Asked by christy
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
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
Answers
Answered by
MathMate
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.
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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