Questions: A box contains assorted chocolates. Some chocolates contain only nuts and the others contain only cherries. A chocolate is randomly selected and eaten. Then another random selection is made from the remaining chocolates. Event A: The first selection contains nuts. Event B: The second selection contains cherries.

Solution Steps

1. Define the total number of chocolates, and the number of chocolates with nuts and cherries.
2. Calculate the probability of Event A, which is selecting a chocolate with nuts first.
3. Calculate the probability of Event B given Event A, which is selecting a chocolate with cherries second after a chocolate with nuts has been removed.
4. Use these probabilities to find the joint probability of both events occurring.

Let the total number of chocolates be \( T = 10 \). Among these, the number of chocolates containing nuts is \( N = 6 \) and the number containing cherries is \( C = 4 \).

The probability of selecting a chocolate with nuts first (Event A) is given by: \[ P(A) = \frac{N}{T} = \frac{6}{10} = 0.6 \]

After selecting a chocolate with nuts, the total number of chocolates remaining is \( T - 1 = 9 \). The probability of selecting a chocolate with cherries second (Event B) given that the first selection was a nut chocolate is: \[ P(B|A) = \frac{C}{T - 1} = \frac{4}{9} \approx 0.4444 \]

The joint probability of both events occurring is calculated as: \[ P(A \cap B) = P(A) \cdot P(B|A) = 0.6 \cdot 0.4444 \approx 0.2667 \]

Final Answer