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A bag contains 4 red , 6 white and 5 bla...

A bag contains 4 red , 6 white and 5 black balls 2 balls are drawn at random . Find the probability of getting one red and one white ball .

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To solve the problem of finding the probability of drawing one red and one white ball from a bag containing 4 red, 6 white, and 5 black balls, we can follow these steps: ### Step 1: Determine the total number of balls in the bag. The bag contains: - 4 red balls - 6 white balls - 5 black balls Total number of balls = 4 + 6 + 5 = 15 **Hint:** Always start by calculating the total number of items in the set. ### Step 2: Calculate the total number of ways to draw 2 balls from the bag. The total number of ways to choose 2 balls from 15 is given by the combination formula \( C(n, r) \), which is defined as: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] For our case, we need to calculate \( C(15, 2) \): \[ C(15, 2) = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2 \times 1} = 105 \] **Hint:** Use the combination formula to find the number of ways to choose items from a set. ### Step 3: Calculate the number of favorable outcomes for drawing 1 red and 1 white ball. To get one red and one white ball, we can choose: - 1 red ball from 4 red balls: \( C(4, 1) \) - 1 white ball from 6 white balls: \( C(6, 1) \) Calculating these: \[ C(4, 1) = 4 \quad \text{and} \quad C(6, 1) = 6 \] The total number of favorable outcomes (1 red and 1 white) is: \[ C(4, 1) \times C(6, 1) = 4 \times 6 = 24 \] **Hint:** Multiply the number of ways to choose each type of item to find the total favorable outcomes. ### Step 4: Calculate the probability of drawing 1 red and 1 white ball. The probability \( P \) of drawing 1 red and 1 white ball is given by the ratio of the number of favorable outcomes to the total number of outcomes: \[ P(\text{1 red and 1 white}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{24}{105} \] ### Step 5: Simplify the probability. To simplify \( \frac{24}{105} \), we can find the greatest common divisor (GCD) of 24 and 105, which is 3. \[ \frac{24 \div 3}{105 \div 3} = \frac{8}{35} \] Thus, the probability of drawing one red and one white ball is: \[ \frac{8}{35} \] **Final Answer:** The probability of getting one red and one white ball is \( \frac{8}{35} \). ---
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