Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

A card is drawn from a well-shuffled standard deck of 52 cards. a) what is the probability that it is a spade? b) Probability that it is a king? c) Probability that it is a king of spades? are the events in part a and b independent?

OpenStudy (ayubie):

a) 1/4 or 25% b) 4/52 or about 7.6% c) 1/52 or 1.9% they are independent

OpenStudy (anonymous):

Almost had it typed out, but yes.

OpenStudy (anonymous):

@DanielLions do you understand the answers that @ayubie gave?

OpenStudy (anonymous):

@SpoonyBard Idk anything about cards, so no.

OpenStudy (anonymous):

There are 13 cards for each of the four suits. So, 13 out of the 52 are either spades, clubs, hearts, or diamonds. 13/52 is simplified to 1/4 There are only 4 kings because there is one king for each suit. 4/52 of the cards are kings. There are only 4 kings, and only one of those kings are spades, so there is only 1/52

OpenStudy (anonymous):

Do you understand now?

OpenStudy (anonymous):

I think we should show how we know them to be independent also @OrangeMaster

OpenStudy (anonymous):

@OrangeMaster yes and @SpoonyBard please.

OpenStudy (anonymous):

do you know the definition of independence in probability?

OpenStudy (anonymous):

yes @SpoonyBard

OpenStudy (anonymous):

ok, then we can use bayes' rule to see if it fits the definition. Let \(P(S)\) be the probability of drawing a spade, and \(P(K)\) be the probability of drawing a king\[P(S\cap K)=P(S)P(K|S)\]we already have \(P(S)\), can you figure out the probability of drawing a king given that it is a spade?, i.e. \(P(K|S)\)

OpenStudy (anonymous):

sorry, not bayes' rule, the multiplication rule*

OpenStudy (anonymous):

or, as an equivalent question on an intuitive level, does knowing the suit tell us anything about the odds of drawing a king?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!