A box contains three apples and two oranges. Two fruits are removed randomly in succession. The probability that the first is an apple and the second is an orange is $\_\_\_\_\_$.

Question

If $ P(A|B) = P(A'|B) $, then which of the following statements is true?

Answer 1

Think of this as picking 2 fruits from the box one at a time, without replacement, and caring about the order they come out in. Instead of chaining two conditional probabilities together, we can count outcomes directly using permutations, since every fruit is a distinct physical object being pulled out in a definite order.

The total number of ways to draw 2 fruits in order from 5 fruits is a permutation of 5 items taken 2 at a time. The first draw can be any of the 5 fruits, and the second draw can be any of the remaining 4, so:

$$ P(5,2) = 5 \times 4 = 20 $$

This is the total number of equally likely ordered outcomes for the two draws, and it forms the denominator of our probability.

Now count the favorable outcomes, where the first fruit is an apple and the second is an orange. There are 3 apples to pick for the first spot, and once an apple is removed, 2 oranges remain in the box for the second spot. By the basic counting principle, we multiply the number of choices at each stage:

$$ \text{favorable outcomes} = 3 \times 2 = 6 $$

So the probability of the desired order is the ratio of favorable outcomes to total outcomes:

$$ P = \frac{6}{20} = \frac{3}{10} $$

Let's summarize:

Total ordered ways to draw 2 fruits out of 5 without replacement: 20.
Ordered ways with apple first and orange second: 6.
Probability = 6/20, which reduces to 3/10, matching option (C).
This counting method gives the same answer as the step by step conditional probability method, since both are just two ways of tracking the same set of equally likely outcomes.

$$ \boxed{\dfrac{3}{10}} $$

Answer 2

Think of this as picking 2 fruits from the box one at a time, without replacement, and caring about the order they come out in. Instead of chaining two conditional probabilities together, we can count outcomes directly using permutations, since every fruit is a distinct physical object being pulled out in a definite order.

The total number of ways to draw 2 fruits in order from 5 fruits is a permutation of 5 items taken 2 at a time. The first draw can be any of the 5 fruits, and the second draw can be any of the remaining 4, so:

$$ P(5,2) = 5 \times 4 = 20 $$

This is the total number of equally likely ordered outcomes for the two draws, and it forms the denominator of our probability.

Now count the favorable outcomes, where the first fruit is an apple and the second is an orange. There are 3 apples to pick for the first spot, and once an apple is removed, 2 oranges remain in the box for the second spot. By the basic counting principle, we multiply the number of choices at each stage:

$$ \text{favorable outcomes} = 3 \times 2 = 6 $$

So the probability of the desired order is the ratio of favorable outcomes to total outcomes:

$$ P = \frac{6}{20} = \frac{3}{10} $$

Let's summarize:

Total ordered ways to draw 2 fruits out of 5 without replacement: 20.
Ordered ways with apple first and orange second: 6.
Probability = 6/20, which reduces to 3/10, matching option (C).
This counting method gives the same answer as the step by step conditional probability method, since both are just two ways of tracking the same set of equally likely outcomes.

$$ \boxed{\dfrac{3}{10}} $$

A box contains three apples and two oranges. Two fruits are removed randomly in succession. The probability that the first is an apple and the second is an orange is \(\_\_\_\_\_\).

Show Hint

The Correct Option is C

Solution and Explanation

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