SOLUTION: Hi there, I am trying to work out the following: 52 card deck without replacement. 4 players 1 card each. First player has a heart. Each of the other 3 players get a card. What

Algebra ->  Probability-and-statistics -> SOLUTION: Hi there, I am trying to work out the following: 52 card deck without replacement. 4 players 1 card each. First player has a heart. Each of the other 3 players get a card. What      Log On


   

Question 838663: Hi there,
I am trying to work out the following: 52 card deck without replacement. 4 players 1 card each. First player has a heart. Each of the other 3 players get a card. What are the chances of ANY of the other 3 cards being a heart (or how do you work it out).
This was where I got to: Second player: probability is 12/51 (twelve hearts left with 51 cards), Third player 12/50, Fourth 12/49. How do I determine the probability of ANY of the other players having a HEART?
What do I do with these numbers - 0.2352941176470588, .24,
0.2448979591836735 ?
If I wanted to find out what was the probability of them ALL being HEARTS I would multiply these numbers (I think)
Would you recommend any text book / accessible material on this type of probability question?
Thanking you in advance
Kyran
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
52 cards, 13 Hearts, 39 Non-Hearts
First player - Heart
51 cards, 12 Hearts, 39 Non-Hearts
Look at the possible draws of the next three cards (H - heart, N - non-heart) and calculate the probabilities of each, they are all independent events.
NNN-%2839%2F51%29%2838%2F50%29%2837%2F49%29=.438848
NNH-%2839%2F51%29%2838%2F50%29%2812%2F49%29=0.142329
NHN-%2839%2F51%29%2812%2F50%29%2838%2F49%29=0.142329
NHH-%2839%2F51%29%2812%2F50%29%2811%2F49%29=0.0412
HNN-%2812%2F51%29%2839%2F50%29%2838%2F49%29=0.142329
HNH-%2812%2F51%29%2839%2F50%29%2811%2F49%29=0.0412
HHN-%2812%2F51%29%2811%2F50%29%2839%2F49%29=0.0412
HHH-%2812%2F51%29%2811%2F50%29%2810%2F49%290.010564
If you sum up all of the probabilities, they do add up to 1. They must, these are the only possible draws.
Now look for the ones that have at least 1 heart.
Or look for the one that has no hearts and subtract from 1.
P(at least 1 heart)=1-0.438848=0.561152
Yes, all of them being hearts would be HHH, P=0.010564