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Mathematics 36 Online
OpenStudy (hesan):

A card is drawn from a deck of well shuffled cards. Determine the probability that the card is either queen or heart.

OpenStudy (anonymous):

How many queens are there in 52 cards??

OpenStudy (hesan):

4

OpenStudy (anonymous):

Yes.. So probability of selecting a queen will be 4 by 52.. Getting??

OpenStudy (hesan):

yes

OpenStudy (anonymous):

And now tell me, how many hearts are there in 52??

OpenStudy (hesan):

4

OpenStudy (anonymous):

Really??

OpenStudy (anonymous):

Total heart cards are :13

OpenStudy (hesan):

oh yes yes sorry

OpenStudy (anonymous):

can you tell me now the probability of choosing a heart??

OpenStudy (hesan):

this i dont understnd, 13 or 12?

OpenStudy (anonymous):

What 13 or 12?? There are 13 hearts are there.. Total cards are 52 now find the probability..

OpenStudy (hesan):

13/52

OpenStudy (anonymous):

Yes you are right.. Now the required probability is sum of these two probabilities.. Just add 4/52 + 13/52 can you add them??

OpenStudy (hesan):

17/52

OpenStudy (anonymous):

Yeah you got it right..

OpenStudy (kinggeorge):

Hold up. There's one thing you forgot. Both of those sums are counting the queen of hearts. That means you also need to subtract 1/52.

OpenStudy (hesan):

no but its the wrong ans

OpenStudy (hesan):

@KingGeorge: i dont get it

OpenStudy (anonymous):

There is queen in the heart too... So @KingGeorge is saying right..

OpenStudy (kinggeorge):

Your solution should be 4/52+13/52-1/52. You need to subtract the queen since you've counted it twice.

OpenStudy (anonymous):

what is the probability of getting a queen of hearts 1/52 you have to subtract it from the result..

OpenStudy (hesan):

dont understand why we subtract

OpenStudy (zarkon):

\[P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\]

OpenStudy (zarkon):

general addition rule

OpenStudy (hesan):

P(A)=4/52

OpenStudy (hesan):

A and B are queen and heart. Right?

OpenStudy (zarkon):

yes

OpenStudy (zarkon):

Queen and Heart=queen of hearts

OpenStudy (anonymous):

What is the problem you are facing??

OpenStudy (anonymous):

See, there are 4 queens in 52 cads in which one queen is of heart.. Firstly we find the probability of queen.. Then we find the probability of heats but in there is a queen in hearts too, so probability of getting a heart will be 13/52 But in both the probabilities there is one intersection that is the queen of hearts.. So, total probability will be : \[= \frac{4}{52} + \frac{13}{52} - \frac{1}{52}\]

OpenStudy (hesan):

@waterineyes: Oh thank you, I get it now!

OpenStudy (anonymous):

Welcome dear..

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