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  2. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack, two cards are drawn at random (without replacement) and both are found to be spades. Find the probability of the lost card being a spade.  
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A card from a pack of 52 playing cards is lost. From the remaining cards of the pack, two cards are drawn at random (without replacement) and both are found to be spades. Find the probability of the lost card being a spade.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

Let E1: a spade is lost
      E2: no spade is lost
     A: two cards drawn are spade
Here P\left ( E_{1} \right )= \frac{13}{52}= \frac{1}{4}\; \; \; \; P\left ( E_{2} \right )= 1-\frac{1}{4}= \frac{3}{4}
P\left ( \frac{A}{E_{1}} \right )= \frac{^{12}C_{2}}{^{51}C_{2}},P\left ( \frac{A}{E_{2}} \right )= \frac{^{13}C_{2}}{^{51}C_{2}}
By baye's Theorem = P\left (\frac{ E_{1}}{A} \right )= \frac{P\left ( E_{1} \right )P\left ( \frac{A}{ E_{1}} \right )}{P\left ( E_{1} \right )P\left ( \frac{A}{ E_{1}} \right )+P\left ( E_{2} \right )P\left ( \frac{A}{ E_{2}} \right )}
\Rightarrow P\left ( \frac{ E_{1}}{A} \right )= \frac{\frac{1}{4}\times \frac{22}{425}}{\frac{1}{4}\times \frac{22}{425}+\frac{3}{4}\times \frac{26}{425}}= \frac{22}{22+78}
\Rightarrow \therefore P\left ( \frac{E_{1}}{A} \right )= \frac{11}{50}

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