Question:
A bag contains \(2\) red, \(3\) green and \(2\) blue balls. Two balls are drawn at random. Then, the probability that none of the balls drawn is blue is
A bag contains \(2\) red, \(3\) green and \(2\) blue balls. Two balls are drawn at random. Then, the probability that none of the balls drawn is blue is
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For probability problems involving selection without replacement,
\[
P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
\]
and combinations are generally used for counting selections.
Updated On: Jun 25, 2026
- \(\dfrac{10}{21}\)
- \(\dfrac{11}{21}\)
- \(\dfrac{2}{7}\)
- \(\dfrac{5}{7}\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the total number of balls.
The bag contains \[ 2\text{ red},\quad 3\text{ green},\quad 2\text{ blue} \] Hence total balls are \[ 2+3+2=7 \]
Step 2: Find the total ways of drawing \(2\) balls.
Number of ways of selecting \(2\) balls from \(7\) balls is \[ {}^7C_2 \] Thus, \[ {}^7C_2=\frac{7\times 6}{2}=21 \]
Step 3: Find the favorable cases.
We need the probability that none of the selected balls is blue.
Non-blue balls are: \[ 2\text{ red}+3\text{ green}=5 \] Number of ways of selecting \(2\) balls from these \(5\) non-blue balls is \[ {}^5C_2 \] Therefore, \[ {}^5C_2=\frac{5\times 4}{2}=10 \]
Step 4: Calculate the required probability.
Hence, \[ P(\text{none is blue}) = \frac{{}^5C_2}{{}^7C_2} \] So, \[ P= \frac{10}{21} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{10}{21}} \]
The bag contains \[ 2\text{ red},\quad 3\text{ green},\quad 2\text{ blue} \] Hence total balls are \[ 2+3+2=7 \]
Step 2: Find the total ways of drawing \(2\) balls.
Number of ways of selecting \(2\) balls from \(7\) balls is \[ {}^7C_2 \] Thus, \[ {}^7C_2=\frac{7\times 6}{2}=21 \]
Step 3: Find the favorable cases.
We need the probability that none of the selected balls is blue.
Non-blue balls are: \[ 2\text{ red}+3\text{ green}=5 \] Number of ways of selecting \(2\) balls from these \(5\) non-blue balls is \[ {}^5C_2 \] Therefore, \[ {}^5C_2=\frac{5\times 4}{2}=10 \]
Step 4: Calculate the required probability.
Hence, \[ P(\text{none is blue}) = \frac{{}^5C_2}{{}^7C_2} \] So, \[ P= \frac{10}{21} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{10}{21}} \]
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