Question:
A box contains 5 white balls, one red ball and 4 black balls. If three balls are drawn at random, then the probability of getting no red ball is
A box contains 5 white balls, one red ball and 4 black balls. If three balls are drawn at random, then the probability of getting no red ball is
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Logic Tip: Alternatively, use the direct sequential probability method without combinations. The probability of the first ball not being red is $\frac{9}{10}$. The second is $\frac{8}{9}$. The third is $\frac{7}{8}$. Multiply them together: $\frac{9}{10} \times \frac{8}{9} \times \frac{7}{8} = \frac{7}{10}$. This avoids calculating factorials entirely!
Updated On: Apr 27, 2026
- $\frac{1}{10}$
- $\frac{3}{10}$
- $\frac{7}{10}$
- $\frac{9}{10}$
- $\frac{3}{5}$
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The Correct Option is C
Solution and Explanation
Concept:
The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes. When drawing multiple items without replacement where order doesn't matter, we use combinations: $^nC_r = \frac{n!}{r!(n-r)!}$. "No red ball" means all 3 balls must be drawn strictly from the pool of non-red balls.
Step 1: Determine the total possible outcomes.
Total balls in the box = $5 \text{ (white)} + 1 \text{ (red)} + 4 \text{ (black)} = 10 \text{ balls}$. We are drawing 3 balls. The total number of ways to choose 3 balls out of 10 is: $$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$$
Step 2: Determine the number of favorable outcomes.
We want to draw exactly 0 red balls. This means all 3 balls must be chosen from the other colors. Number of non-red balls = $5 \text{ (white)} + 4 \text{ (black)} = 9 \text{ balls}$. The total number of ways to choose 3 balls out of these 9 non-red balls is: $$^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$$
Step 3: Calculate the probability.
$$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$$ $$\text{Probability} = \frac{84}{120}$$ Divide both the numerator and the denominator by their greatest common divisor, 12: $$\text{Probability} = \frac{7}{10}$$
The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes. When drawing multiple items without replacement where order doesn't matter, we use combinations: $^nC_r = \frac{n!}{r!(n-r)!}$. "No red ball" means all 3 balls must be drawn strictly from the pool of non-red balls.
Step 1: Determine the total possible outcomes.
Total balls in the box = $5 \text{ (white)} + 1 \text{ (red)} + 4 \text{ (black)} = 10 \text{ balls}$. We are drawing 3 balls. The total number of ways to choose 3 balls out of 10 is: $$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$$
Step 2: Determine the number of favorable outcomes.
We want to draw exactly 0 red balls. This means all 3 balls must be chosen from the other colors. Number of non-red balls = $5 \text{ (white)} + 4 \text{ (black)} = 9 \text{ balls}$. The total number of ways to choose 3 balls out of these 9 non-red balls is: $$^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$$
Step 3: Calculate the probability.
$$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$$ $$\text{Probability} = \frac{84}{120}$$ Divide both the numerator and the denominator by their greatest common divisor, 12: $$\text{Probability} = \frac{7}{10}$$
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