A card is selected from a deck of 52 playing cards. Find the probability of selecting

· an even number ten or less given the card is a not a diamond.
· an Ace, given that the card is black.
· a spade given the card is red.

Show step by step work. Give all solutions exactly in reduced fraction form.

Question

A card is selected from a deck of 52 playing cards. Find the probability of selecting
 
·         an even number ten or less given the card is a not a diamond. 
·         an Ace, given that the card is black. 
·         a spade given the card is red.
 
Show step by step work. Give all solutions exactly in reduced fraction form.

anonymous · Answer

Well, the first step is to find your total number of desired outcomes.  Are you familiar with a deck of cards?

anonymous · Answer

yes

anonymous · Answer

Well simple probability is desired outcomes over total number of outcomes.  How many even numbers are there, including 10?

anonymous · Answer

20

anonymous · Answer

Ok, but we're taking out one of the faces...so no diamonds.  How many desired outcomes do we have now?

anonymous · Answer

15

anonymous · Answer

Ok, so your probability is going to be a fraction 15/52 because it is the desired outcome (15 cards are even and NOT a diamond) and 52 is the total number of cards.  Can you simplify 15/52?

anonymous · Answer

what about the rest of the question. Can help with the other parts

anonymous · Answer

ok for the second one, how many total aces are there in a deck?

zarkon · Answer

why divide by 52?

anonymous · Answer

4

anonymous · Answer

@Zarkon - because you express probability as [desired outcomes]/[total outcomes].  There are 52 cards total in a a deck.

zarkon · Answer

it is not 52 in this case

anonymous · Answer

@black12 ok so there are 4 aces.  How many are going to be black aces?

anonymous · Answer

2

anonymous · Answer

You got it! so now put in in fraction form:  There are 2 black aces and 52 total cards.  What is the fraction?

anonymous · Answer

2/52=1/26

zarkon · Answer

that is not correct either

zarkon · Answer

these are conditional probabilities...

anonymous · Answer

Zarkon, this is simple probability.

zarkon · Answer

doesn't seem that way...since I have yet to see a correct answer

anonymous · Answer

@black12 What grade are you in, if you don't mind me asking?

anonymous · Answer

Zarkon, I didn't mean that this is simple in a sarcastic way.  I'm just saying that there is only one even going on.  Would you care to explain your way of doing it?

anonymous · Answer

even = event

anonymous · Answer

COLLEGE

zarkon · Answer

again this is conditional probability..

@MissMai 
I am not being rude

zarkon · Answer

I'm giving you a fact

anonymous · Answer

zarkon can u show me ur way please

zarkon · Answer

$$P(A|B)=\frac{P(A\text{ and }B)}{P(B)}$$

zarkon · Answer

A=an even number ten or less
B=card is a not a diamond.

anonymous · Answer

can you input it cause Iam leaning that but its confusing

zarkon · Answer

P(A and B)=15/52
P(B)=39/52

zarkon · Answer

P(A and B)/P(B)=15/39

anonymous · Answer

I honestly think that it is simple probability because you are only selecting one card.

zarkon · Answer

"an even number ten or less $\underline{\text{given}}$ the card is a not a diamond."

it is conditional probability

anonymous · Answer

Explain how, please.

zarkon · Answer

the word 'GIVEN' tells us that this is a conditional probability problem

anonymous · Answer

b12 are you getting this or do you need it broken down?

anonymous · Answer

@Z - I get it now.  It's been a while.