I was solving some questions of probability and I came across this one. I have already made the following calculation for now.
Question: In a basket there are fifteen apples and ten oranges. Knowing that there are in the basket there are four rotten apples and two rotten oranges, determine the probability of a person withdrawing from the basket, at random, two fruits, get two apples or two fruits in good condition.
The probability of obtaining an apple is given by $\dfrac{C_{15,2}}{C_{25,2}}$, and the probability of obtaining two fruits is given by $\dfrac{C_{6,2}}{C_{25,2}}$. I think there is an intersection, but I do not see how to calculate such a probability at the intersection. I think the probability in intersection is $\dfrac{C_{11,2}}{C_{25,2}}$. Threfore, $$\dfrac{C_{15,2}+C_{6,2}-C_{11,2}}{C_{25,2}}$$ is a probability.
Or I may be solving the question wrong.
Thanks for the comments and the solutions.
Best Answer
Use the inclusion-exclusion principle:
$$\Bbb P(A \cup B) = \Bbb P(A) + \Bbb P(B) - \Bbb P(A \cap B)$$
In this case, $A$ is the event of obtaining two apples, and $B$ the event of obtaining two good fruits.
$$\Bbb P(A) = \frac{C^{15}_2}{C^{25}_2}$$
$$\Bbb P(B) = \frac{C^{19}_2}{C^{25}_2}$$
The intersection $A \cap B$ means that you get two good apples, and there are $11$ of them
$$\Bbb P(A \cap B) = \frac{C^{11}_2}{C^{25}_2}$$
Hence, the final answer is
$$\frac{C^{15}_2 + C^{19}_2 - C^{11}_2}{C^{25}_2}$$