Choose the correct answer from the given four options. Two cards …

MCQ

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:

A
$\frac{1}{13}\times\frac{1}{13}$
B
$\frac{1}{13}\times\frac{1}{13}$
C
$\frac{1}{13}\times\frac{1}{17}$
D
$\frac{1}{13}\times\frac{4}{15}$

✓

Answer

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

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

Explore more

More "M.C.Q (1 Marks)" from Probability→Probability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $f: \rightarrow R \rightarrow(0, \infty)$ be strictly increasing function such that $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$. Then, the value of $\lim _{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$ is equal to

→

For $x y=e^{x-y}, \frac{d y}{d x}=$ __________ .

→

Let $ \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right. $ $ \left.+\ldots \ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right) \text { be } \frac{\pi}{k},$ using only the principal values of the inverse trigonometric functions. Then $\mathrm{k}^2$ is equal to..............

→

If ${x^3} + {y^3} - 3axy = 0$, then ${{dy} \over {dx}}$ equals

→

The minimum value of the function $f(x) = {x^{10}} + {x^2} + \frac{1}{{{x^{12}}}} + \frac{1}{{\left( {1\ +\ {{\sec }^{ - 1}}\ x} \right)}}$ is

→

For linear programming problem the objective function $Z =10500 x+9000 y$, if the corner points of the bounded feasible region are $(0,0),(40,0),(30,20)$ and $(0,50)$, then the maximum value of $Z$ is _________ .

→

One hundred identical coins each with probability $p$ of showing up heads are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then the value of $p$ is

→

Let $\quad P_1=I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$,

$P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$

where $P _{ K }^{ T }$ denotes the transpose of the matrix $P _{ K }$. Then which of the following options is/are correct?

$(1)$ $X -30 I$ is an invertible matrix

$(2)$ The sum of diagonal entries of $X$ is 18

$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$

$(4)$ $X$ is a symmetric matrix

→

A unit vector perpendicular to both $\hat{\text{i}}+\hat{\text{j}}$ and $\hat{\text{j}}+\hat{\text{k}}$ is: