An urn contains 10 Coupons marked 1,2,3,4,5,6,7,8,9,10 are drawn. Then find the chance that the difference of the coupons exceeds 3 ?
I will assume the difference to be between pairs of coupons
possible cases:
5,1
6,1 6,2
7,1 7,2 7,3
8,1 8,2 8,3 8,4
9,1 9,2 9,3 9,4 9,5
10,1 10,2 10,3 10,4 10,5 10,6
total 21
number of pairs = C(10,2) = 45
prob(of your event) = 21/45 = 7/15
To find the chance that the difference between the coupons drawn exceeds 3, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
First, let's determine the total number of possible outcomes. Since there are 10 coupons in the urn, the total number of ways to draw them is given by 10 factorial (10!).
We need to calculate the number of favorable outcomes, where the difference between the coupons drawn exceeds 3. To do this, we can consider two cases:
Case 1: The first coupon drawn is less than 7
If the first coupon drawn is less than 7, then the coupons that have a difference greater than 3 can be selected from 7, 8, 9, and 10. There are 4 coupons in this case.
Case 2: The first coupon drawn is 7 or greater
If the first coupon drawn is 7 or greater, then the coupons that have a difference greater than 3 can be selected from 1, 2, 3, and 4. There are 4 coupons in this case as well.
Therefore, the total number of favorable outcomes is 4 (from Case 1) + 4 (from Case 2) = 8.
Finally, we can calculate the probability by dividing the number of favorable outcomes (8) by the total number of possible outcomes (10!):
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
= 8 / (10!)
Thus, the chance that the difference between the coupons drawn exceeds 3 is 8 divided by 10 factorial.