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:

Think of picking the two balls together as a single group instead of one after another. Work out how many 2-ball groups you can form from just the red balls, and how many 2-ball groups you can form from all the balls, then compare the two counts.

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: \[ \boxed{\dfrac{5}{33}} \]

Answer 2

Concept:

• Instead of multiplying two conditional draw-by-draw probabilities, the same answer can be reached by counting how many ways red pairs can be formed out of all possible pairs, using combinations.
• Since order does not matter when picking two balls together, dividing the count of favourable red pairs by the count of all possible pairs gives the probability directly.

Step 1: Count the total number of ways to choose 2 balls out of all 12.
Total balls $= 5 + 7 = 12$. The number of ways to choose any 2 balls from 12 is $\binom{12}{2} = \dfrac{12 \times 11}{2} = 66$.

Step 2: Count the number of ways to choose 2 red balls out of the 5 red balls.
$\binom{5}{2} = \dfrac{5 \times 4}{2} = 10$.

Step 3: Divide the favourable count by the total count.
$P(\text{both red}) = \dfrac{\binom{5}{2}}{\binom{12}{2}} = \dfrac{10}{66}$.

Step 4: Simplify the fraction.
$\dfrac{10}{66} = \dfrac{5}{33}$.

Final Answer: The probability that both balls are red is $\dfrac{5}{33}$.