Question

A bag contains 5 red balls, 4 blue balls and 3 green balls. Two balls are drawn at random without replacement. The probability that both balls are red is:

• A. \(\frac{5}{33}\)
• B. \(\frac{10}{33}\)
• C. \(\frac{20}{33}\)
• D. \(\frac{5}{12}\)

Hint:

For "without replacement" questions, both the numerator and denominator change after each draw.

Answer 1

The Correct Option is A

Concept: When objects are drawn without replacement, the probability of successive events is found using multiplication of conditional probabilities.

Step 1: Find the probability that the first ball is red Total balls: \[ 5+4+3=12 \] Therefore, \[ P(\text{First red}) = \frac{5}{12} \]

Step 2: Find the probability that the second ball is red After drawing one red ball, remaining red balls \(=4\). Remaining total balls \(=11\). Hence, \[ P(\text{Second red} \mid \text{First red}) = \frac{4}{11} \]

Step 3: Apply multiplication rule \[ P(\text{Both red}) = \frac{5}{12}\times\frac{4}{11} \] \[ = \frac{20}{132} \] \[ = \frac{5}{33} \] Final Answer: \[ \boxed{\frac{5}{33}} \]