Question:medium

A bag contains \(5\) red and \(7\) blue balls. If two balls are drawn at random, what is the probability that both are red?

Show Hint

For drawing without replacement: \[ P(A \cap B) = P(A) \times P(B|A) \] Alternatively, use combinations: \[ P(\text{both red}) = \frac{\binom{5}{2}}{\binom{12}{2}} \]
Updated On: Mar 16, 2026
  • \( \dfrac{5}{66} \)
  • \( \dfrac{10}{66} \)
  • \( \dfrac{5}{33} \)
  • \( \dfrac{25}{66} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question
We need to calculate the probability of drawing two red balls in succession, without replacement, from a bag containing 5 red and 7 blue balls.
Step 2: Key Formula or Approach
We can solve this using the multiplication rule for dependent events:
\[ P(\text{Event 1 and Event 2}) = P(\text{Event 1}) \times P(\text{Event 2} | \text{Event 1 has occurred}) \] Alternatively, we can use combinations:
\[ P(\text{Both red}) = \frac{\text{Number of ways to choose 2 red balls}}{\text{Total number of ways to choose 2 balls}} = \frac{\binom{5}{2}}{\binom{12}{2}} \] Step 3: Detailed Explanation (Using Multiplication Rule)
First, determine the total number of balls:
Total balls = 5 Red + 7 Blue = 12 balls.
The probability that the first ball drawn is red is:
\[ P(\text{1st is Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{12} \] After drawing one red ball, there are 4 red balls left and a total of 11 balls remaining.
The probability that the second ball drawn is also red, given the first was red, is:
\[ P(\text{2nd is Red} | \text{1st is Red}) = \frac{\text{Remaining red balls}}{\text{Remaining total balls}} = \frac{4}{11} \] Now, we multiply these probabilities to get the probability of both events happening:
\[ P(\text{Both are Red}) = P(\text{1st is Red}) \times P(\text{2nd is Red} | \text{1st is Red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} \] Simplifying the fraction gives:
\[ \frac{20}{132} = \frac{10}{66} \] Step 4: Final Answer
The probability that both balls drawn are red is \( \frac{10}{66} \).
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