Maths M.C.Q (1 Marks)

{As there are total number of $n\,!$ ways in which letters can take envelopes and just one way in which they have corresponding envelopes}.

Therefore the probability that the problem is not solved by any one of them $ = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$

Hence the probability that problem is solved$ = 1 - \frac{2}{5} = \frac{3}{5}$.

$P({E_1}) = \frac{1}{4}$

so $P({\bar E_1}) = 1 - \frac{1}{4} = \frac{3}{4}$ and $P({E_2}) = \frac{1}{3}$

So $P({\bar E_2}) = \frac{2}{3}$

Clearly ${E_1}$ and ${E_2}$ are independent events.

So, $P({\bar E_1} \cap {\bar E_2}) = P({\bar E_1}) \times P({\bar E_2}) = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$.

$ = \frac{2}{3}.\frac{3}{4}.\frac{4}{5} = \frac{2}{5}.$

as $(10,\,20,\,30\,\,.......)$ have been considered.

Hence required probability $ = \frac{{50 + 10}}{{100}} = \frac{3}{5}.$

$ = 1 - \frac{1}{{4\,!}} = \frac{{23}}{{24}}.$

The probability that wife is not selected $ = 1 - \frac{1}{5} = \frac{4}{5}$

The probability that only husband selected $ = \frac{1}{7} \times \frac{4}{5} = \frac{4}{{35}}$

The probability that only wife selected $ = \frac{1}{5} \times \frac{6}{7} = \frac{6}{{35}}$

Hence required probability $ = \frac{6}{{35}} + \frac{4}{{35}} = \frac{{10}}{{35}} = \frac{2}{7}$.

$ \Rightarrow {q_1} = \frac{2}{3},$ ${q_2} = \frac{5}{7}$ and ${q_3} = \frac{5}{8}$

Required probability $ = {p_1}{q_2}{q_3} - {q_1}{p_2}{q_3} + {q_1}{q_2}{p_3}.$

Hence required probability

$ = P(A)P(\bar B)P(\bar C) + P(\bar A)P(B)P(\bar C) + P(\bar A)P(\bar B)P(C).$

$ = \frac{2}{9} \times \frac{1}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times 1 \times 1 \times 1 = \frac{1}{{1260}}$.

Therefore the required probability

$ = \frac{{48 \cdot 47 \cdot .... \cdot 39}}{{52 \cdot 51 \cdot .... \cdot 43}}.\frac{4}{{42}} = \frac{{164}}{{4165}}.$

Therefore, the probability is greater than $\frac{1}{2}$ which is given in only one option.

$A = $ India wins the toss and wins the match.

$B = $ India losses the toss and wins the match

$\therefore$  Required probability

$ = P(A) + P(B) = \frac{3}{4} \times \frac{4}{5} + \frac{1}{4} \times \frac{1}{2} = \frac{{29}}{{40}}$.

$B$ denote the event getting $II$;

$C$ denote the event getting $III$;

and $D$ denote the event getting fail.

Obviously, these four event are mutually exclusive and exhaustive,

therefore $P(A) + P(B) + P(C) + P(D) = 1 $

$\Rightarrow P(D) = 1 - 0.95 = 0.05$.

Sample space = $\left\{ \begin{array}{l}(6,\,\,1)\,\,(6,\,\,2)\,\,(6,\,\,3)\,\,(6,\,\,4)\$6,\,\,5)\,\,(1,\,\,6)\,\,(2,\,\,6)\,\,(3,\,\,6)\$4,\,\,6)\,\,(5,\,\,6)\,\,(6,\,\,6)\end{array} \right\}$

Probability of at least one $6$

$ = P$(one $6$) $ + P$(both $6$)

$ = \frac{{10}}{{36}} + \frac{1}{{36}} = \frac{{11}}{{36}}.$

$\therefore$ Required probability $ = \frac{6}{{{4^3}}} = \frac{6}{{64}} = \frac{3}{{32}}$.

Sunday -Monday, Monday -Tuesday, Tuesday -Wednesday, Wednesday -Thursday, Thursday -Friday, Friday -Saturday, Saturday -Sunday

$P(53$ Fridays) = $\frac{2}{7}$; $P(53$ Saturdays) $ = \frac{2}{7}$

$P(53$ Fridays and $53$ Saturdays) $ = \frac{1}{7}$

$\therefore$ $P(53$ Fridays or Saturdays) = $P(53$ Fridays$) +  P(53$ Saturdays) $-P(53$ Fridays and Saturdays)

$ = \frac{2}{7} + \frac{2}{7} - \frac{1}{7}$ $ = \frac{3}{7}$.

