00:01
In this problem, it is said that a carton contains 12 eggs, three are cracked, we randomly select five of the eggs and we need to find the required probabilities.
00:09
First of all, we want to find the probability that all of the cracked eggs are selected.
00:13
So all of the cracked eggs are selected.
00:16
So out of the three cracked eggs, three are selected.
00:19
We can do this in 3c3 ways.
00:21
And here we use c, which represents combination.
00:24
And we use combination and not permutation in this case because the order of selection of the eggs does not matter.
00:29
Now, all of the cracked eggs are selected, so three of the cracked eggs are selected, and since we select five eggs, that means that the remaining 5 minus 3, that is the remaining two eggs are not cracked, and out of the 12 eggs, three are cracked, so 9 are not cracked, we can select two eggs which are not cracked out of the 9 eggs which are not cracked in 9c2 ways.
00:49
And using the multiplication rule of counting, if we multiply these two numbers, then this expression will give us our required number of favorable outcomes.
00:57
And the total number of outcomes will be the number of ways we can select five eggs out of 12, that will be 12c5.
01:03
So this is what we get.
01:05
3c3 is 1 because ncn is 1.
01:07
9c2 is 9 factorial over 2 factorial times the factorial of 9 minus 2, that's 7.
01:13
And 12c5 is 12 factorial over 5 factorial times the factorial of 12 minus 5, which is 7.
01:19
And if we calculate this, we are going to end up with 1 over 22, so that will be a required answer.
01:24
Next, we want to find the probability that none of the cracked eggs...