You are planning to draw 2 cards at random without replacement. Your goal is to pull
out a face card on the first draw and a heart in the second draw.

Question

You are planning to draw 2 cards at random without replacement. Your goal is to pull 
out a face card on the first draw and a heart in the second draw.

istim · Answer

(52P4)*(52P13)
I am not sure thought.

istim · Answer

Do you know if that is how "Permutations" work?

anonymous · Answer

im not sure...

anonymous · Answer

i think i am supposed to use the multiplication rule

istim · Answer

I do not remember then.

anonymous · Answer

If the occurence of Event A changes the probability of Event B, then Events A and B are dependent. On the other hand, if the occurence of Event A does not change the probability of Event B, then Events A and B are independent.

istim · Answer

Well, the even does affect the second event. With one card drawn, the total number of cards in the deck decrease from 52 to 51.

anonymous · Answer

yeah that is right

istim · Answer

Yayy! Dumbcow is with us.

istim · Answer

52P4 Means that we took out 4 cards in order from 52, so that can't be right.

dumbcow · Answer

are we trying to find probability of drawing a face card 1st and a heart 2nd  ?

anonymous · Answer

yes

istim · Answer

I think so.

anonymous · Answer

i really don't know how to do this one...

dumbcow · Answer

there are 2 cases then:
case 1 -> the face card is a heart
there are only 3 cards that fit (jack hearts, queen hearts, king hearts)
then that leaves only 12 hearts remaining for next card
$$P = \frac{3}{52}*\frac{12}{51}$$
case 2-> face card is not a heart
$$P = \frac{9}{52}*\frac{13}{51}$$
to find combined probability, add the probabilities from both cases
$$\rightarrow \frac{3*12}{52*51}+\frac{9*13}{52*51} = \frac{9}{15,028}$$

dumbcow · Answer

oh my final answer is off..i did some bad computation there 
sorry

anonymous · Answer

that makes sense

dumbcow · Answer

final answer:
$$\rightarrow \frac{3}{52}$$

anonymous · Answer

great how about the quarter section...

dumbcow · Answer

well there is only 1 special quarter in the roll so probability of selecting it is 1/n where n is number of quarters
$$\rightarrow \frac{1}{n} = \frac{3}{52}$$

anonymous · Answer

thanks you have been such a great help!!!!

dumbcow · Answer

your welcome
it doesn't come out to a whole number though so i may have done something wrong, not sure

anonymous · Answer

thanks any ways...