Probability

Total $3 \times 3$ matrix using

$ \{-1,0,1\}=3^9=19683 $

Total $3 \times 3$ non-zero matrix

$ =3^9-1=19682 $

Now, skew-symmetric matrices

$ a_{11}=a_{22}=a_{33}=0 $

$a_{12}, a_{13}, a_{23}$ can be filled in $3^3=27$ ways

But all cannot be zero, so 26 ways.

$ P=\frac{26}{19682}=\frac{1}{757} $

$ \begin{aligned} &\text { 1st throw 2nd 3rd 4th 5th 6th 7th }\\ &\begin{array}{lll} 1 \rightarrow \frac{1}{6} & 6 \rightarrow \frac{1}{6} & 7 \rightarrow\left(\frac{1}{6}\right)^2 \\ 1 \rightarrow \frac{1}{6} & 1 \rightarrow \frac{1}{6} 5 \rightarrow \frac{1}{6} & 7 \rightarrow\left(\frac{1}{6}\right)^3 \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & 1 \rightarrow \frac{1}{6} 1 \rightarrow \frac{1}{6} 1 \rightarrow \frac{1}{6} 4 \rightarrow \frac{1}{6} \\ & 7 \rightarrow\left(\frac{1}{6}\right)^4 \\ & \text { Total }=\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^3 \ldots\left(\frac{1}{6}\right)^6 \end{aligned}\\ &\text { Given series is GP, } a=\frac{1}{36}, r=\frac{1}{6} \text { and } n=5\\ &P=\frac{1}{36}\left(\frac{1-\frac{1}{6^5}}{1-\frac{1}{6}}\right)=\frac{1}{30}\left(1-\frac{1}{6^5}\right) \end{aligned} $

$ \begin{aligned} &\text { Given, there are } 10 \text { coins in a box. }\\ &\begin{aligned} & 8 \text { normal }+2 \text { biased } \\ & P(\text { biased coin/appear head }) \\ & \quad=\frac{P(\text { biased coin } \cap \text { appear head })}{P(\text { appear head })} \\ & \quad=\frac{\frac{2}{10} \times\left(\frac{2}{2}\right)^6}{\frac{2}{10} \times\left(\frac{2}{2}\right)^6+\frac{8}{10} \times\left(\frac{1}{2}\right)^6} \\ & \quad=\frac{2}{2+8 \times \frac{1}{64}}=\frac{2 \times 8}{16+1}=\frac{16}{17} \end{aligned} \end{aligned} $

$ \begin{array}{ccccccc} \hline \boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}} & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{P}\left(\boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}}\right) & 0.1 & k & 0.2 & 2 k & 3 k & k \\ \hline \end{array} $

$ \begin{aligned} &\text { Total probability }\\ &\begin{aligned} & 0.1+k+0.2+2 k+3 k+k=1 \\ & \Rightarrow \quad 7 k+0.3=1 \\ & \Rightarrow \quad k=0.1 \\ & \therefore \sigma^2=\sum x^2 p(x)-\left(\sum x p(x)\right)^2 \\ & \text { Now, } \sum x^2 p(x)=4 \times(0.1)+1(k)+0(0.2) +1(2 k)+4(3 k)+9(k) \\ & =2.8 \\ & \text { and } \sum x p(x)=-0.2-0.1+0+0.2 +0.6+0.3=0.8 \\ & \Rightarrow \quad \sigma^2=2.8-(0.8)^2=2.16 \end{aligned} \end{aligned} $

Total balls $=4$

Out of 4 balls, 2 balls are white.

$E_1=$ Event that all balls white.

$E_2=$ Event that 3 balls are white.

$E_3=$ Event that 2 balls are white.

