Question

A box contains 5 red balls and 7 blue balls. Two balls are drawn at random without replacement. What is the probability that both balls are red?

• A. $\frac{11}{33}$
• B. $\frac{10}{35}$
• C. $\frac{12}{33}$
• D. $\frac{5}{33}$

Hint:

For "without replacement" problems, probabilities change after each draw.

Answer 1

The Correct Option is D

Step 1: Understand the Problem 
We need to find the probability of drawing two red balls in succession without replacement. The total number of balls is \(5 + 7 = 12\). Since the draws are without replacement, the probability of the second draw depends on the first. 

Step 2: Calculate the Probability 
The probability that both balls are red is the product of: - The probability that the first ball is red. - The probability that the second ball is red, given the first is red. 
- First draw: There are 5 red balls out of 12 total balls. \[ P(\text{first red}) = \frac{5}{12} \] 
- Second draw: After drawing one red ball, 4 red balls and 11 total balls remain. \[ P(\text{second red} \mid \text{first red}) = \frac{4}{11} \] The joint probability is: \[ P(\text{both red}) = P(\text{first red}) \times P(\text{second red} \mid \text{first red}) \] \[ P(\text{both red}) = \frac{5}{12} \times \frac{4}{11} \] \[ P(\text{both red}) = \frac{5 \times 4}{12 \times 11} = \frac{20}{132} \] Simplify the fraction: \[ \frac{20 \div 4}{132 \div 4} = \frac{5}{33} \] 

Step 3: Final Answer 
The probability that both balls are red is: \[ \boxed{\dfrac{5}{33}} \]