart OR an even number is therefore obtained by the addition rule:

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A)=13 hearts out of 52=13/52=1/4

P(B)=6(assuming the queen is 12) /13 (for each suit)=6/13

P(A∩B)=#even hearts/52=6/52=3/26

So

P(A∪B)=P(A)+P(B)-P(A∩B)

=1/4+6/13-3/26

=(14+24-6)/52

=8/13

The other ones are all similar.

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