$ \because P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3} $

So, $P\left(E_1 / A\right)$

$ \begin{gathered} =\frac{P\left(A / E_1\right) \cdot P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)} \\ +P\left(A / E_3\right) P\left(E_3\right) \\ \because P\left(A / E_1\right)={ }^4 C_2 /{ }^4 C_2=1, \\ P\left(A / E_2\right)=\frac{{ }^3 C_2}{{ }^4 C_2}=1 / 2 \\ P\left(A / E_3\right)={ }^2 C_2 /{ }^4 C_2=1 / 6 \\ \therefore P\left(E_1 / A\right)=\frac{1 \times 1 / 3}{1 / 3+1 / 2 \cdot 1 / 3+1 / 6 \cdot 1 / 3} \\ =3 / 5 \end{gathered} $

If the equation $x^2+a x+b=0$ has real roots, then

$ a^2-4 b \geq 0 \Rightarrow a^2 \geq 4 b $

$\because a \in A$ and $A=\{3,4,5\}$

and $b \in B$ and $B=\{1,2,3,4\}$

When, $a=3 \Rightarrow a^2=9$

Then, possible values of $b=1,2$

When, $a=4 \Rightarrow a^2=16$

Then, possible values of $b=1,2,3,4$ and when $a=5 \Rightarrow a^2=25$

Then, possible values of $b=1,2,3,4$

Therefore, total possible cases

$ =2+4+4=10 $

and total combinations of $a$ and

$ b=3 \times 4=12 $

Hence, probability that the equation

$ x^2+a x+b=0 $

has real roots $=\frac{10}{12}=\frac{5}{6}$

$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{X}=\boldsymbol{x} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \boldsymbol{P} & 0.15 & 0.23 & k & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\ (\boldsymbol{X}=\boldsymbol{x}) & & & & & & & & \\ \hline \end{array} $

$ \begin{aligned} & \because \sum P(X=x)=1 \\ & \Rightarrow 0 \cdot 15+0 \cdot 23+k+0 \cdot 10+ \\ & 0 \cdot 20+0 \cdot 08+0 \cdot 07+0 \cdot 05=1 \\ & \Rightarrow \quad k+0.88=1 \\ & \Rightarrow \quad k=0.12 \\ & \because E=\{x / x, \text { is a prime number }\} \\ & =\{2,3,5,7\} \\ & F=\{x \mid x<4\}=\{1,2,3\} \\ & \text { Then, } E \cup F=\{1,2,3,5,7\} \\ & \text { So, } P(E \cup F)=P(X=1) \\ & +P(X=2)+P(X=3)+P(X=5)+P(X=7) \\ & =0.15+0.23+0.12+0.20+0.07 \\ & =0.77 \end{aligned} $

$\because$ Total floor $=7$,

at first floor, total persons $=5$

Then, remaining floor $=6$

Now, total number of ways in which each of the 5 persons can leave cabin at any of the 6 floors $=6^5$ (total)

And favourable number of ways, that is number of ways in which 5 persons leave at different floors $={ }^6 P_5=6$

Therefore, required probability $=\frac{{ }^6 P_5}{6^5}$

$ =\frac{\left(\frac{6!}{1!}\right)}{6^5}=\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 6 \cdot 6 \cdot 6 \cdot 6}=\frac{5}{54} $

∵ A bag contains 3 red balls, 5 black balls and 7 blue balls

$ \text { Total outcomes }={ }^{15} \mathrm{C}_3 $

∴ Probability of getting atleast two blue balls

$ \begin{aligned} & =\frac{\text { Favourable outcomes }}{\text { Total outcomes }} \\ & =\frac{{ }^7 C_2 \cdot{ }^3 C_1+{ }^7 C_2 \cdot{ }^5 C_1+{ }^7 C_3}{{ }^{15} C_3} \\ & =\frac{\frac{7 \times 6}{2} \times 3+\frac{7 \times 6}{2} \times 5+\frac{7 \times 6 \times 5}{3 \times 2}}{\frac{15 \times 14 \times 13}{3 \times 2}} \\ & =\frac{203}{455}=\frac{29}{65} \end{aligned} $

We have, two independent events in which two dice are thrown simultaneously by person A and two cards are drown at random simultaneously from a pack of 52 playing cards by a person $B$.

Consider the first event when $A$ thrown two dice.

Total possible outcomes, $n(s)=36$

Favourable outcomes when sum is prime

$ [(1,1)]\left[\begin{array}{l} (1,2) \\ (2,1) \end{array}\right] \ldots \ldots \ldots $

So, number of favourable outcome $=15 \Rightarrow P(A$ win the game $)=15 / 36$

And, now second event when $B$ drawns two cards. Total possible outcomes $={ }^{52} C_2$

$B$ win the game when $B$ gets a face card, a card having prime number.