Favourable cases $= [12, 24, 32, 44, 52]$

$\therefore$ Required probability $ = \frac{5}{{25}} = \frac{1}{5}.$

$(i)$ When first is an ace of heart and second one is non-ace of heart

$ = \frac{1}{{52}} \times \frac{{51}}{{52}}\, \Rightarrow \frac{1}{{52}}$

$(ii)$ When first is non-ace of heart and second one is an ace of heart

$ = \frac{{51}}{{52}} \times \frac{1}{{51}} = \frac{1}{{52}}$

$\therefore$ Required probability $ = \frac{1}{{52}} + \frac{1}{{52}} = \frac{1}{{26}}.$

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

$P($black ball)$ = {p_2} = \frac{5}{8}$

From bag $B,$ $P$ (red ball) $ = {p_3} = \frac{6}{{10}}$

$P$(black ball) = ${p_4} = \frac{4}{{10}}$

Required probability

$ = P$[(red ball from bag $A$ and black from $B)$ or

(red from bag $B$ and black from $A)]$

$ = {p_1} \times {p_4} + {p_2} \times {p_3} = \frac{3}{8} \times \frac{4}{{10}} + \frac{5}{8} \times \frac{6}{{10}} = \frac{{21}}{{40}}$.

$\therefore $ Required probability $= P\,(X) \cdot P(\bar Y) + P(\bar X)P(Y)$

$=\left( {\frac{3}{5}} \right)\,\left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{2}{5}} \right)\,\left( {\frac{1}{2}} \right)$

$=\frac{3}{5}\,.\,\frac{1}{2} + \frac{2}{5}.\frac{1}{2} = \frac{1}{2}$.

$= \left( {1 - \frac{1}{2}} \right)\,\left( {1 - \frac{1}{3}} \right) = \frac{1}{3}$ and

probability that the problem is solved $=  1 - \frac{1}{3} = \frac{2}{3}$.

$P(\bar A) = 1 - P(A) =  1 - \frac{1}{2} = \frac{1}{2}$

$P(\bar B) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$

$\therefore $ $P(\bar A).P(\bar B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.

$= \frac{{{}^{39}{C_1}}}{{{}^{52}{C_1}}} \times \frac{{{}^{39}{C_1}}}{{{}^{52}{C_1}}} \times \frac{{{}^{13}{C_1}}}{{{}^{52}{C_1}}} $

$= \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{9}{{64}}.$

The sum of the selected numbers is odd if exactly one of them is given and one is odd.

$\therefore $ Favourable number of outcomes $ = {}^{10}{C_1} \times {}^{10}{C_1}$

$\therefore $ Required probability $ = \frac{{{}^{10}{C_1} \times {}^{10}{C_1}}}{{{}^{20}{C_2}}} = \frac{{10}}{{19}}.$

$ = \frac{{1 + 35 + 4}}{{14\,.\,13\,.\,2}} = \frac{{40}}{{14\,.\,26}} = \frac{{10}}{{91}}.$

$2 $ children are chosen in ${}^4{C_2}$ ways and other $2$ persons are chosen in ${}^5{C_2}$ ways.

Hence required probability $= \frac{{{}^4{C_2} \times {}^5{C_2}}}{{{}^9{C_4}}} = \frac{{10}}{{21}}.$

$ = 1 - \frac{{{}^{13}{C_2}}}{{{}^{25}{C_2}}} = 1 - \frac{{13\,.\,12}}{{25\,.\,24}} = \frac{{37}}{{50}}.$

Number of total ways $ = {}^{12}{C_3} = 220$

Favourable number of ways to be failure $ = {}^9{C_3} = 84$

Hence required probability $ = 1 - \frac{{84}}{{220}} = \frac{{34}}{{55}}.$

Probability that both balls are black $ = \frac{{{}^8{C_2}}}{{{}^{15}{C_2}}} = \frac{4}{{15}}$

Probability that one ball is white and one is black $ = \frac{{{}^7{C_1} \times {}^8{C_1}}}{{{}^{15}{C_2}}} = \frac{8}{{15}}$.

$ = {}^4{C_1} \times {}^6{C_4} + {}^4{C_2} \times {}^6{C_3} + {}^4{C_3} \times {}^6{C_2} + {}^4{C_4} \times {}^6{C_1} + {}^6{C_5}$

$ = 60 + 120 + 60 + 6 + 6 = 252$

No. of ways in which at least one woman exist are

$ = {}^4{C_1} \times {}^6{C_4} + {}^4{C_2} \times {}^6{C_3} + {}^4{C_3} \times {}^6{C_2} + {}^4{C_4} \times {}^6{C_1} = 246$

Hence required probability $ = \frac{{246}}{{252}} = \frac{{41}}{{42}}$.

