Maths M.C.Q (1 Marks)

Since $A$ and $B$ are mutually exclusive,

so $P(A \cup B) = P(A) + P(B) = 0.25 + 0.05 = 0.30$.

We have to find $P\left( {\frac{A}{{A \cup B}}} \right),$ which is equal to

$P\frac{{[A \cap (A \cup B)]}}{{P(A \cup B)}} = \frac{{P(A)}}{{P(A \cup B)}} = \frac{{0.25}}{{0.30}} = \frac{5}{6}$.

$P\{ A/(A \cup B)\} = \frac{{P[A \cap (A \cup B)]}}{{P(A \cup B)}}$

$\{$Since $A \cap (A \cup B) = A \cap [A - B - A \cap B]$$ = A - A \cap B - A \cap B = A\} $

$ \Rightarrow P\left( {\frac{A}{{A \cup B}}} \right) = \frac{{P(A)}}{{P(A \cup B)}}$

$= \frac{{\frac{1}{2}}}{{\frac{1}{2} - \frac{1}{5} - \frac{1}{{10}}}} = \frac{{\frac{1}{2}}}{{\frac{6}{{10}}}} = \frac{5}{6}$

and similarly $P\left( {\frac{{A \cap B}}{{A' \cup B'}}} \right)$.

$P\left( {\frac{{A'}}{B}} \right) = 1 - \frac{1}{4} = \frac{3}{4}$ as $A,\,\,B$ are independent

$ \Rightarrow A',\,B$ are independent.

$P\left( {\frac{{B'}}{{A'}}} \right) = P(B') = 1 - \frac{1}{2} = \frac{1}{2}.$

$A \to $ Ball drawn is black; ${E_1} \to $ Bag $ I$ is chosen;

${E_2} \to $ Bag $II$ is chosen and ${E_3} \to $ Bag $III$ is chosen.

Then $P({E_1}) = ({E_2}) = P({E_3}) = \frac{1}{3},\,\,P\left( {\frac{A}{{{E_1}}}} \right) = \frac{3}{5}.$

$P\left( {\frac{A}{{{E_2}}}} \right) = \frac{1}{5},\,\,P\left( {\frac{A}{{{E_3}}}} \right) = \frac{7}{{10}}$

Required probability $ = P\left( {\frac{{{E_3}}}{A}} \right)$

$ = \frac{{P({E_3})P(A/{E_3})}}{{P({E_1})P(A/{E_1}) + P({E_2})P(A/{E_2}) + P({E_3})P(A/{E_3})}} = \frac{7}{{15}}$.

${A_1}:$ He knows the answer.

${A_2}:$ He does not know the answer.

$E:$ He gets the correct answer.

Then $P({A_1}) = \frac{9}{{10}},\,\,P({A_2}) = 1 - \frac{9}{{10}} = \frac{1}{{10}},$

$\,P{\rm{ }}\left( {\frac{E}{{{A_1}}}} \right) = 1,\,\,P\left( {\frac{E}{{{A_2}}}} \right) = \frac{1}{4}$

$\therefore $ Required probability $ = P\left( {\frac{{{A_2}}}{E}} \right) $

$= \frac{{P({A_2})P(E/{A_2})}}{{P({A_1})P(E/{A_1}) + P({A_2})P(E/{A_2})}} = \frac{1}{{37}}.$

$ \Rightarrow \,\frac{{P(B \cap A)}}{{P(A)}} = \frac{1}{2}$

$ \Rightarrow P(B \cap A) = \frac{1}{8}$

$P\left( {\frac{A}{B}} \right) = \frac{1}{4}$

$ \Rightarrow \,\frac{{P(A \cap B)}}{{P(B)}} = \frac{1}{4}$

$ \Rightarrow P(B)\, = \frac{1}{2}$

$P(A \cap B) = \frac{1}{8}\, = P(A).\,P(B)\,$

$\therefore $ Events $A$ and $B$ are independent.

Now, $P\,\left( {\frac{{A'}}{B}} \right) = \frac{{P(A' \cap B)}}{{P(B)}} $

$= \frac{{P(A')\,P(B)}}{{P(B)}} = \frac{3}{4}$

and $P\,\left( {\frac{{B'}}{{A'}}} \right) = \frac{{P(B' \cap A')}}{{P(A')}} $

$= \frac{{P(B')\,P(A')}}{{P(A')}} = \frac{1}{2}$.

