In 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.
A card is drawn from a standard deck of cards. Find P(A U B) in each part.
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
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