Question:
Four cards are drawn successively with replacement from well shuffled deck of 52 cards, then the probability that only two cards are club cards is ...........
Four cards are drawn successively with replacement from well shuffled deck of 52 cards, then the probability that only two cards are club cards is ...........
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Each draw is a club with probability 1/4; use the binomial formula for exactly two successes in four trials.
Updated On: Oct 1, 2026
- \(\frac{26}{128}\)
- \(\frac{24}{128}\)
- \(\frac{27}{128}\)
- \(\frac{28}{128}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Drawing with replacement makes the four draws independent. The chance of a club in one draw is \(\dfrac{13}{52}=\dfrac14\), and of a non-club is \(\dfrac34\).
Step 2: Binomial formula
\[ P(X=2)={}^4C_2\left(\frac14\right)^2\left(\frac34\right)^2 \]
Step 3: Calculate
\[ {}^4C_2=6,\quad\left(\frac14\right)^2\left(\frac34\right)^2=\frac{9}{256} \]
\[ P=6\times\frac{9}{256}=\frac{54}{256}=\frac{27}{128} \]
Step 4: Check the options
The value \(\frac{27}{128}\) is option (C). The others have different numerators and do not follow from this formula.
Final Answer:
The probability is 6 times (1/4)^2 times (3/4)^2, which is 27/128, option (C). \[ \boxed{\frac{27}{128}} \]
Drawing with replacement makes the four draws independent. The chance of a club in one draw is \(\dfrac{13}{52}=\dfrac14\), and of a non-club is \(\dfrac34\).
Step 2: Binomial formula
\[ P(X=2)={}^4C_2\left(\frac14\right)^2\left(\frac34\right)^2 \]
Step 3: Calculate
\[ {}^4C_2=6,\quad\left(\frac14\right)^2\left(\frac34\right)^2=\frac{9}{256} \]
\[ P=6\times\frac{9}{256}=\frac{54}{256}=\frac{27}{128} \]
Step 4: Check the options
The value \(\frac{27}{128}\) is option (C). The others have different numerators and do not follow from this formula.
Final Answer:
The probability is 6 times (1/4)^2 times (3/4)^2, which is 27/128, option (C). \[ \boxed{\frac{27}{128}} \]
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