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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I got a wierd problem today I wasn't sure how to solve. It seems to be a wierd combination of probability and permutations. (or combinations)

Three cards are drawn at random from a deck of 52 cards. (without replacement) What is the probability that two cards will be black and one will be red?

Now obviously there are 26 blacks and 26 reds. The chance of getting a black on the first try is 26/52 or 1/2, on the second draw, since one black card is missing the chances are 25/51, on the third draw there are 50 cards in the deck, but still 26 reds, so 26/50

So 26/52 * 25/51 * 26/50 The answer reduces to 13/102.

But obviously this is not correct. Since it merely said the probability of getting two blacks and one red, regardless of what order they appear in.

Now in permuations, I learned to eliminate repeated combinatons (regardless of order) by dividing by the number of objects in each combination. In This case 3. But I didn't learn how to do it in probability questions. :-/

If I multiply 13/102 by 1/3 (dividing by 3) then we have 13/306. This is obviously not the answer since the demoninator just got larger. The odds should get greater if order does not matter, not less. I checked the anawer to the problem and it says 13/34. Now the only way to get this is to divide the denominator by 3. However I realized I can achieve the same result by dividing the whole expression by 3: (13/102)/3. This will multiply the numerator by 3 and it can be reduced to 13/34.

Is this the correct way to solve the problem?

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

well obviously it works for this problem but is this the solution for all problems of this type?

A logarithm is just a misspelled algorithm.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

B = black, R= red

BBB .1176470588 Find this one first: (26/52) * (25/51) * (24/50)

BBR

RRB

RRR .1176470588

Then All four combinations must be 100% so double .1176470588, subtract it from 1, and cut it in half!

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I would be interested in knowing how to solve this directly, however, without iterating the possibilities and choosing the

easy ones: 3 black or 3 red

**igloo** **myrtilles** **fourmis**

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**Pi Man****Guest**

How many different ways can you select 3 cards from 52? 52 choose 3. Or (52 * 51 * 50) / (3 * 2 * 1). That's equal to 22100.

How many ways can you draw 2 Red and 1 Black (or 2 Black and 1 Red)? (26 choose 2) * (26 choose 1). Or, (26*25) / (2*1) * (26 / 1). Thats equal to 8450. So the probability of drawing 2 Read and 1 black is 8450 / 22100 = 13/34.

To double-check our work, we can calculate the odds of drawing 3 red (or 3 blue). That's 26 choose 3. Which is equal to (26 * 25 * 24) / (3 * 2 *1) = 2600. The probabilty is then 2600 / 22100 = 2/17.

2 Red and 1 black ==> 13/34

1 Red and 2 black ==> 13/34

3 Red ==> 2/17

3 Black ==> 2/17

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,319

Pi Man wrote:

How many different ways can you select 3 cards from 52? 52 choose 3. Or (52 * 51 * 50) / (3 * 2 * 1). That's equal to 22100.

How many ways can you draw 2 Red and 1 Black (or 2 Black and 1 Red)? (26 choose 2) * (26 choose 1). Or, (26*25) / (2*1) * (26 / 1). Thats equal to 8450. So the probability of drawing 2 Read and 1 black is 8450 / 22100 = 13/34.

To double-check our work, we can calculate the odds of drawing 3 red (or 3 blue). That's 26 choose 3. Which is equal to (26 * 25 * 24) / (3 * 2 *1) = 2600. The probabilty is then 2600 / 22100 = 2/17.

2 Red and 1 black ==> 13/34

1 Red and 2 black ==> 13/34

3 Red ==> 2/17

3 Black ==> 2/17

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Great! The ways how 3 is chosen from 52 have the same probability. That's the meaning of Classical Probility Model. The next thing is to select ways to meet the question, no matter how them are generated. Looking at the problem from a different angle is often where innovations come from!

**X'(y-Xβ)=0**

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