水果篮

我们将 $a$、$o$ 和 $p$ 称为你从篮子中取出相应水果所需的圈数。如果我们在摘橙子和桃子之前摘完苹果，我们就知道 $a < o$ 和 $a < p$。有两种方法可以满足这个联合条件。 $a < o < p$ 或 $a < p < o$。让我们计算一下每个的概率。

如果我们先摘苹果，后摘橙子，先摘橙子，后摘桃子，那么桃子就是最后摘完的。因为我们是随机采摘的，所以每个水果成为最后一个水果的概率都是相等的。在 $60$ 的所有水果中，$30$ 是桃子。因此，最后一个水果是桃子的概率是 $\frac{30}{60}$ 或 $\frac12$。现在，我们需要计算在桃子是最后一个从篮子中采摘的水果的条件下，在苹果之后完成采摘橙子的概率。苹果和橙子之间总共有 $30$ 水果，其中橙子是 $20$，因此我们的条件概率是 $P(a < o | p = 60) = \frac23$。这些结果的联合概率为： $$\frac12 \cdot \frac23 = \frac13$$ 对于我们在桃子之前完成苹果采摘、在橘子之前完成桃子采摘的情况进行类似的操作，我们得到最后一个水果是橘子的概率 $\frac13$，以及当橘子最后一个时在桃子之前完成苹果的概率 $\frac34$ 给我们一个联合概率： $$\frac13 \cdot \frac34 = 1/4$$ 其中任何一个发生的概率是这些概率的总和，因此我们的答案是： $$\frac13 + \frac14 = \boxed{\frac7{12}}$$

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Original Explanation

Let's call $a$, $o$, and $p$ the number of turns it takes you to finish taking that respective fruit out of the basket. If we finish picking apples before oranges and peaches, we know $a < o$ and $a < p$. There exist two ways where we can satisfy this joint condition. Either $a < o < p$ or $a < p < o$. Let's compute the probability of each.

In the case where we finish picking apples before oranges and oranges before peaches, peaches are what would be finished last. Because we are picking at random, each individual fruit has an equal probability of being the last fruit. Out of the $60$ total fruits, $30$ of them are peaches. Thus the probability the very last fruit is a peach is $\frac{30}{60}$ or $\frac12$. We now need to find the probability that oranges are finished being picked after apples conditioned upon peaches being the last fruit picked from the basket. Between apples and oranges there are a total of $30$ fruits, oranges being $20$ of them, thus our conditional probability is $P(a < o | p = 60) = \frac23$. The joint probability of these outcomes is: $$\frac12 \cdot \frac23 = \frac13$$

Doing something similar for the case where we finish picking apples before peaches and peaches before oranges we get a probability of $\frac13$ of the last fruit being an orange and a probability of $\frac34$ of finishing apples before peaches when oranges are last giving us a joint probability of: $$\frac13 \cdot \frac34 = 1/4$$

The probability that either of these occur is the sum of these probabilities and so our answer is: $$\frac13 + \frac14 = \boxed{\frac7{12}}$$