Question

A bag contains \(5\) red balls, \(4\) blue balls, and \(3\) green balls. Two balls are drawn at random without replacement. What is the probability that both balls drawn are red?

• A. \(\frac{5}{33}\)
• B. \(\frac{10}{33}\)
• C. \(\frac{2}{11}\)
• D. \(\frac{5}{22}\)

Hint:

For "without replacement" problems, either use conditional probability or combinations. Both methods should give the same result.

Answer 1

The Correct Option is A

Concept: Probability questions in CUET GAT often involve drawing objects without replacement. In such cases, the probability changes after each draw because the composition of the bag changes. The key principle is: \[ P(A \cap B)=P(A)\times P(B|A) \] where the second probability is conditional on the first event already occurring.

Step 1: Determine the total number of balls. \[ 5+4+3 = 12 \] Total balls: \[ 12 \]

Step 2: Calculate the probability of drawing a red ball first. Number of red balls: \[ 5 \] Total balls: \[ 12 \] Therefore: \[ P(\text{First red}) = \frac{5}{12} \]

Step 3: Calculate the probability of drawing a red ball second. After one red ball has been drawn: Remaining red balls: \[ 4 \] Remaining total balls: \[ 11 \] Thus: \[ P(\text{Second red}|\text{First red}) = \frac{4}{11} \]

Step 4: Multiply the probabilities. \[ P(\text{Both red}) = \frac{5}{12} \times \frac{4}{11} \] \[ = \frac{20}{132} \] \[ = \frac{5}{33} \]

Step 5: Verification using combinations. Favourable selections: \[ {}^5C_2 = 10 \] Total selections: \[ {}^{12}C_2 = 66 \] Therefore: \[ \frac{10}{66} = \frac{5}{33} \] Same answer obtained.

Step 6: Final conclusion. Hence: \[ \boxed{\frac{5}{33}} \] Therefore: \[ \boxed{\text{Option (A)}} \]