Total face card = 12 and total card having prime number $=16$

$ \begin{aligned} \Rightarrow \text { Favourable outcomes } & =\frac{{ }^{12} C_1 \times{ }^{16} C_1}{{ }^{52} C_2} \\ & =\frac{12 \times 16}{\frac{52 \times 51}{2}}=\frac{192}{1326} \end{aligned} $

Hence, $P$ (both $A$ and $B$ win)

$ =\frac{15}{36} \times \frac{192}{1326}=\frac{40}{663} $

[ ∵ both events are independent]

Sample space when three coin tossed simultaneously {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Winning probability in each toss

$ (A \text { or } B)=\frac{5}{8} $

And, loosing probability in each toss

$ (A \text { or } B)=\frac{3}{8} $

∴ Required probability

$ \begin{aligned} & =\frac{5}{8} \times \frac{3}{8}+\frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{3}{8} \\ & \quad+\frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{3}{8}+\ldots \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{5}{8} \times \frac{3}{8}+\left(\frac{5}{8}\right)^3 \times \frac{3}{8}+\left(\frac{5}{8}\right)^5 \times \frac{3}{8}+\ldots \\ & =\frac{5}{8} \times \frac{3}{8}\left[1+\left(\frac{5}{8}\right)^2+\left(\frac{5}{8}\right)^4+\ldots\right] \\ & =\frac{15}{64} \times \frac{1}{1-\frac{25}{64}} \end{aligned}\\ &\text { [ ∵ Sum of infinite, GP }=\frac{a}{1-r} \text { ] }\\ &=\frac{15}{64} \times \frac{64}{39}=\frac{15}{39} \end{aligned} $

Number of ways to select two cards

$ \begin{aligned} & ={ }^{52} C_2 \\ & =\frac{52 \times 51}{2}=1326 \end{aligned} $

Number of ways of getting a face card and a spade card other than the face card

$ \begin{aligned} & ={ }^{12} C_1 \times{ }^{10} C_1 \\ & =12 \times 10=120 \end{aligned} $

$ \begin{aligned} &\text { Required probability }\\ &=\frac{120}{1326}=\frac{20}{221} \end{aligned} $

$ \begin{aligned} & \text { Total outcomes }=6 \times 6 \times 6=216 \\ & \text { Favourable outcomes }=6 \times 5 \times 4=120 \\ & \text { Required probability }=\frac{120}{216}=\frac{5}{9} \end{aligned} $

Chance of $A, B, C$ in getting through the test $=1: 2: 3$

Chance of $A: P(A)=\frac{1}{6}$

Chance of $B: P(B)=\frac{2}{6}$

Chance of $C: P(C)=\frac{3}{6}$

If $I$ denotes the facing interview successfully, then

$ \begin{aligned} P(I \mid A) & =0.8 \\ P(I \mid B) & =0.7 \\ P(I \mid C) & =0.6 \\ P(I)= & P(A) P(I \mid A)+P(B) P(I \mid B)+P(C) P(I \mid C) \\ = & \frac{1}{6} \times 0.8+\frac{2}{6} \times 0.7+\frac{3}{6} \times 0.6=\frac{4}{6}=\frac{2}{3} \\ \text { Now, } P(A \mid I)= & \frac{P(A) P(I \mid A)}{P(I)}=\frac{0.8 \times \frac{1}{6}}{\frac{4}{6}}=\frac{1}{5} \end{aligned} $

Let $X$ be the number of spades, then $X=0,1,2$

$ \begin{gathered} P(X=0)=\frac{39}{52} \times \frac{39}{52}=\left(\frac{39}{52}\right)^2 \\ P(X=1)=\frac{13}{52} \times \frac{39}{52} \times 2=\frac{1014}{(52)^2} \\ P(X=2)=\frac{13}{(52)} \times \frac{13}{(52)}=\frac{169}{(52)^2} \\ \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ =\left(0^2 \times \frac{39^2}{(52)^2}+1^2 \times \frac{1014}{(52)^2}+2^2 \times \frac{169}{(52)^2}\right) \\ \quad-\left(0 \times \frac{39^2}{(52)^2}+1 \times \frac{1014}{(52)^2}+2 \times \frac{169}{(52)^2}\right) \\ =\frac{1690}{52^2}-\frac{(1352)^2}{(52)^4}=\frac{1014}{52^2}=\frac{3}{8} \end{gathered} $

Given, $P(A \cup B)=P(A \cap B)$

We know,

$ \begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(A)+P(B)=2 P(A \cap B) \neq 1 \end{aligned} $

Hence, option (d) is not true.

