Question

A bag contains \( 4 \frac{2}{3} \) dozens of candies, out of which \( a \) are lemon candies. If one candy is drawn at random from the bag, there is some probability that it will be a lemon candy. Now, \( 4 \frac{2}{3} \) dozen more lemon candies are put in the bag, and the probability of drawing a lemon candy will be \( 2 \frac{1}{2} \) times higher than in the first case. Find \( a \).

• A. 16
• B. 14
• C. 20
• D. 18

Hint:

In probability problems, always carefully define the total number of outcomes and the favorable outcomes, then use the given conditions to form and solve equations.

Answer 1

The Correct Option is B

Step 1: Define the total number of candies and lemon candies.
The total number of candies in the bag is \( 4 \frac{2}{3} \) dozen, which is equal to \( 4 \frac{2}{3} \times 12 = 56 \) candies.
Let the number of lemon candies in the bag be \( a \).
Thus, the probability of drawing a lemon candy in the first case is: \[ P(\text{lemon candy}) = \frac{a}{56} \] Step 2: Add more lemon candies and calculate the new probability.
Now, \( 4 \frac{2}{3} \) dozen more lemon candies are added, i.e., \( 56 \) lemon candies are added. So, the new total number of candies becomes: \[ \text{New total number of candies} = 56 + 56 = 112 \] The new number of lemon candies becomes: \[ \text{New number of lemon candies} = a + 56 \] The new probability of drawing a lemon candy is: \[ P(\text{lemon candy}) = \frac{a + 56}{112} \] Step 3: Use the given condition.
We are told that the new probability is \( 2 \frac{1}{2} \) times higher than the first probability: \[ \frac{a + 56}{112} = \frac{5}{2} \times \frac{a}{56} \] Step 4: Solve the equation.
Multiply both sides by 112: \[ a + 56 = \frac{5}{2} \times 2a \] Simplify: \[ a + 56 = 5a \] \[ 56 = 4a \] \[ a = 14 \] Step 5: Conclusion.
The number of lemon candies in the bag is \( \boxed{14} \), which corresponds to option (2).