Also $(X = 9) \cap (Y = 0) = 09,\,\,90,$ we have

$P(Y = 0) = \frac{{19}}{{100}}$ and $P(X = 9) \cap (Y = 0) = \frac{2}{{100}}$

Hence required probability

$ = P\left\{ {(X = 9)/(Y = 0)} \right\} = \frac{{\left\{ {P(X = 9) \cap (Y = 0)} \right\}}}{{P(Y = 0)}} = \frac{2}{{19}}$.

$\Rightarrow A \cap B=P(A)$

Also, $\mathrm{P}(\mathrm{A})<\mathrm{P}(\mathrm{B})$

Consider $\mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{B})} \neq \frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{A})}$            ............. $(1)$

Consider $P(A | B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}$          ............. $(2)$

It is known that, $\mathrm{P}(\mathrm{B}) \leq 1$ $\Rightarrow \frac{1}{P(B)} \geq 1$

$\Rightarrow \frac{P(A)}{P(B)} \geq P(A)$

From $( 2 )$, we obtain

$\Rightarrow \mathrm{P}(\mathrm{A} | \mathrm{B}) \geq \mathrm{P}(\mathrm{A})$       ............. $(3)$

$\therefore \mathrm{P}(\mathrm{A} | \mathrm{B})$ is not less than $\mathrm{P}(\mathrm{A})$.

Thus, from $( 3 )$, it can be concluded that the relation given in alternative $C$ is correct.

then $P(A) = 1/2,\,P(B) = 1/2$, $P(E/A) = 2/5\, ,$  $P(E/B) = 4/6 = 2/3$.

$P(E) = P(A)P(E/A) + P(B)P(E/B) = \frac{1}{2} \cdot \frac{2}{5} + \frac{1}{2} \cdot \frac{2}{3} = \frac{8}{{15}}$.

$P(E) = \frac{1}{6},\,\,P(E') = \frac{5}{6},\,\,P(A/E) = \frac{3}{4}$ and $P(A/E') = \frac{1}{4}$

From Baye’s theorem, $P(E/A) = \frac{{P(E).P(A/E)}}{{P(E).P(A/E) + P(E').P(A/E')}}$

$ = \frac{{\frac{1}{6} \times \frac{3}{4}}}{{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}}} = \frac{3}{8}.$

Probability of picked bag $A$ $P(A) = \frac{1}{2}$

Probability of picked bag $B$ $P(B) = \frac{1}{2}$

Probability of green ball picked from bag $A$

$ = P(A).P\left( {\frac{G}{A}} \right)$$ = \frac{1}{2} \times \frac{4}{7} = \frac{2}{7}$

Probability of green ball picked from bag $B$

$ = P(B).P\left( {\frac{G}{B}} \right)$$ = \frac{1}{2} \times \frac{3}{7} = \frac{3}{{14}}$

Total probability of green ball = $\frac{2}{7} + \frac{3}{{14}} = \frac{1}{2}$

Probability of fact that green ball is drawn from bag $B$

$P\left( {\frac{G}{B}} \right) = \frac{{P(B)P\left( {\frac{G}{B}} \right)}}{{P(A)P\left( {\frac{G}{A}} \right) + P(B)P\left( {\frac{G}{B}} \right)}} $

$= \frac{{\frac{1}{2} \times \frac{3}{7}}}{{\frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{3}{7}}} = \frac{3}{7}$.

$E_{1}:$ A speaks truth

$E_{2}:$ A speaks false

Let $\mathrm{X}$ be the event that a head appears.

$\mathrm{P}\left(E_{1}\right)=\frac{4}{5}$

$\therefore \mathrm{P}\left(E_{2}\right)=1-\mathrm{P}\left(E_{1}\right)=1-\frac{4}{5}=\frac{1}{5}$

If a coin is tossed, then it may result in either head $(H)$ or tail $(T)$.

The probability of getting a head is $\frac{1}{2}$ whether $A$ speaks truth or not.

