Question

a box of chocolates contains 5 milk chocolates, 3 caramels, and 8 strawberry creams. you randomly select and eat 3 of them. the first is a milk chocolate, the second is a caramel and the third is a strawberry cream. what is the probability that this event happens? show all of your steps used to solve and reduce only the final answer fraction.

Explanation:

Step1: Find total number of chocolates

Total chocolates = 5 (milk) + 3 (caramels) + 8 (strawberry) = 16.

Step2: Probability of first (milk)

Probability first is milk: $\frac{5}{16}$.

Step3: Probability of second (caramel)

After removing 1 milk, total left = 15. Probability second is caramel: $\frac{3}{15}$.

Step4: Probability of third (strawberry)

After removing 1 milk and 1 caramel, total left = 14. Probability third is strawberry: $\frac{8}{14}$.

Step5: Multiply probabilities

Total probability = $\frac{5}{16} \times \frac{3}{15} \times \frac{8}{14}$. Simplify: $\frac{5\times3\times8}{16\times15\times14} = \frac{120}{3360} = \frac{1}{28}$.

Answer:

$\frac{1}{28}$