Math help please?

[LuDaCriS66]

1) Four cherry pies and two apple pies are to be divided between two families, the Whittys and the Rasmussens. If the pies are distributed randomly and each family gets the same number of pies, what is the probability that the Rasmussens will get three cherry pies?

2) A card is cut from a shuffled deck and removed; then a second card is cut from the deck. What is the probability of cutting two face cards, given that the first card is a face card?

These questions are on probability and the chapter "Conditional Probability". These questions confuse the heck of out me because they involve using statements like "Let A be the event that there are 2 face cards". I never know what to apply A or B to exactly.

I know that the answer is 1/5 for question #1 and 11/51 for question #2 but I can't quite figure out why yet. This is the only section that confuses me in this chapter..

Any ideas?

 

[LuDaCriS66]

anyone?

 

[Legendary]

<< 2) A card is cut from a shuffled deck and removed; then a second card is cut from the deck. What is the probability of cutting two face cards, given that the first card is a face card? >>



I don't think the answer for #2 is 11/51...because that's just the probability of getting the 2nd face card...not both face cards like the #2 you wrote.

If the answer is 11/51 from the back of the book, then your question is "What is the probability of picking a second face card if the first card picked is a face card?"

However, if the question you wrote is exactly out of the book...then I think this is how it's done.
You can look at it as P(Face card on first) AND P(Face card on second)
Now if there are 3 face cards per suit, 12 per deck...12/52 for first face card and since you removed one face card...there's only 11 left and since you removed one card....there's only 51 total left...11/51 for the second face card.

And = multiplication
12/52 * 11/51 = 132 / 2652

I don't know which one you mean, but the explanation for the first one is contained in the 2nd solution.

 

[SUBxWRX]

O great.....statistics and probability.....

 

[SUBxWRX]

Ok for #2, we know a deck constains 52 cards, 12 of which are face cards. The probability of picking a face card in the first try is 12/52. Now there are 51 cards and 11 which are face cards. the probability of picking a face card in the second try is 15/51. You multiply the two probabilities (12/52) * (11/51) = 11/221 = .0497733

But it says you are given that the first card is face card. So you have to divide the product of the two probabilities over (12/52) because you know the first card is a face card. You get 11/51.

#1 ........just draw pictures or make a quick diagram.....not that hard

 

[LuDaCriS66]

<< Ok for #2, we know a deck constains 52 cards, 12 of which are face cards. The probability of picking a face card in the first try is 12/52. Now there are 51 cards and 11 which are face cards. the probability of picking a face card in the second try is 15/51. You multiply the two probabilities (12/52) * (11/51) = 11/221 = .0497733

But it says you are given that the first card is face card. So you have to divide the product of the two probabilities over (12/52) because you know the first card is a face card. You get 11/51.

#1 ........just draw pictures or make a quick diagram.....not that hard >>



Thanks! Now when you say that the probability of picking the first card is 12/52... is that permutations or combinations? Is order important? If it is then it's permutations.

 

[LuDaCriS66]

bump

 

[bizmark]

1) How many ways of drawing 3 from the 6 pies are there? There's 6 Choose 3 = 6!/(3!*3!)=20. How many ways are there of choosing 3 pies? There's 4 pies, you gotta choose 3, so that's 4 Choose 3 = 4!/(1!*3!)=4. All events are equally likely, there are 4 ways of doing what you want out of 20 ways of doing things total, there's a 4/20=1/5 chance of doing it as you want.

Alternatively look at it like this. What are the chances that you'll draw a Chery pie on the first draw? 4/6. The 2nd draw, given that you drew one on the first draw? 3/5. The 3rd draw, given that you drew one on the first and second draws? 2/4. Multiply: 4/6*3/5*2/4=1/5.

2) There's a 12 in 52 chance of drawing a face card on the first draw, since all card are equally likely drawn. However, this is unimportant. The important thing is to find out what is the probability is that you'll draw a 2nd face card on the 2nd draw. You've already taken out one face card, so there's 11 left. There's 51 cards left in the deck since you drew one already. So the probability is 11/51. No need to worry about combinations or permutations or anything. It's like saying I have 12 red balls and 40 blue ones in a jar. I reach in and pick one, what's the chance that I draw a red one? 12/52.

 

[SUBxWRX]

Combination .... order is not important because it doesnt matter which face card

 

[LuDaCriS66]

<< 1) How many ways of drawing 3 from the 6 pies are there? There's 6 Choose 3 = 6!/(3!*3!)=20. How many ways are there of choosing 3 pies? There's 4 pies, you gotta choose 3, so that's 4 Choose 3 = 4!/(1!*3!)=4. All events are equally likely, there are 4 ways of doing what you want out of 20 ways of doing things total, there's a 4/20=1/5 chance of doing it as you want.

Alternatively look at it like this. What are the chances that you'll draw a Chery pie on the first draw? 4/6. The 2nd draw, given that you drew one on the first draw? 3/5. The 3rd draw, given that you drew one on the first and second draws? 2/4. Multiply: 4/6*3/5*2/4=1/5.

2) There's a 12 in 52 chance of drawing a face card on the first draw, since all card are equally likely drawn. However, this is unimportant. The important thing is to find out what is the probability is that you'll draw a 2nd face card on the 2nd draw. You've already taken out one face card, so there's 11 left. There's 51 cards left in the deck since you drew one already. So the probability is 11/51. No need to worry about combinations or permutations or anything. It's like saying I have 12 red balls and 40 blue ones in a jar. I reach in and pick one, what's the chance that I draw a red one? 12/52. >>



DAMN!! That's perfect! Thank you so much! I really appreciate it.