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In shuffling a pack of playing cards, four are accidently dropped. The probability that the missing card should be one from each suit.
A) $\dfrac{1}{{256}}$
B) $\dfrac{1}{{270725}}$
C) $\dfrac{{{2^9}}}{{20825}}$
D) None of these

Last updated date: 13th Jun 2024
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Hint: These types of questions where cards are missing are similar to the questions to which we have to choose the cards so generally we apply the same concept in both of these type of questions i.e. to select card from each and every suit and to find the probability. In these kinds of questions we must have knowledge of cards that there are a total of $52$ cards in the pack of cards.

Random experiments $\therefore $
$ \to $ Use combination concept
Formula $ = {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$ \to $ 4 suits (Club 13, Spade 13, Heart 13, Diamond 13)

Complete step by step answer:
Here in the question, it is given that in shuffling a pack of playing cards, four are accidently dropped and we need to find the probability of the event that the missing card should be one from each suit.

The total number of ways in which 4 cards dropped out of 52 cards which is ${}^{52}{C_4}$
$\therefore $Exhaustive number of cases $ = {}^{52}{C_4}$
There are four suits and each suit contains 13 cards.
$\therefore $The number of ways of selecting one card from each suit is \[{}^{13}{C_1} \times {}^{13}{C_1} \times {}^{13}{C_1} \times {}^{13}{C_1} = {\left( {{}^{13}{C_1}} \right)^4}\]
So, the required probability is
$\dfrac{{{{\left( {{}^{13}{C_1}} \right)}^4}}}{{{}^{52}{C_4}}}$
By using formula $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
= $\dfrac{{\dfrac{{13!}}{{1!\left( {13 - 1} \right)!}} \times \dfrac{{13!}}{{1!\left( {13 - 1} \right)!}} \times \dfrac{{13!}}{{1!\left( {13 - 1} \right)!}} \times \dfrac{{13!}}{{1!\left( {13 - 1} \right)!}}}}{{\dfrac{{52!}}{{4!\left( {52 - 4} \right)!}}}}$
On solving this we got,
 $ = \dfrac{{\dfrac{{13 \times 12!}}{{1 \times 12!}} \times \dfrac{{13 \times 12!}}{{1 \times 12!}} \times \dfrac{{13 \times 12!}}{{1 \times 12!}} \times \dfrac{{13 \times 12!}}{{1 \times 12!}}}}{{\dfrac{{52!}}{{4! \times 48!}}}}$
So, $\dfrac{{13 \times 13 \times 13 \times 13}}{{270725}} = 0.105498$
The required probability is 0.105498.

So the correct option is (D).

Note: In this type of questions students will face problems while solving or using formula. So be careful while using or solving formulas. Through formula questions become very simple and easy. Also one should make the mistake in calculation factorial so we are careful while solving this. One should also remember the knowledge of how many cards and suits are there in the pack of cards and about the number of cards in each suit because without this knowledge we will not be able to solve such problems.