Math help! Algebra 2?

Question

Hey I need some help with a math problem here. It has to do with probability.

2 cards are randomly selected from a standard deck of 52 playing cards. What is the probability of getting 2 hearts or 2 numbers LESS than 9 (count aces as 1)?

thanks in advance!

REMEMBER TO COUNT ACES AS ONES!

Answer 1

Look at it this way:

There are 52 cards in a deck of card. Whatever card you pick first has the probability of 1/52 to be chosen. Now lets say you have picked a card with the probability of 1/52 and you want to pick the second one. The second one will have the probability of 1/51 to be a heart. So now add those probabilities: 1/52 + 1/51 = 0.0192 + 0.0196 = 0.0388 = The probability of two hearts showing.

I think you can do the second one yourself. Just think of how many cards will be left after you pick a card and what are the chances that those card will be in your selected range.

Good luck.

Answer 2

**** NOTE: Since you have specified that aces do count as being below 9, I have removed all my references to the possibility that Aces don't count.*****

Probability is expressed as a fraction "favorable/possible."

For each situation, "possible" is the number of ways of drawing 2 cards out of 52 (which one we draw first doesn't matter).

We call that 52-choose-2, and it's equal to (52! / (2! *(52-2)!))

That's equal to 52*51 / 2, or, 1326

So there are 1326 possible ways to draw two cards.

What about favorable? Well, since there are 13 hearts, the number of ways to draw two hearts is 13-choose-2.

It's equal to (13! / (2! *(13-2)!))

or, 78. Therefore, the probability of drawing 2 hearts is
78 / 1326, or 39/663, which is about a 6% chance.

For 2 numbers less than 9, we need to know how many there are in the deck. If Aces count (as "1"), then there are 32 such cards.
So the favorable outcomes are 32-choose-2.

(32! / (2! *(32-2)!)) = 496

So the probability is 496/1326 = 248/663. That's about a 37% chance.

PS 52! is 52 factorial and it means 52 * 51 * 50......* 2 * 1

****
Ah, so do you want to know the chance of getting EITHER of these outcomes? Many have told you to add the two probabilities together. But to do so would be to ignore the fact that many of the desirable outcomes are BOTH two hearts AND two numbers less than 9! How many outcomes overlap? Well, there are 8 hearts below 9. So the ways to draw two of them are 8-choose-2.
That's (8! / (2! *(8-2)!)), or 28.

To find the probability of EITHER outcome, we count favorable outcomes like this:

Number of ways to draw 2 hearts PLUS ways to draw 2 numbers less than 9 MINUS number of ways that overlap.

So: 78 + 496 - 28, or 546 favorable outcomes.

The probability is 546/1326, or 91/221, which is about a 41% chance, NOT 43%.

Answer 3

The probability of getting hearts is 12/52. The probability of getting a number less than 9 is 45/52. I'm positive that 12/52 is the probability of getting hearts but not so sure about getting a number less than 9.

Answer 4

There are 13 hearts in a 52 card deck. (1/4 of deck). If you select one there are 12 left. The odds of picking another is 12/51 since there is one less card in total and one less heart to choose. So the odds of picking 2 hearts is (1/4) times (12/51). This reduces to 3/51. There are 8 cards that are less than 9 in each suit. So 8/13 of each suit fit this condition. The first draw has odds 8/13. The second draw there are only 51 cards and only 31 cards less than 9. So we have (8/13) times (31/51). Since the question asked this OR this, we add probabilities. (3/51) + (8/13)(31/51) =
(287/663) = 43%

Answer 5

1st part:
13 cards per suit, 4 suits. Odds [probability] of getting a heart on the first card are 13/52 or 25.0%. But that reduces the number of available hearts and the number of available cards to 12 hearts / 51 cards (or 23.5%). Multiply the two percentages to determine the probability of drawing two cards and getting two hearts: 5.875%.

2nd part:

1, 2, 3, 4, 5, 6, 7, and 8 are lower, 9, 10, J, Q, K are higher. There are four cards of each denomination. So the probability of getting a lower-than-nine card for the first card is 32/52 (or 61.5%). Once that card is successfully drawn there are only 51 cards remaining, of which 20 are 9-or-higher and 31 are 8-or-lower. So the probability for that draw becomes 31/51 (or 60.8%). To find the answer you simply multiply the two together and get 37.4%.

Answer 6

Because there is only 1 card with 2 hearts and there are 52 cards in a deck, the probability of getting it is 1/52.

There are 4 cards with 2 numbers on them and there are 52 cards in a deck, the probability of getting them is 4/52.

So the probability of getting 2 hearts or 2 number is..

1/52 x 4/52 = 4/2704 or 0.15%

Answer 7

Hello

13 hearts two at a time 13C2
32 cards less than 9 two at a time 32C2
And 8 hearts that are less than 9 which we already counted 8C2

So we have (13C2 + 32C2 - 8C2)/ 52C2

I'll let you do the math

Hope This Helps

Answer 8

Alexis' answer is absurd because it doesn't take into account that there are 13 hearts. There is no 13 in his calculation. Surely, if there were only two suits in a deck of cards, and 26 hearts, the result should be different, but not in Alexis' world!!

detektibgapo must be from another planet because he thinks "getting two hearts" means getting a card that has two hearts on it. The rest of us know that "getting two hearts" means getting one of the 13 heart cards in the deck, and then getting another one. I think in fact there are three cards that have two hearts on them: the J, Q, and K of hearts. If detektibgapo was thinking of the duece of hearts, he is wrong. It has four hearts on it, two in the body of the card and one each in the upper-left and lower-right corners.

mrgets' answer seems reasonable.

Answer 9

(13/52)*(12/51)

(36/52)*(35/51)

Answer 10

(13C2 + 32c2)/52c2 = 0.4328