Question:
A bag contains \(21\) toys numbered \(1\) to \(21\). A toy is drawn and then another toy is drawn without replacement. The probability that both toys will show even numbers is
A bag contains \(21\) toys numbered \(1\) to \(21\). A toy is drawn and then another toy is drawn without replacement. The probability that both toys will show even numbers is
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In probability questions involving drawing without replacement, reduce both the total number of objects and the favorable number of objects after the first draw.
Updated On: Jun 25, 2026
- \(\dfrac{5}{21}\)
- \(\dfrac{3}{14}\)
- \(\dfrac{11}{42}\)
- \(\dfrac{4}{21}\)
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The Correct Option is B
Solution and Explanation
Step 1: Count the total number of toys.
The toys are numbered from \(1\) to \(21\).
Therefore, total number of toys is \[ 21 \]
Step 2: Count the even-numbered toys.
The even numbers from \(1\) to \(21\) are \[ 2,4,6,8,10,12,14,16,18,20 \] So, the number of even-numbered toys is \[ 10 \]
Step 3: Find the probability that the first toy is even.
Probability that the first toy shows an even number is \[ \frac{10}{21} \]
Step 4: Find the probability that the second toy is even.
Since the drawing is without replacement, after drawing one even-numbered toy, remaining total toys are \[ 20 \] Remaining even-numbered toys are \[ 9 \] So, the probability that the second toy also shows an even number is \[ \frac{9}{20} \]
Step 5: Multiply the probabilities.
Therefore, \[ P(\text{both even}) = \frac{10}{21}\times \frac{9}{20} \] Simplifying, \[ P(\text{both even}) = \frac{90}{420} \] \[ = \frac{3}{14} \]
Step 6: Final conclusion.
Hence, the required probability is \[ \boxed{\frac{3}{14}} \]
The toys are numbered from \(1\) to \(21\).
Therefore, total number of toys is \[ 21 \]
Step 2: Count the even-numbered toys.
The even numbers from \(1\) to \(21\) are \[ 2,4,6,8,10,12,14,16,18,20 \] So, the number of even-numbered toys is \[ 10 \]
Step 3: Find the probability that the first toy is even.
Probability that the first toy shows an even number is \[ \frac{10}{21} \]
Step 4: Find the probability that the second toy is even.
Since the drawing is without replacement, after drawing one even-numbered toy, remaining total toys are \[ 20 \] Remaining even-numbered toys are \[ 9 \] So, the probability that the second toy also shows an even number is \[ \frac{9}{20} \]
Step 5: Multiply the probabilities.
Therefore, \[ P(\text{both even}) = \frac{10}{21}\times \frac{9}{20} \] Simplifying, \[ P(\text{both even}) = \frac{90}{420} \] \[ = \frac{3}{14} \]
Step 6: Final conclusion.
Hence, the required probability is \[ \boxed{\frac{3}{14}} \]
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