Question

In shuffling a pack of cards $3$ are accidentally dropped, then the chance that the missing card should be of different suits is
A. $\dfrac{169}{425}$
B. $\dfrac{261}{425}$
C. $\dfrac{104}{425}$
D. $\dfrac{425}{169}$

Answer 1

First we will calculate the number of ways of picking/dropping $3$ cards from a pack of $52$ cards and treat it as the number of events in the sample space. Now we need to calculate the probability that the missing card should be of a different suit. For that, we will calculate the number of possible ways to pick $3$ cards from $4$ suits of each $13$ card. Now we will treat it as the number of possible outcomes for the event. We know that the probability of an event is the ratio of the number of possible outcomes to an event to the total number of events in sample space.

Complete step-by-step solution
Given that, $3$ cards are dropped from the pack of $52$ cards.
Then the number of possible ways to pick/drop $3$ cards from $52$ cards is ${}^{52}{{C}_{3}}$.
$\therefore$ Total number of outcomes in the sample space is given by $n\left( S \right)={}^{52}{{C}_{3}}$
Let $A$ be the event of getting a missed card from a different suit.
We have $4$ different suits of $13$ cards in the pack of $52$ cards. Now we need to pick/drop $3$ cards in $4$ suits of each $13$ card in ${}^{4}{{C}_{3}}$ ways. We then can drop 1 card from 13 cards in ${}^{13}{{C}_{1}}$ ways. So by rule of product we can drop 3 cards from 3 different suits each with 13 cards at the same time in ${}^{13}{{C}_{1}}\times {}^{13}{{C}_{1}}\times {}^{13}{{C}_{1}}={{\left( {}^{13}{{C}_{1}} \right)}^{3}}$ ways. Then again by rule of product we can drop 3 cards from 4 different suits in ${}^{4}{{C}_{3}}\times {{\left( {}^{13}{{C}_{3}} \right)}^{3}}$ way.
$\therefore$ Number of possible outcomes for the event $A$ is $n\left( A \right)={}^{4}{{C}_{3}}.{{\left( {}^{13}{{C}_{1}} \right)}^{3}}$.
Now we are going to calculate the required probability by taking the ratio of the number of possible outcomes of event $A$ to the total number of outcomes in the sample space.
$\begin{aligned} & \Rightarrow P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)} \\\ & \Rightarrow P\left( A \right)=\dfrac{{}^{4}{{C}_{3}}.{{\left( {}^{13}{{C}_{1}} \right)}^{3}}}{{}^{52}{{C}_{3}}} \\\ & \Rightarrow P\left( A \right)=\dfrac{4\times {{\left( 13 \right)}^{3}}}{52\times 51\times 50} \\\ & \Rightarrow P\left( A \right)=\dfrac{169}{425} \\\ \end{aligned}$
$\therefore$ Required probability is $\dfrac{169}{425}$.

Note: While solving the problems related to the pack of cards, we need to know about the categories/suits of cards. There are four categories/suites of $13$ cards in the pack of $52$ cards. They are spade, heart, club, and diamond. Spade and club are black colored while the heart and diamond are red-colored. The 13 cards in each suit include - ace, king, queen, jack, and cards numbered from 2 to 10.