Question:
A bag contains 5 red and 7 blue balls. Find the probability of drawing 2 red balls without replacement.
A bag contains 5 red and 7 blue balls. Find the probability of drawing 2 red balls without replacement.
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For probabilities without replacement:
- The total number of items decreases after each draw.
- Use conditional probability \(P(A \cap B) = P(A)P(B|A)\).
Updated On: May 1, 2026
- \( \frac{5}{33} \)
- \( \frac{10}{33} \)
- \( \frac{25}{144} \)
- \( \frac{7}{33} \)
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The Correct Option is A
Solution and Explanation
Concept:
For probability without replacement, the outcome of the first event affects the probability of the second event. The multiplication rule of probability is used:
\[
P(A \cap B) = P(A) \times P(B|A)
\]
Step 1: Total number of balls. \[ 5 \text{ red} + 7 \text{ blue} = 12 \text{ balls} \]
Step 2: Probability of drawing the first red ball. \[ P(\text{First Red}) = \frac{5}{12} \]
Step 3: Probability of drawing the second red ball. After drawing one red ball, \(4\) red balls remain out of \(11\) total balls. \[ P(\text{Second Red}) = \frac{4}{11} \]
Step 4: Calculate the total probability. \[ P(\text{Two Reds}) = \frac{5}{12} \times \frac{4}{11} \] \[ = \frac{20}{132} = \frac{5}{33} \] Thus, the required probability is: \[ \boxed{\frac{5}{33}} \]
Step 1: Total number of balls. \[ 5 \text{ red} + 7 \text{ blue} = 12 \text{ balls} \]
Step 2: Probability of drawing the first red ball. \[ P(\text{First Red}) = \frac{5}{12} \]
Step 3: Probability of drawing the second red ball. After drawing one red ball, \(4\) red balls remain out of \(11\) total balls. \[ P(\text{Second Red}) = \frac{4}{11} \]
Step 4: Calculate the total probability. \[ P(\text{Two Reds}) = \frac{5}{12} \times \frac{4}{11} \] \[ = \frac{20}{132} = \frac{5}{33} \] Thus, the required probability is: \[ \boxed{\frac{5}{33}} \]
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