If the numbers are divisible by three then their sum of digits must be $3, 6$ or $9$

But sum $3$ is impossible. Then for sum $6$, digits are $1, 2, 3$

Number of ways $ = 3\,!$

Similarly for sum $9$, digits are $2, 3, 4$. Number of ways =$3\, !$

Thus number of favourable ways $ = 3\,! + 3\,!$

Hence required probability $ = \frac{{3\,!\, + \,3\,!}}{{4\,!}} = \frac{{12}}{{24}} = \frac{1}{2}.$

Favourable number of ways for $'I'$ come together is $9\,!$

Thus probability that $'I'$ come together

$ = \frac{{9\,!\, \times \,2\,!}}{{10\,!}} = \frac{2}{{10}} = \frac{1}{5}$.

Hence required probability $ = 1 - \frac{1}{5} = \frac{4}{5}.$

Hence required probability $ = \frac{{2(n - 1)\,\,!}}{{n\,\,!}} = \frac{2}{n}$.

$ = \frac{{{}^5{C_3} \times {}^{10}{C_8}}}{{{}^{15}{C_{11}}}} + \frac{{{}^5{C_4} \times {}^{10}{C_7}}}{{{}^{15}{C_{11}}}} + \frac{{{}^5{C_5} \times {}^{10}{C_6}}}{{{}^{15}{C_{11}}}} = \frac{{12}}{{13}}$.

{Since $A$ is a finite set, therefore every injective map from $A$ to itself is bijective also}.

$\therefore $ The required probability $ = \frac{{n\,\,!}}{{{n^n}}} = \frac{{(n - 1)\,\,!}}{{{n^{n - 1}}}}.$

Since the drawn balls can’t be of red colour, as these are $2$ in numbers.

Therefore favourable number of ways $ = {}^3{C_3} + {}^4{C_3}.$

Hence the required probability $ = \frac{5}{{84}}.$

If all the $6$ girls are together, then the number of arrangement are $7\,\,!\, \times \,6\,\,!$.

Hence required probability$ = \frac{{7\,\,!\,.\,6\,\,!}}{{12\,\,!}}$

$ = \frac{{6 \times 5 \times 4 \times 3 \times 2}}{{12 \times 11 \times 10 \times 9 \times 8}} = \frac{1}{{132}}$.

$= \frac{{2 \times 3 \times 4}}{{\left( {\frac{{9 \times 8 \times 7}}{{3 \times 2}}} \right)}} = \frac{2}{7}$.

And number of ways to form the numbers, they have even digit at both ends $ = 4 \times 3{ \times ^5}{P_3}$

$\therefore $ Probability $ = \frac{{4 \times 3{ \times ^5}{P_3}}}{{^7{P_5}}} = \frac{2}{7}$.

The number of ways to select $2$ socks out of $9$ = $^9{C_2}$

If number of ways to select both brown socks $ = {\,^5}{C_2}$

And number of ways to select both white socks $ = {\,^4}{C_2}$

$\therefore \,P$ (Both brown or white) $ = \frac{{^5{C_2} + {\,^4}{C_2}}}{{^9{C_2}}} = \frac{{16}}{{36}} = \frac{{48}}{{108}}$.

$\therefore $ The favourable cases $ = 4\,!\, \times \,3\,!$

Hence the required probability $=\frac{{  4\,!\, \times 3\,!}}{{7\,!}} = \frac{6}{{7 \times 6 \times 5}} = \frac{1}{{35}}$.

The number of favourable ways $= {}^{100}{C_1} + {}^{100}{C_3} + \,.......\, + {}^{100}{C_{99}} = {2^{100 - 1}} = {2^{99}}$

Hence required probability $ = \frac{{{2^{99}}}}{{{2^{100}}}} = \frac{1}{2}.$

Two socks drawn from the drawer will match if either both are brown of both are blue.

Therefore favourable number of cases is ${}^5{C_2} + {}^4{C_2}.$

Hence the required probability $ = \frac{{{}^5{C_2} + {}^4{C_2}}}{{{}^9{C_2}}} = \frac{4}{9}.$

Favourable number of ways $2\,.\,4\,!$

Hence required probability $ = \frac{{2\,.\,4\,!}}{{5\,!}} = \frac{2}{5}$.