bubbliiiiiiii wrote:Excerpts from your post from the URL above:
First we will the number of ways of picking 4 cards with distinct values.
Number of ways of picking 1st card is 12
2nd card with value distinct from 1st can be picked from 10(12 minus 1st card and card with same value as 1st) in 10 ways
Similarly, 3rd card in 8 ways
and 4th card in 6 ways.
So, number of ways is 12.10.8.6
Total number of ways of picking 4 cards one after the other i 12.11.10.9
So, probability of picking 4 cards with distinct values is 12.10.8.6/12.11.10.9 = 16/33
So, probability of getting at least 1 pair = 1-(probability of getting 4 distinct values) = 1-16/33 = 17/33
I did exactly the same way till the total no. of favourable outcomes.
What I did for total no. of outcomes, i.e., to select four cards from set of 12 cards was 12C4 ways..
which is ..
12!/(8!4!)
and you did it as
Total number of ways of picking 4 cards one after the other i 12.11.10.9
Is the difference in our approaches due to the phrase 'one after the other', which is used by you and not by me?
Hi,
Got your point..While calculating the first part(favourable outcomes) we have considered the order right, by which I mean:
We are counting the combination 1,2,3,4 different from 2,3,4,1 4,3,2,1 and so on..
So, when counting the total number of ways also, we need to consider the order hence we write it as 12P4.
Is the formula nCr=n!/[n!(n-r)!] applicable only when r items are picked from n items and replaced?
No, we use nCr when we select r items from n items and the order is
not important.
When r items are picked from n items one after other and replaced, then the number of ways is n^r.
This particular problem can be done using the formula 12C4 as you used, but then the number of ways we calculated for the 4 picked to be different should be divided by 4! as well because we are considering that order doesn't matter while calculating the total number of ways.
Please let me know if you still have a problem with my logic.
