A factory makes three different types of clothes: clothes A, clothes B, clothes C. The factory produces hundreds of clothes each year, with twice as many clothes B as clothes A. The number of clothes C is twice the amount of clothes A combined clothes B. Four clothes made by the factory are selected at random from all types of clothes produced by the factory in a given year. Determine the probability that the sample will contain two shirts B and two shirts C.
The answer should be 32/243. But I don't know how to get that answer. Kindly help :)
First, you can consider the probability that you select one clothing from all clothes and it turns out to be A or B or C. We call them $P_A$, $P_B$ and $P_C$. Apparently, $P_A=\frac{1}{9}$, $P_B=\frac{2}{9}$ and $P_C=\frac{2}{3}$. You want to select four pieces containing two B and two C, so there are several possibilities. For example, the first is B, the second is B, the third is C and the fourth is C. Let's denote this by BBCC. The probability for BBCC is $P_B\times P_B\times P_C\times P_C=\frac{16}{729}$. Apart from BBCC, there are more possibilities like BCBC, BCCB, CBBC and so on. Notice that it's like place two B and two C in four cells and once we determine the places for two B, we determine one possibility. So we know the number of possibilities is $C_4^2=6$. The probability for all possibility is the same, that is, $\frac{16}{729}$. So $\frac{16}{729}\times6=\frac{32}{243}$. We get the answer. We consider all orderly possibilities so the result is right for selecting four pieces randomly.
My English is terrible and I'm so sorry about that.