Question

Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then, P(X=1)+P(X=2) equals
(a) $\dfrac{52}{169}$
(b) $\dfrac{25}{169}$
(c) $\dfrac{49}{169}$
(d) $\dfrac{24}{169}$

Answer 1

Hint: First, before proceeding for this, we must know that the here the probability given is for the success that we get an ace which is p where from the card drawn from 52 cards, the probability of success means getting an ace is $\dfrac{1}{13}$. Then, we also know that in probability, the total probability is always 1, which gives the probability of failure q as $\dfrac{12}{13}$.Then, for getting the value of P(X=1)+P(X=2), we use the formula of the binomial where n is number of times cards drawn and r is the success rate as $P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$, we get the final result.

Complete step-by-step answer:
In this question, we are supposed to find the value of P(X=1)+P(X=2) when two cards are drawn successively with replacement from a well shuffled deck of 52 cards where X denote the random variable of number of aces obtained in the two drawn cards.
So, before proceeding for this, we must know that the here the probability given is for the success that we get an ace which is p.
So, from the card drawn from 52 cards, the probability of success means getting an ace is:
$\dfrac{4}{52}=\dfrac{1}{13}$
Now, we also know that in probability, the total probability is always 1, which gives the probability of failure q as:
$\begin{align}
  & q=1-p \\
 & \Rightarrow q=1-\dfrac{1}{13} \\
 & \Rightarrow q=\dfrac{13-1}{13} \\
 & \Rightarrow q=\dfrac{12}{13} \\
\end{align}$
Then, for getting the value of P(X=1)+P(X=2), we use the formula of the binomial where n is number of times cards drawn and r is the success rate as:
$P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$
So, firstly, we need the value of P(X=1) which gives value or r as 1 to get:
$\begin{align}
  & P\left( X=1 \right)={}^{2}{{C}_{1}}{{\left( \dfrac{1}{13} \right)}^{1}}{{\left( \dfrac{12}{13} \right)}^{2-1}} \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{2!}{\left( 2-1 \right)!1!}{{\left( \dfrac{1}{13} \right)}^{1}}{{\left( \dfrac{12}{13} \right)}^{1}} \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{2\times 1!}{1!1!}\left( \dfrac{12}{169} \right) \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{2}{1}\left( \dfrac{12}{169} \right) \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{24}{169} \\
\end{align}$
Similarly, we can get the value of P(X=2) in which we get value of r as 2 to get:
$\begin{align}
  & P\left( X=2 \right)={}^{2}{{C}_{2}}{{\left( \dfrac{1}{13} \right)}^{2}}{{\left( \dfrac{12}{13} \right)}^{2-2}} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{2!}{\left( 2-2 \right)!2!}{{\left( \dfrac{1}{13} \right)}^{2}}{{\left( \dfrac{12}{13} \right)}^{0}} \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{2!}{0!2!}\left( \dfrac{1}{169} \right) \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{1}{169} \\
\end{align}$
Now, by adding the two probabilities, we get the result required as:
$\begin{align}
  & P\left( X=1 \right)+P\left( X=2 \right)=\dfrac{24}{169}+\dfrac{1}{169} \\
 & \Rightarrow P\left( X=1 \right)+P\left( X=2 \right)=\dfrac{25}{169} \\
\end{align}$
So, we get the value of P(X=1)+P(X=2) as $\dfrac{25}{169}$.

So, the correct answer is “Option b”.

Note: Now, to solve these types of questions we need to know some of the basic formulas of factorial and combination. Now, to get the value of ${}^{n}{{C}_{r}}$ is given by the formula of the combination:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$