Question
Choose the correct answer from the given four options.
Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability, that both cards are queens, is:
  1. $\frac{1}{13}\times\frac{1}{13}$
  2. $\frac{1}{13}\times\frac{1}{13}$
  3. $\frac{1}{13}\times\frac{1}{17}$
  4. $\frac{1}{13}\times\frac{4}{15}$

Answer

  1. $\frac{1}{13}\times\frac{13}{13}$

Solution:

Required probability $=\frac{4}{52}\cdot\frac{4}{52}$

$=\frac{1}{13}\times\frac{1}{13}$ [with replacement]

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $A = \left[ {\begin{array}{*{20}{c}}\alpha &0\\1&1\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}1&0\\5&1\end{array}} \right]$, then value of $\alpha $for which ${A^2} = B$, is
Let $\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{V}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ then $\frac{\text{dU}}{\text{dV}}=$
  1. $\frac{1}{2}$
  2. $\text{x}$
  3. $\frac{1-\text{x}^2}{\text{x}^2-4}$
  4. $1$
The value of $\int_1^{{e^2}} {\frac{{dx}}{{x{{(1 + \ln x)}^2}}}} $ is
If $\mid\text{a}\mid=4$ and $-3\underline{<}\lambda\underline{<}2$ then the range of $\mid\lambda\text{a}\mid$ is:
  1. [0, 8]
  2. [-12, 8]
  3. [0, 12]
  4. [8, 12]
Statement $1$ : If $A$ and $B$ be two sets having $p$ and $q$ elements respectively, where $q > p$. Then the total number of functions from set $A$ to set $B$ is $q^P$.
Statement $2$ : The total number of selections of $p$ different objects out of $q$ objects is ${}^q{C_p}$.
If $\Delta (x) = \left| {\,\begin{array}{*{20}{c}}{{x^n}}&{\sin x}&{\cos x}\\{n!}&{\sin \frac{{n\pi }}{2}}&{\cos \frac{{n\pi }}{2}}\\a&{{a^2}}&{{a^3}}\end{array}\,} \right|,$ then the value of $\frac{{{d^n}}}{{d{x^n}}}[\Delta (x)]$ at $x = 0$ is
If ${x^2}{e^{y\,}}\, + \,2xy{e^{x\,}}\, + 13\, = \,0$, then $\frac{{dy}}{{dx}}$ equals
Let $l = \mathop {Lim}\limits_{x \to {0^ + }} x^m (ln\,\, x)^n$ where $m, n \in N$ then :
The function $f(x)=\left\{\begin{array}{cc}|x-3|, & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$ is
If $\left[\begin{array}{cc}3 c+6 & a-d \\ a+d & 2-3 b\end{array}\right]=\left[\begin{array}{cc}12 & 2 \\ -8 & -4\end{array}\right]$ then find $a b-c d$ :