A box contains 7 white and 5 black balls. If 3 balls are drawn simultaneously at random, what is the probability that they are not all of the same colour? Calculate the probability of the same event for the case where the balls are drawn in succession with replacement between drawings.
Probability that they are not all of the same colour $$= \frac{^7C_2\times ^5C_1}{^{12}C_3}+\frac{^5C_2\times ^7C_1}{^{12}C_3}=\frac{35}{44}$$
For the second case, I did it like:
Probability that they are not all of the same colour where the balls are drawn in succession with replacement between drawings $$= \frac{7^2\times 5}{12^3}+\frac{5^2\times 7}{12^3}=\frac{35}{144}$$
But in my book the answer is $\frac{35}{48}$.
Best Answer
Your second answer with replacement is out by a factor of $3$, because order matters in the $12^3$ denominator and so needs to be taken into account in the numerator
The probability of one black ball and two white balls is ${3 \choose 1} \frac{7^2\times 5}{12^3}$ and the probability of two black balls and a white ball is ${3 \choose 2} \frac{7\times 5^2}{12^3}$