$\therefore \mathrm{P}\left(\mathrm{X} | \mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{X} | \mathrm{E}_{2}\right)=\frac{1}{2}$

The probability that there is actually a head is given by $\mathrm{P}\left(\mathrm{E}_{1} | \mathrm{X}\right)$ 

$\mathrm{P}\left(\mathrm{E}_{1} | \mathrm{X}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{X} | \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{X} | \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{X} | \mathrm{E}_{2}\right)}$

$=\frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{4}{5} \cdot \frac{1}{2}+\frac{1}{5} \cdot \frac{1}{2}}$

$=\frac{\frac{1}{2} \cdot \frac{4}{5}}{\frac{1}{2}\left(\frac{4}{5}+\frac{1}{5}\right)}$

$=\frac{\frac{4}{5}}{1}$

$=\frac{4}{5}$

Therefore, the correct answer is $B$

The total number of observations is six.

$\therefore \mathrm{P}(\mathrm{X}=1)=\frac{3}{6}=\frac{1}{2}$

$\mathrm{P}(\mathrm{X}=2)=\frac{2}{6}=\frac{1}{3}$

$\mathrm{P}(\mathrm{X}=5)=\frac{1}{6}$

Therefore, the probability distribution is as follows.

Mean $=\mathrm{E}(\mathrm{X})=\Sigma_{p_i} x_{i}$

$=\frac{1}{2} \times 1+\frac{1}{3} \times 2+\frac{1}{6} .5$

$=\frac{1}{2}+\frac{2}{3}+\frac{5}{6}$

$=\frac{3+4+5}{6}$

$=\frac{12}{6}$

$=2$

The correct answer is $B$.

$P({\rm{rotten}}\,{\rm{egg}}) = \frac{{10}}{{100}} = \frac{1}{{10}} = q$; $n = 5,$ $r = 5$

So the probability that none egg is rotten

$ = {}^5{C_5}{\left( {\frac{9}{{10}}} \right)^5}.{\left( {\frac{1}{{10}}} \right)^0} = {\left( {\frac{9}{{10}}} \right)^5}$.

$P$(defected) $ = \frac{2}{{10}} = \frac{1}{5} = q$ and $n = 2,$ $r = 2$

Hence required probability $ = {}^n{C_r}{p^r}.{q^{n - r}}$

$ = {}^2{C_2}{\left( {\frac{4}{5}} \right)^2}.{\left( {\frac{1}{5}} \right)^0} = \frac{{16}}{{25}}.$

probability that head occurs $8$ times $ = {}^n{C_8}{\left( {\frac{1}{2}} \right)^8}{\left( {\frac{1}{2}} \right)^{n - 8}}$

$\therefore \,\,\,{}^n{C_6}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^{n - 6}} = {}^n{C_8}{\left( {\frac{1}{2}} \right)^8}{\left( {\frac{1}{2}} \right)^{n - 8}}$

${}^n{C_6} = {}^n{C_8} \Rightarrow n = 14$.

and variance $(X) = npq = 1$ $⇒ q = \frac{1}{2}$ or $p = \frac{1}{2}$ and $n = 4$

Thus $p(X \ge 1) = 1 - p(X = 0) = 1 - {}^4{C_0}{\left( {\frac{1}{2}} \right)^4} = \frac{{15}}{{16}}$.

Putting $q = 1 - p,$ we get required result.

Mean $ = np = 3 \times \frac{1}{6} = \frac{1}{2}$

Variance $ = nqp = 3 \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{{12}}.$

$\Rightarrow q = 1 - \frac{1}{3} = \frac{2}{3},$ also $n = 2.$

Therefore, variance $ = npq = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}.$

$ = {}^3{C_2}.{\left( {\frac{2}{6}} \right)^2}\left( {\frac{4}{6}} \right) + {}^3{C_3}{\left( {\frac{2}{6}} \right)^3} = \frac{2}{9} + \frac{1}{{27}} = \frac{7}{{27}}$.

$ = {}^8{C_1}{\left( {\frac{1}{{20}}} \right)^1}{\left( {\frac{{19}}{{20}}} \right)^7} + {}^8{C_0}{\left( {\frac{1}{{20}}} \right)^0}{\left( {\frac{{19}}{{20}}} \right)^8} $

$= \frac{{27}}{{20}}{\left( {\frac{{19}}{{20}}} \right)^7}$

Therefore required probability

$ = {}^5{C_3}{\left( {\frac{3}{4}} \right)^3}{\left( {\frac{1}{4}} \right)^2} + {}^5{C_4}{\left( {\frac{3}{4}} \right)^4}\left( {\frac{1}{4}} \right) + {}^5{C_5}{\left( {\frac{3}{4}} \right)^5}$

