Question:
From a pack of \(52\) cards, \(3\) cards are drawn at random. Then, the probability that one is ace, one is queen and one is jack is:
From a pack of \(52\) cards, \(3\) cards are drawn at random. Then, the probability that one is ace, one is queen and one is jack is:
Show Hint
In card probability problems, calculate:
\[
\text{Probability}
=
\frac{\text{Favourable outcomes}}{\text{Total outcomes}}
\]
using combinations wherever selections are unordered.
Updated On: Jun 24, 2026
- \(\dfrac{19}{5525}\)
- \(\dfrac{21}{5525}\)
- \(\dfrac{17}{5525}\)
- \(\dfrac{16}{5525}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Find the total number of ways.
Total number of ways to draw \(3\) cards from \(52\) cards is \[ ^{52}C_3 \] \[ ^{52}C_3 = \frac{52\times 51\times 50}{3\times 2\times 1} = 22100 \]
Step 2: Find the favourable outcomes.
We need: \[ 1 \text{ ace},\quad 1 \text{ queen},\quad 1 \text{ jack} \] Number of aces \[ =4 \] Number of queens \[ =4 \] Number of jacks \[ =4 \] Thus, favourable ways are \[ 4\times 4\times 4 = 64 \]
Step 3: Find the probability.
Therefore, \[ P = \frac{64}{22100} \] Simplifying, \[ P = \frac{16}{5525} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\frac{16}{5525}} \]
Total number of ways to draw \(3\) cards from \(52\) cards is \[ ^{52}C_3 \] \[ ^{52}C_3 = \frac{52\times 51\times 50}{3\times 2\times 1} = 22100 \]
Step 2: Find the favourable outcomes.
We need: \[ 1 \text{ ace},\quad 1 \text{ queen},\quad 1 \text{ jack} \] Number of aces \[ =4 \] Number of queens \[ =4 \] Number of jacks \[ =4 \] Thus, favourable ways are \[ 4\times 4\times 4 = 64 \]
Step 3: Find the probability.
Therefore, \[ P = \frac{64}{22100} \] Simplifying, \[ P = \frac{16}{5525} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\frac{16}{5525}} \]
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