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.

  • $\begingroup$ What do you mean by "and two or rotten"? $\endgroup$ – Fareed AF Apr 23 at 12:24
  • $\begingroup$ Sorry. In the basket there are four rotten apples and two rotten oranges. $\endgroup$ – user98236 Apr 23 at 12:26
  • $\begingroup$ You wrote $$\text{the probability of obtaining two fruits is given by} \frac{C_{6,2}}{C_{25,2}}$$ In my opinion, Apples and Oranges are both fruits... Unless you mean the probability of obtaining a rotten fruit? But the question asks for the probability of obtaining fruit in good condition $\endgroup$ – glowstonetrees Apr 23 at 12:33
  • $\begingroup$ Yes. I wrote in fact wrong. The probability of obtaining two fruit in good condition is $\dfrac{C_{19,2}}{C_{25,2}}$ $\endgroup$ – user98236 Apr 23 at 12:36
  • $\begingroup$ Otherwise, your answer should be correct $\endgroup$ – glowstonetrees Apr 23 at 12:39

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

$$\frac{C^{15}_2 + C^{19}_2 - C^{11}_2}{C^{25}_2}$$


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