$ = \frac{{10\,.\,27}}{{{4^5}}} + \frac{{5\,.\,81}}{{{4^5}}} + \frac{{243}}{{{4^5}}} = \frac{{270 + 405 + 243}}{{1024}} = \frac{{459}}{{512}}.$

$P$ ($7$ heads) $ = {}^n{C_7}{\left( {\frac{1}{2}} \right)^7}{\left( {\frac{1}{2}} \right)^{n - 7}} = {}^n{C_7}{\left( {\frac{1}{2}} \right)^n}$

and $P$ ($9$ heads) $ = {}^n{C_9}{\left( {\frac{1}{2}} \right)^9}{\left( {\frac{1}{2}} \right)^{n - 9}} = {}^n{C_9}{\left( {\frac{1}{2}} \right)^n}$

$P$ ($7$ heads) $ = P$($9$ heads) $ \Rightarrow {}^n{C_7} = {}^n{C_9} \Rightarrow n = 16$

$\therefore \,\,\,P$($3$ heads) $ = {}^{16}{C_3}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^{16 - 3}} $

$= {}^{16}{C_3}{\left( {\frac{1}{2}} \right)^{16}} = \frac{{35}}{{{2^{12}}}}.$

For exactly $4$ right predictions

Probability $ = {}^7{C_4}{\left( {\frac{1}{3}} \right)^4}.\,{\left( {\frac{2}{3}} \right)^3} = \frac{{280}}{{{3^7}}}.$

$n = 3,$ $p = \frac{2}{6} = \frac{1}{3},$ $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$

So, mean $(\mu ) = 3 \times \frac{1}{3} = 1$

Variance $({\sigma ^2}) = 3 \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{3}$.

$ \Rightarrow 4{p^2} = {q^2} \Rightarrow 4{p^2} = {(1 - p)^2}$

$ \Rightarrow 3{p^2} + 2p - 1 = 0 \Rightarrow p = \frac{1}{3}$.

The required probability $ = {}^5{C_5}{\left( {\frac{{19}}{{20}}} \right)^5}.{\left( {\frac{1}{{20}}} \right)^0} = {\left( {\frac{{19}}{{20}}} \right)^5}.$

$p = \frac{{10}}{{100}} = \frac{1}{{10}},\,\,q = \frac{9}{{10}}$

Number of patients = $6$, Number of die cases = $3$

$\therefore $ Probability that $3$ will die

$ = {}^6{C_3}{\left( {\frac{1}{{10}}} \right)^3}{\left( {\frac{9}{{10}}} \right)^3} = 1458 \times {10^{ - 5}}$.

$\therefore$ Probability for getting success $ = \frac{2}{3}$

$\therefore$ Required probability

$ = {\,^4}{C_4}{\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^0} + {\,^4}{C_3}{\left( {\frac{2}{3}} \right)^3}\left( {\frac{1}{3}} \right)$

$ = {\left( {\frac{2}{3}} \right)^4} + 4{\left( {\frac{2}{3}} \right)^3}\left( {\frac{1}{3}} \right) = \frac{{16}}{{27}}$.

$\therefore $ Required probability $ = {\,^5}{C_3}{\left( {\frac{1}{2}} \right)^3}\,{\left( {\frac{1}{2}} \right)^2}$$ = 10.\frac{1}{{{2^5}}} = \frac{5}{{16}}.$

Probability for black ball $ = \frac{4}{6} = \frac{2}{3}$

$\therefore $ Required probability

$ = {\,^5}{C_5}{\left( {\frac{1}{3}} \right)^5}\,{\left( {\frac{2}{3}} \right)^0}\, + {\,^5}{C_4}{\left( {\frac{1}{3}} \right)^4}\left( {\frac{2}{3}} \right)$

$ = {\left( {\frac{1}{3}} \right)^4}\,\left[ {\frac{1}{3} + 5.\,\frac{2}{3}} \right]$$ = \frac{{11}}{{{3^5}}} = \frac{{11}}{{243}}$.

Mean $= np$,  Variance $ = 3 = npq,$  Mean $ = \,4 = \,np$

Now, $q = \frac{3}{4},\,\,p = \frac{1}{4}$ and $n = 16$

Probability of success $ = \,{\,^{16}}{C_6}\,{\left( {\frac{3}{4}} \right)^{10}}\,{\left( {\frac{1}{4}} \right)^6}$.

