MCQ
Two dice are thrown simultaneously. The probability that sum is odd or less than $7$ or both, is
  • A
    $\frac{2}{3}$
  • B
    $\frac{1}{2}$
  • $\frac{3}{4}$
  • D
    $\frac{1}{3}$

Answer

Correct option: C.
$\frac{3}{4}$
c
(c) Required probability
$ = P({\rm{less}}\,\,{\rm{than}}\,\,7) + P({\rm{odd}}) + P({\rm{both}}) - P(7 \cap {\rm{odd}})$
$ - P(7 \cap {\rm{both}}) - P({\rm{odd}} \cap {\rm{both}}) + P({\rm{odd}} \cap 7 \cap {\rm{both}})$
But $P({\rm{both}}) = P(7 \cap {\rm{odd}}) = P(7 \cap {\rm{both}}) = P(o{\rm{dd}} \cap {\rm{both}})$
$ = P({\rm{odd}} \cap 7 \cap {\rm{both}})$
Therefore required probability
$ = P({\rm{Less\, than 7)}} + P({\rm{odd) -- }}P({\rm{7 }} \cap {\rm{odd}})$
$P({\rm{odd}}) = \frac{{18}}{{36}} = \frac{1}{2}$
$P({\rm{less}}\,\,{\rm{than}}\,\,7) = \frac{{15}}{{36}} = \frac{5}{{12}},\,\,P({\rm{both}}) = \frac{6}{{36}} = \frac{1}{6}$
Hence required probability $ = \frac{5}{{12}} + \frac{1}{2} - \frac{2}{{12}} = \frac{9}{{12}} = \frac{3}{4}.$

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

The three lines $lx + my + n = 0$, $mx + ny + l = 0$, $nx + ly + m = 0$ are concurrent if
If $\alpha $ and $\beta $ are roots of the equation ${x^2} - x + 1 = 0$  then ${\alpha ^{2009}} + {\beta ^{2009}} = $ . . . .
$8-$ digit numbers are formed using the digits $1, 1, 2, 2, 2, 3, 4, 4.$ The number of such numbers in which the odd digits do no occupy odd places, is
Let $w$ $(Im\, w \neq 0)$ be a complex number. Then the set of all complex number $z$ satisfying the equation $w - \overline {w}z  = k\left( {1 - z} \right)$ , for some real number $k$, is
If $m$ arithmetic means $( A . Ms )$ and three geometric means $(G.Ms)$ are inserted between $3$ and $243$ such that $4^{\text {th }}$ $A.M.$ is equal to $2^{\text {nd }}$ $G.M.$, then $m$ is equal to
Let $\mathrm{X}$ and $\mathrm{Y}$ be two events such that $\mathrm{P}(\mathrm{X})=\frac{1}{3}, \mathrm{P}(\mathrm{X} \mid \mathrm{Y})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{Y} \mid \mathrm{X})=\frac{2}{5}$. Then

$[A]$ $\mathrm{P}\left(\mathrm{X}^{\prime} \mid \mathrm{Y}\right)=\frac{1}{2}$   $[B]$ $\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{1}{5}$    $[C]$ $\mathrm{P}(\mathrm{X} \cup \mathrm{Y})=\frac{2}{5}$    $[D]$ $\mathrm{P}(\mathrm{Y})=\frac{4}{15}$

If $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in $H.P.$, then
A set of parallel chords of the parabola ${y^2} = 4ax$ have their mid-point on
The value of $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} $is
If the difference between the roots of the equation $\text{x}^2 +\text{ax}+1=0$ is less than $\sqrt{5},$ then the set of possible values of a is: