Hello, I am new in this forum and I am stuck in a small problem:
Suppose you have an 18-card deck. 9 of the cards show the letter "A" and the rest 9 the letter "B". If you shuffle the deck and then draw 6 cards then what are the chances...
1) ...of having 6 "A"s?
2) ...of having 1 "A" and 5 "B"s?
3) ...of having 2 "A"s and 4 "B"s?
4) ...of having 3 "A"s and 3 "B"s?
Please answer me with a solution because I don't know much to do.
Thanks so much
Hello, Kostas!
Welcome aboard!
We have an 18-card deck. 9 cards show the letter "A" and 9 cards have the letter "B".
If you shuffle the deck and then draw 6 cards. then what are the chances of:
1) having 6 A's?
2) having 1 A and 5 B's?
3) having 2 A's and 4 B's?
4) having 3 A's and 3 B's?
Plato is absolutely correct!
I'll baby-step through one example for you . . .
3) 2 A's and 4 B's
First of all, there are: .ways to draw 6 cards.
How many of them have 2 A's and 4 B's?
. . There are: .ways to get 2 A's.
. . . . There are: .ways to get 4 B's.
Hence, there are: .ways to get 2 A's and 4 B's.
Therefore: .
Get the idea?
of the 18 cards you have 9 A's.Originally Posted by Kostas
thus your chance of getting an A on the first on the first draw is 9 out of 18. or 9/18
2nd card there are only 8 A's and 17 cards. 8/17
3rd card there are only 7 A's left with a total of 16 cards: 7/16
4th: 6/15
5th: 5/14
6th: 4/13
Multiply: 9/18 * 8/17 * 7/16 * 6/15 * 5/14 * 4/13
That will give you the probability of getting 6 A's on the first 6 draws.
---
Similiar procedure for the other questions.
Um, I'll try aidan's way for a bit...
3) 2 "A"s and 4 "B"s:
This goes: 9/18 * 8/17 * 9/16 * 8/15 * 7/14 * 6/13 = around 1/62 = around 1.6% which is too low I think... Soroban results in a 4,536/18,564 = around 25% (which I think it's correct). This way works for 6 "A"s but in this I think it's wrong, except if I am mistaken somewhere...
P.S. Soroban where you write 4,536/16,564 I think it's 18,564
aidan's way only works for the case where the number of A’s is six.
This shows the danger of an inexperience person replying.
While all replies are welcome, you should be a aware of the rep of the poster.
Hi. I'd just like to unnecessarily make this thread longer by saying that this is an example of the hypergeometric distribution.Hello, I am new in this forum and I am stuck in a small problem:
Suppose you have an 18-card deck. 9 of the cards show the letter "A" and the rest 9 the letter "B". If you shuffle the deck and then draw 6 cards then what are the chances...
1) ...of having 6 "A"s?
2) ...of having 1 "A" and 5 "B"s?
3) ...of having 2 "A"s and 4 "B"s?
4) ...of having 3 "A"s and 3 "B"s?
Please answer me with a solution because I don't know much to do.
Thanks so much
Aidan's method can be used here but you need to add one generalization.
To begin with, there are 9 As out of 18 cards: probability the first card drawn is an A is 9/18. Now there are only 8 As left out of 17: probability the second card is drawn is an A is 8/17. Now there are 9 Bs out of 16 cards so the probability the third card is drawn is 9/16. Now there are 8 Bs out of 15 cards so the probability the fourth card is drawn is 8/15. Now there are 7 Bs out of 14 cards so the probability the fifth card is drawn is 7/14. Now there are 6 Bs out of 13 cards so the probability the sixth card is 6/13.
The probability of 2As and 4Bs in that order is (9/18)(8/17)(9/16)(8/15)(7/14)(6/13).
Now the important point is that if we looked at 2As and 4Bs in a different order, we would get different fractions but the denominators would be exactly the same as in that first calculation whikle the numerators would be the same numbers in a different order! In other words, the product of the fractions, and so the probability, would be exactly the same. All we need to do to allow for all the different orders is to multiply that by the number of ways 2 As and 4Bs can be order: that is [tex]\frac{6!}{2! 4!}[/itex], the binomial coefficient. The probability of 2 As and 4 Bs in any order is
Notice that the denominators of all the fractions, 17 down to 13, all multiplied together, is the same as 17!/12!. The numerator involves 9(8)= 9!/7! and 9(8)(7)(6)= 9!/5!. Putting those together should give the same answer Soroban got, corrected.
And on that note I declare the problem solved and the thread closed.Aidan's method can be used here but you need to add one generalization.
To begin with, there are 9 As out of 18 cards: probability the first card drawn is an A is 9/18. Now there are only 8 As left out of 17: probability the second card is drawn is an A is 8/17. Now there are 9 Bs out of 16 cards so the probability the third card is drawn is 9/16. Now there are 8 Bs out of 15 cards so the probability the fourth card is drawn is 8/15. Now there are 7 Bs out of 14 cards so the probability the fifth card is drawn is 7/14. Now there are 6 Bs out of 13 cards so the probability the sixth card is 6/13.
The probability of 2As and 4Bs in that order is (9/18)(8/17)(9/16)(8/15)(7/14)(6/13).
Now the important point is that if we looked at 2As and 4Bs in a different order, we would get different fractions but the denominators would be exactly the same as in that first calculation whikle the numerators would be the same numbers in a different order! In other words, the product of the fractions, and so the probability, would be exactly the same. All we need to do to allow for all the different orders is to multiply that by the number of ways 2 As and 4Bs can be order: that is [tex]\frac{6!}{2! 4!}[/itex], the binomial coefficient. The probability of 2 As and 4 Bs in any order is
Notice that the denominators of all the fractions, 17 down to 13, all multiplied together, is the same as 17!/12!. The numerator involves 9(8)= 9!/7! and 9(8)(7)(6)= 9!/5!. Putting those together should give the same answer Soroban got, corrected.
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