MCQ
The probability that a leap year selected randomly will have $53$ Sundays is
  • A
    $\frac{1}{7}$
  • $\frac{2}{7}$
  • C
    $\frac{4}{{53}}$
  • D
    $\frac{4}{{49}}$

Answer

Correct option: B.
$\frac{2}{7}$
b
(b) A leap year contain $366$ days i.e. $52$ weeks and $2$ days, clearly there are $52$ Sundays in $52$ weeks
For the remaining two days, we may have any of the following two days.
$(i)$ Sunday and Monday, $(ii)$ Monday and Tuesday,
$(iii)$ Tuesday and Wednesday, $(iv)$ Wednesday and
Thursday, $(v)$ Thursday and Friday, $(vi)$ Friday and
Saturday and $(vii)$ Saturday and Sunday.
Now for $53$ Sundays, one of the two days must be Sunday

Hence required probability $ = \frac{2}{7}.$

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

Let $\alpha ,\beta $ be the roots of ${x^2} - x + p = 0$ and $\gamma ,\delta $ be the roots of ${x^2} - 4x + q = 0$. If $\alpha ,\beta ,\gamma ,\delta $ are in $G.P.$, then integral values of $p,\,q$ are respectively
Let $x_1,x_2,.........,x_{100}$ are $100$ observations such that  $\sum {{x_i} = 0,\,\sum\limits_{1 \leqslant i \leqslant j \leqslant 100} {\left| {{x_i}{x_j}} \right|} }  = 80000\,\& $ mean deviation from their mean is $5,$ then their standard deviation, is-
If $y=m x+4$ is a tangent to both the parabolas, $\mathrm{y}^{2}=4 \mathrm{x}$ and $\mathrm{x}^{2}=2 \mathrm{by},$ then $\mathrm{b}$ is equal to 
The number of all possible triplets $(a_1 , a_2 , a_3)$ such that $a_1+ a_2 \,cos \, 2x + a_3 \, sin^2 x = 0$ for all $x$ is
Equation of the line which passes through the point $( - 4,\;3)$ and the portion of the line intercepted between the axes is divided internally in the ratio $5 : 3$ by this point, is
If $\alpha ,\beta $ are the roots of $(x - a)(x - b) = c,$ $c \ne 0,$ then the roots of $(x - \alpha )(x - \beta ) + c = 0$ shall be
$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $
The sum of the series

$\frac{3}{{1! + 2! + 3!}} + \frac{4}{{2! + 3! + 4!}} + \frac{5}{{3! + 4! + 5!}} + ...... + \frac{{2008}}{{\left( {2006} \right)! + \left( {2007} \right)! + \left( {2008} \right)!}}$ is equal to

The solution of $\frac{{dy}}{{dx}} = {2^{y - x}}$ is
Consider two curves $C_1 : y^2 = 2x$ and $C_2 : x^2 + y^2 -3x + 2 = 0$, then