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In this problem, it is said that in a certain shipment of seven phones, three are defective.
00:06
Five of the phones are selected at random without replacement, and we need to find the probability that three of the five phones are defective.
00:12
So our required probability will be the number of favorable outcomes divided by the total number of outcomes.
00:18
First of all, consider the total number of outcomes out of seven phones.
00:22
Any five are selected.
00:24
This can be done in seven c -5 ways.
00:26
Here we use c, which represents combination, and we use combination and our permutation.
00:30
In this case because the order of selection of the phones does not matter.
00:35
Next, consider the number of favorable outcomes.
00:37
That will be the number of ways, so that three of the five phones are defective.
00:40
So out of the three defective phones, we need to select any three at random.
00:44
This can be done in three c3 ways.
00:46
And the remaining 5 minus 3, that's the remaining two phones, must be not defective...