$\therefore $ $q = \frac{1}{2}$ in a toss.

Required probability ${ = ^{10}}{C_5}{\left( {\frac{1}{2}} \right)^5}{\left( {\frac{1}{2}} \right)^5} = \frac{{63}}{{256}}$.

Probability of unsuccess $q = \frac{3}{4}$

Mean $= np$

Standard deviation $= \sqrt {{\rm{Variance }}} $

$ \Rightarrow $ Variance $= 9$

$⇒ npq = 9 ⇒ n.\,\frac{1}{4}.\,\frac{3}{4} = 9 ⇒ n = 48$

Mean $ = np$ $ = \frac{1}{4} \times 48 = 12$.

Probability of inoccurrence of '$4$' $ = \frac{5}{6}$

$\therefore $ Required probability

$ = {\,^2}{C_1}\left( {\frac{1}{6}} \right)\,\left( {\frac{5}{6}} \right)\, + {\,^2}{C_2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^0} = \frac{{11}}{{36}}$.

{Here at least two heads means two heads or three heads}.

$\therefore$ $p = \frac{4}{{52}} = \frac{1}{{13}},$ $q = 1 - \frac{1}{{13}} = \frac{{12}}{{13}}$

$P(X = 1) = 2 \times \left( {\frac{1}{{13}}} \right) \times \left( {\frac{{12}}{{13}}} \right) = \frac{{24}}{{169}}$

$P(X = 2) = 2.{\left( {\frac{1}{{13}}} \right)^2}{\left( {\frac{{12}}{{13}}} \right)^0} = \frac{1}{{169}}$

Mean = $\sum\limits_{}^{} {{P_i}{X_i}} = \frac{{24}}{{169}} + \frac{2}{{169}} = \frac{{26}}{{169}} = \frac{2}{{13}}$.

$\therefore$ $P(X = x) = \frac{{^3{C_x}{ + ^7}{C_{4 - x}}}}{{^{10}{C_4}}}$

Putting $x = 1,\,2$ we have

$P(0 < x < 3)$$ = \frac{{^3{C_1}{ \times ^7}{C_3}}}{{210}} + \frac{{^3{C_2}{ \times ^7}{C_2}}}{{210}}$

$ = \frac{{3 \times 35 + 3 \times 21}}{{210}} = \frac{{105 + 63}}{{210}} = \frac{{168}}{{210}} = \frac{4}{5}$.

Probability of getting a defective bulb, $\mathrm{p}=\frac{10}{100}=\frac{1}{10}$

$\therefore q=1-p=1-\frac{1}{10}=\frac{9}{10}$

Clearly, $\mathrm{X}$ has binomial distribution with $\mathrm{n}=5$ and $\mathrm{p}=\frac{1}{10}$

$\therefore \mathrm{P}(\mathrm{X}=\mathrm{x})=^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} q^{n-x} p^{x}=^{5} \mathrm{C}_{\mathrm{x}}\left(\frac{9}{10}\right)^{5-x} \cdot\left(\frac{1}{10}\right)^{x}$

$\mathrm{P}$ (none of the bulbs is defective) $=\mathrm{P}(\mathrm{X}=0)$

$=^{5} \mathrm{C}_{0} \cdot\left(\frac{9}{10}\right)^{5}$

$=1 .\left(\frac{9}{10}\right)^{5}$

$=\left(\frac{9}{10}\right)^{5}$

The correct answer is $D$.

Probability of students who are not swimmers, $q=\frac{1}{5}$

$\therefore p=1-q=1-\frac{1}{5}=\frac{4}{5}$

Clearly, $\mathrm{X}$ has a binomial distribution with $\mathrm{n}=5$ and $\mathrm{p}=\frac{4}{5}$

$\therefore \mathrm{P}(\mathrm{X}=\mathrm{x})=^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} q^{n-\mathrm{x}} p^{x}$ $=^{5} \mathrm{C}_{\mathrm{x}} \cdot\left(\frac{1}{5}\right)^{5-x} \cdot\left(\frac{4}{5}\right)^{x}$

$\mathrm{P}$ (four students are swimmers) $=\mathrm{P}(\mathrm{X}=4)=^{5} \mathrm{C}_{4}\left(\frac{1}{5}\right) \cdot\left(\frac{4}{5}\right)^{4}$

Therefore, the correct answer is $A$.

