Finding Probablity Distribution

[Lancelot59]

From a box containing 4 dimes and 2 nickels,3 coins are selected at random without replacement. Find the propability distribution for the total T of the 3 coins.

Well, this was a simple matter of drawing out at tree diagram and writing everything out.

The problem is that none of the statements the book make sense. They pulled these calculations from somewhere:

3 dimes: 4/6 * 3/5 * 2/4 = 1/5

2 dimes and a nickel: 3 * (4/6 * 3/5 * 2/4) = 3/5

1 dime and two nickels: 3 * (4/6 * 2/5 * 1/4) = 1/5

This just assumes all of the statements for a given outcome are the same. For instance, one of the 2D1N statements is incorrect. If you follow the path DDN you hit 4/6 for D, 2/5 for N, and then 3/4 for the final D. This makes even less sense because there are seven possible valid combinations, so therefore the distribution should be out of 7. Am I missing something here?

 

[Ray Vickson]

From a box containing 4 dimes and 2 nickels,3 coins are selected at random without replacement. Find the propability distribution for the total T of the 3 coins.

Well, this was a simple matter of drawing out at tree diagram and writing everything out.

The problem is that none of the statements the book make sense. They pulled these calculations from somewhere:

3 dimes: 4/6 * 3/5 * 2/4 = 1/5

2 dimes and a nickel: 3 * (4/6 * 3/5 * 2/4) = 3/5

1 dime and two nickels: 3 * (4/6 * 2/5 * 1/4) = 1/5

This just assumes all of the statements for a given outcome are the same. For instance, one of the 2D1N statements is incorrect. If you follow the path DDN you hit 4/6 for D, 2/5 for N, and then 3/4 for the final D. This makes even less sense because there are seven possible valid combinations, so therefore the distribution should be out of 7. Am I missing something here?

The author is just using the hypergeometric distribution, but if you have not seen this before, here is the 25 cent tour.

Denote the events by Ni = {coin i is a nickel} and Di = {coin i is a dime}. The 3 dimes case is easy enough: P(D1) = 4/6 because 4 of the 6 coins are dimes. Then P(D2|D1) = 3/5, because given D1 there are 5 coins left, of which 3 are dimes. Similarly, P(D3|D1 D2) = 2/4. Altogether, P(D1 D2 D3) = P(D1)*P(D2|D1)*P(D3|D1 D2) = (4/6)(3/5)(2/4). Now look at the 2 dimes, 1 nickel case. We have P(D1 D2 N3) = (4/6)(3/5)(2/4). Also, P(D1 N2 D3) = (4/6)(2/5)(3/4) and P(N1 D2 D3) = (2/6)(4/5)(3/4). Notice that D1D2N3, D1N2D3 and N1D2D3 all have the same probability, because in all three cases the numerator = 4*3*2 (but perhaps written in a different order, such as 4*2*3 or 2*4*3) and all three denominators are the same: 6*5*4. Therefore, the 2 dimes, 1 nickel case has probability = 3*(4/6)(3/5)(2/4). Basically, the multiplier '3' is the number of different positions of N in the string of 2 D's and 1 N.

RGV

 

[Lancelot59]

The author is just using the hypergeometric distribution, but if you have not seen this before, here is the 25 cent tour.

Denote the events by Ni = {coin i is a nickel} and Di = {coin i is a dime}. The 3 dimes case is easy enough: P(D1) = 4/6 because 4 of the 6 coins are dimes. Then P(D2|D1) = 3/5, because given D1 there are 5 coins left, of which 3 are dimes. Similarly, P(D3|D1 D2) = 2/4. Altogether, P(D1 D2 D3) = P(D1)*P(D2|D1)*P(D3|D1 D2) = (4/6)(3/5)(2/4). Now look at the 2 dimes, 1 nickel case. We have P(D1 D2 N3) = (4/6)(3/5)(2/4). Also, P(D1 N2 D3) = (4/6)(2/5)(3/4) and P(N1 D2 D3) = (2/6)(4/5)(3/4). Notice that D1D2N3, D1N2D3 and N1D2D3 all have the same probability, because in all three cases the numerator = 4*3*2 (but perhaps written in a different order, such as 4*2*3 or 2*4*3) and all three denominators are the same: 6*5*4. Therefore, the 2 dimes, 1 nickel case has probability = 3*(4/6)(3/5)(2/4). Basically, the multiplier '3' is the number of different positions of N in the string of 2 D's and 1 N.

RGV

Makes sense, turns out I can't count and put an incorrect number on my diagram. Thanks for the explanation!

 

[HallsofIvy]

From a box containing 4 dimes and 2 nickels,3 coins are selected at random without replacement. Find the propability distribution for the total T of the 3 coins.

Well, this was a simple matter of drawing out at tree diagram and writing everything out.

The problem is that none of the statements the book make sense. They pulled these calculations from somewhere:

3 dimes: 4/6 * 3/5 * 2/4 = 1/5

There are initially 6 coins. 4 are dimes, 2 are nickels. The probability that the first coin drawn is a dime is 4/6. After drawing a dime, there are 5 coins, 3 dimes, 2 nickels. The probability that the second coin drawn is a dime is 3/5. After drawing a dime there are 4 coins, 2 dimes and two nickels. The probability the third coin drawn is a dime is 2/4. The probability all three coins are dimes is the product of those, 4/6*3/5*2/4.

2 dimes and a nickel: 3 * (4/6 * 3/5 * 2/4) = 3/5
There are initially 6 coins. 4 are dimes, 2 are nickels. The probability that the first coin drawn is a dime is 4/6. After drawing a dime, there are 5 coins, 3 dimes, 2 nickels. The probability that the second coin drawn is a dime is 3/5. After drawing a dime there are 4 coins, 2 dimes and two nickels. The probability the third coin drawn is a nickel is 2/4. The probability the three coins drawn are "dime, dime, nickel" in that order is 4/6*3/5*2/4. By the same reasoning it is easy to see that the probability of "dime, nickel, dime" is 4/6*2/5*3/5 and of "nickel, dime, dime" is 2/6*4/5*3/4. Those are different fractions but have the same numbers in numerator and denominator so are the same. Because there are three orders: DDN, DND, and NDD, the probability of two dimes and one nickel, in any order, is 3 times that value: 3*(4/6*3/5*2/4).

[quoter]1 dime and two nickels: 3 * (4/6 * 2/5 * 1/4) = 1/5

You mean "3/5". Same reasoning as before- just swap "dime" and "nickel".

This just assumes all of the statements for a given outcome are the same. For instance, one of the 2D1N statements is incorrect. If you follow the path DDN you hit 4/6 for D, 2/5 for N, and then 3/4 for the final D. This makes even less sense because there are seven possible valid combinations, so therefore the distribution should be out of 7. Am I missing something here?

Yes, it does. Each of the calculations involves three fractions. While the fractions are different, the numerators are always 1, 2, and 4, the number of dimes or nickels left, and the denominators are always 6, 5, and 4, the number of coins left, after every draw. Although the fractions are different, their products are the same.