probability question regarding two cards?

Question

I have been struggling with a probability question tonight and hope someone can help me.  The actual question reads:

"Two cards are randomly selected without replacement from a shuffled deck.  Find the probability of getting an Ace on the first card and a spade on the second card."

So as I understand it, the probability of getting an ace on the first draw is 4/52, but what is the probability of getting a spade on the second card?  Because I think it could be 13/51, but what if I picked the ace of spades on the first draw, then it would be 12/51?

Please help by explaining why (giving me the answer only does not help me to learn).

Thank you!

Michael O. · Accepted Answer

Your reasoning is spot on, and here's how you deal with it:

A = Ace was drawn on first draw
S = Spade was drawn on second draw

Only two possibilities work: A, then S...... or  notA, then S.

Pr(A and S) = 1/52 * 12/51 = 1/221
Pr(notA and S) = 3/52 * 13/51 = 1/68

We add these two numbers together, since one OR the other can occur.  ("and" = multiply, "or" = add)

1/221 + 1/68 = 1/52.

Anonymous · Answer

I think the probability is 4/52 for the ace, and 12.75/51 for the second.  This is because there is a 13/52 chance you will get a spade (doesn't have to be an ace) on the first card.  Therefore there is a 3/4 chance that you will have all 13 spades remaining.   With only 12 remaining the chance is 12/52, but with all 13 it i 13/52.  Because it is a 3/4 chance that you will not get a spade first, I think the correct number is 3/4 way between 12 and 13.

Good luck.

frieda · Answer

I decide to remedy this via looking the possibility of no longer getting a sort of taking part in cards and then subtracting it from a million. Your hazard of no longer picking a sort of taking part in cards is 8/fifty two. in case you probably did no longer get a sort of taking part in cards your hazard of no longer getting yet another is 7/fifty one, as quickly as back in case you probably did no longer get a sort of taking part in cards your hazard may well be 6/50, as quickly as back in case you probably did no longer get a sort of taking part in cards your hazard of no longer picking yet another may well be 5/40 9, in case you probably did no longer get a sort of taking part in cards your hazard of no longer picking yet another may well be 4/40 8. Now multiply the probabilities into one yet another: 8/fifty two x 7/fifty one x 6/50 x 5/40 9 x 4/40 8 whilst multiplying fractions you may simplify the two vertically or diagonally, in no way horizontally. So simplify vertically as much as conceivable: 2/13 x 7/fifty one x 3/25 x 5/40 9 x a million/12 Now diagonally simplify the 7 and 40 9: 2/13 x a million/fifty one x 3/25 x 5/7 x a million/12 Diagonally simplify the 5 and 25: 2/13 x a million/fifty one x 3/5 x a million/7 x a million/12 Simplify the three and 12: 2/13 x a million/fifty one x a million/5 x a million/7 x a million/4 Simplify the two and four: a million/13 x a million/fifty one x a million/5 x a million/7 x a million/2 Now which you will no longer simplify any greater effective multiply the fractions into one yet another: a million/46410 that's your hazard of no longer getting a sort of taking part in cards, to discover the possibility of having a minimum of a sort of taking part in cards, subtract it from a million: 46410/46410 - a million/46410 = 46409/46410 Your hazard of having a minimum of a sort of taking part in cards is 46409/46410 (approximately ninety 9.9978%) --------------------- word: in case you do no longer use a calculator to multiply the fractions then simplifying vertically and diagonally is a might desire to.

probability question regarding two cards?

3 Answers