Question:
A box has \(6\) black, \(4\) red, \(2\) white and \(3\) blue shirts. When \(2\) shirts are picked at random, the probability that either both are white or both are blue is:
A box has \(6\) black, \(4\) red, \(2\) white and \(3\) blue shirts. When \(2\) shirts are picked at random, the probability that either both are white or both are blue is:
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For selection problems, first count total combinations using \(^{n}C_r\). Then count each favourable case separately and add them if they are mutually exclusive.
Updated On: Jun 18, 2026
- \(\frac{4}{105}\)
- \(\frac{1}{35}\)
- \(\frac{1}{105}\)
- \(\frac{1}{15}\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the total number of shirts.
The box contains \[ 6+4+2+3=15 \] shirts.
Hence, the total number of ways of selecting \(2\) shirts is \[ {}^{15}C_2 = \frac{15\times 14}{2} = 105. \]
Step 2: Find the number of ways of selecting two white shirts.
There are \(2\) white shirts.
Therefore, \[ {}^{2}C_2=1. \]
Step 3: Find the number of ways of selecting two blue shirts.
There are \(3\) blue shirts.
Therefore, \[ {}^{3}C_2 = \frac{3\times 2}{2} = 3. \]
Step 4: Find the total favourable cases.
The events “both white” and “both blue” are mutually exclusive.
Hence, \[ \text{Favourable cases} = 1+3 = 4. \]
Step 5: Calculate the probability.
\[ P = \frac{\text{Favourable cases}} {\text{Total cases}} = \frac{4}{105}. \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{4}{105}} \]
The box contains \[ 6+4+2+3=15 \] shirts.
Hence, the total number of ways of selecting \(2\) shirts is \[ {}^{15}C_2 = \frac{15\times 14}{2} = 105. \]
Step 2: Find the number of ways of selecting two white shirts.
There are \(2\) white shirts.
Therefore, \[ {}^{2}C_2=1. \]
Step 3: Find the number of ways of selecting two blue shirts.
There are \(3\) blue shirts.
Therefore, \[ {}^{3}C_2 = \frac{3\times 2}{2} = 3. \]
Step 4: Find the total favourable cases.
The events “both white” and “both blue” are mutually exclusive.
Hence, \[ \text{Favourable cases} = 1+3 = 4. \]
Step 5: Calculate the probability.
\[ P = \frac{\text{Favourable cases}} {\text{Total cases}} = \frac{4}{105}. \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{4}{105}} \]
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