$\therefore $ Their probability $= p \cdot p  =  {p^2}$

Now current will flow in lower part of $c$, if $c$ is closed, its probability is $p$.

Thus current will flow from $A$ to $B$ if current flows either in upper part or flow in lower part.

$\therefore$ Required probability $={p^2} + p$.

Let $E$ be the event that $x$ has disease $A =$ person is tested positive $P ( A )=\frac{2}{3} \times \frac{2}{3}+\frac{1}{3} \times \frac{1}{3}=\frac{5}{9}$ From Baye's theorem

$P\left(\frac{E}{A}\right)=\frac{\frac{2}{3} \times \frac{2}{3}}{\frac{5}{9}}=\frac{4}{5}$

For number not divisible by $4$ and not having zero can be formed as

Number of ways to form

All digits odd numbers $=5^4$

$3$ digits odd numbers $+1$ even numbers $(4$ or 8$)=5^3 .2 .4$

$2$ digits odd numbers $+2$ even numbers $(4$ or 8$)=0$

$1$ digits odd numbers $+3$ even numbers $=0$

All even numbers $=2^4$ (only $24 or $64 can be used)

$\therefore$ Required probability $=\frac{2^4}{5^4+5^3 \cdot 8+2^4}$

$=\frac{16}{1641}$

lt is given that, Ravi and Rashmi are each holding $2$ red cards and $2$ black cards (all four red and all four black cards are identical).

Let Ravi picks a Red card from Rashmi and Rashmi picks black card from Ravi and in second time also Ravi picks a red card from Rashmi and Rashmi picks

black card from Ravi, then finally Ravi have four red cards and Rashmi have four black cards, then probability of such event is

$\frac{{ }^2 C_1}{{ }^4 C_1} \times \frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^4 C_1} \times \frac{{ }^1 C_1}{{ }^5 C_1}=\frac{4}{400}=\frac{1}{100}$

Similarly, the probability if at the end Ravi have four black and Rashmi have four red cards is $\frac{1}{100}$

So, required probability

$=\frac{1}{50}=2 \%=p \leq 5 \%$

The quadratic equation

$4 x^2+b x+c =0$

has equal roots if $b^2-16 c=0\,\,b^2=2^4 c$

Now,$c$ should be chosen from the set $S=\{1,2,3, \ldots, 100\}$, such that it is a perfect square number, so

$c=1,4,9,16,25,36,49,64,81,100$

$\therefore$ number of ordered pair $(b, c)$ will be $10$.

$\text { So, required probability } =\frac{10}{100 \times 100}$

$=\frac{1}{1000}=0.001$

We have $30\,males$ and $20\,females$.

$P(\text { males }) =\frac{30}{50}=\frac{3}{5}$

$P(\text { females }) =\frac{20}{50}=\frac{2}{5}$

Required probability $=\frac{(1)+(3)}{(1)+(3)+(5)+(7)}$

$=\frac{\frac{3}{5} \times \frac{1}{2} \times \frac{4}{5}+\frac{3}{5} \times \frac{1}{2} \times \frac{1}{5}}{\frac{3}{5} \times \frac{1}{3} \times \frac{4}{5}+\frac{3}{5} \times \frac{1}{2} \times \frac{1}{5}}$

$+\frac{2}{5} \times \frac{1}{5} \times \frac{4}{5}+\frac{2}{5} \times \frac{4}{5} \times \frac{1}{5}$

$=\frac{\frac{15}{50}}{\frac{15}{50}+\frac{16}{125}}=\frac{75}{107}$

We have, $A$ is $3 \times 3$ matrix consisting of integer entries from $\{-1000,-999, \ldots 999,1000\}$

Total entries $=2001$

Total number of outcomes $=(2001)^9$

Given, $A^2=-I, A=-A^{-1}$ (which is not possible)

$A$ is diagonal matrix.

$A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$

Total number of favourable outcomes

$=(2001)^3$

Required probability

$=\frac{(2001)^3}{(2001)^9}(2001)^{-6}=(2000+1)^{-6}$

$p=(2000)^{-6}\left(1+\frac{1}{2000}\right)^{-6}$

$p=\frac{1}{2^6 \times 10^{18}}\left(1+\frac{1}{2000}\right)^{-6}$

$P < \frac{1}{2^6 \times 10^{18}} \quad\left[\because 1+\frac{1}{2000}\right.$ is decreasing $]$

$P < \frac{1}{10^{18}}$