Question

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 \(\_\_\_\_\_\).

• A. \(\dfrac{3}{5}\)
• B. \(\dfrac{3}{20}\)
• C. \(\dfrac{3}{10}\)
• D. \(\dfrac{2}{5}\)

Hint:

Use the multiplication rule for dependent events: multiply the probability of the first draw by the probability of the second draw given the first fruit was not replaced.

Answer 1

The Correct Option is C

Step 1: Set up the problem.
The box has 3 apples and 2 oranges, so 5 fruits in all. Two fruits are taken out one after another, without putting the first one back. We need the chance that the first fruit is an apple and the second fruit is an orange, in that exact order.

Step 2: Find the chance the first fruit is an apple.
Out of the 5 fruits, 3 are apples, so
\[ P(\text{1st is apple}) = \frac{3}{5} \]

Step 3: Find the chance the second fruit is an orange, given the first was an apple.
After one apple is removed, 4 fruits are left in the box: 2 apples and 2 oranges. So the chance the second draw is an orange is
\[ P(\text{2nd is orange} \mid \text{1st was apple}) = \frac{2}{4} = \frac{1}{2} \]

Step 4: Multiply the two probabilities.
The second event depends on what happened in the first draw, so we use the multiplication rule for dependent events:
\[ P(\text{apple then orange}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10} \]

Step 5: Check the other options.
Option (A), \(\frac{3}{5}\), is just the chance the first fruit is an apple, it ignores the second draw. Option (B), \(\frac{3}{20}\), comes from wrongly taking the second probability as \(\frac{1}{4}\) instead of \(\frac{2}{4}\). Option (D), \(\frac{2}{5}\), is the chance the first fruit is an orange, which is not what the question asks for.

Final Answer:
\[ \boxed{\frac{3}{10}} \]