00:01
Hello students, let x1, x2, x3, x4, x5 and xs denote the number of bagels of six different kinds in the randomly chosen 15 bagels.
00:10
The total number of ways the different bagels could be selected is given by the number of non -negative integer solutions to the equation x1 plus x2 plus x3 plus x4 plus x5 plus xs is equal to 15 where xi greater than or equal to 0 and xi element of z.
00:26
This happen in 15 plus 6 minus 1 c 6 minus 1 ways which is 20 c 5 ways.
00:39
Now if at least one bagel of each kind should be selected, the number of ways of doing so is given by the positive integer solutions to the equation that is x1 plus x2 plus x3 plus x4 plus x5 plus xs is equal to 15 where xi greater than 0 xi element of z.
01:05
This happens in 15 minus 1 c 6 minus 1 ways which is 14 c 5 ways.
01:21
Hence the probability that the probably randomly selected 15 bagels will contain at least one bagel of each kind is probability that at least one bagel of each kind is 14 c 5 divided by 20 c 5 is equal to 14 into 13 into 12 into 11 into 10 divided by 1 into 2 into 3 into 4 into 5 divided by 20 into 19 into 18 into 17 into 16 divided by 1 into 2 into 3 into 4 into 5 is equal to 1 0 0 1 by 7 7 7 5 2 which is 0 .129.
02:31
Therefore, probability that at least one bagel of each kind is 0 .129...