Let us consider, $A, B, C$, where $C$ is between $A$ and $B$. Now choice for $C$ is ${ }^4 C_1$ and $A, B$ interchange their placer in 2 ways.

Now, $A, B, C$ and other 3 students stand in a row in 4! ways.

So, total number of ways to stand in a row such that two $A, B$ and separated by one students in between them is

$ 4 \times 2 \times 4! $

Total number of ways to stand without restriction $=6!$

$ \begin{aligned} \text { Probability } & =\frac{4 \times 2 \times 4!}{6!}=\frac{4 \times 2 \times 4!}{6 \times 5 \times 4!} \\ & =\frac{8}{30}=\frac{4}{15} \end{aligned} $

Let ' $E$ ' be the event of any one cutting a spade in one out first, and ' $S$ ' be the sample space.

Then, $n(E)={ }^{13} C_1$ and $n(S)={ }^{52} C_1$

Now, $p(E)=\frac{n(E)}{n(S)}=\frac{{ }^{13} C_1}{{ }^{52} C_1}=\frac{13}{52}=\frac{1}{4}$

$ \therefore \quad q=1-p=1-\frac{1}{4}=\frac{3}{4} $

Now, probability of $A$ wins the game is

$ \begin{aligned} & =P+q^4 P+q^8 P+\ldots \infty \\ & =\frac{P}{1-q^4}=\frac{1 / 4}{1-\left(\frac{3}{4}\right)^4}=\frac{64}{256-81}=\frac{64}{175} \end{aligned} $

Since, Their are total two bad eggs. So, while drawing 3 eggs we can draw 0 bad eggs, or 1 bad eggs or 2 bad eggs.

Let ' $x$ ' be the random variable denoting number of bad eggs that can be drawn in each draw.

Thus, $P(x=0)$

$ =\frac{{ }^2 C_0 \times{ }^{10} C_3}{{ }^{12} C_3}=\frac{120}{220}=\frac{6}{11} $

Similarly, $P(x=1)=\frac{{ }^2 C_1 \times{ }^{10} C_2}{{ }^{12} C_3}=\frac{9}{22}$

Now,

$ \begin{array}{cccc} \hline \boldsymbol{x} & \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{x} \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{x}^2 \boldsymbol{P}_{\boldsymbol{i}} \\ \hline 0 & 6 / 11 & 0 & 0 \\ \hline 1 & 9 / 22 & 9 / 22 & 9 / 22 \\ \hline 2 & 1 / 22 & 1 / 11 & 2 / 11 \\ \hline \end{array} $

$ \begin{aligned} &\begin{aligned} \therefore \mu & =\sum_{i=0,1,2} x P_i=0+\frac{9}{22}+\frac{1}{11}=\frac{11}{22}=\frac{1}{2} \\ \sigma^2 & =\sum x^2 P_i-\mu^2 \\ & =\frac{9}{22}+\frac{2}{11}-\frac{1}{4}=\frac{18+8-11}{44}=\frac{15}{44} \end{aligned}\\ &\text { Thus, variance is } \frac{15}{44} \text {. } \end{aligned} $

$n=6$

$P=$ Probability of a question marked with correct answer $=1 / 2$

$ \begin{aligned} q & =\frac{1}{2} \\ P(X \geq 4) & ={ }^6 C_4\left(\frac{1}{2}\right)^6+{ }^6 C_5\left(\frac{1}{2}\right)^6+{ }^6 C_6\left(\frac{1}{2}\right)^6 \\ & =\frac{1}{64}(15+6+1)=\frac{22}{64}=\frac{11}{32} \end{aligned} $

To find the value of the constant $c$, we need to ensure that the probability distribution function sums to 1. This means:

Write the sum of the probability mass function:

$\sum\limits_{x=1}^{\infty} c\left(\frac{2}{3}\right)^x = 1.$

Factor out the constant $c$:

$c \sum\limits_{x=1}^{\infty} \left(\frac{2}{3}\right)^x = 1.$

Recognize that the series

$\sum\limits_{x=1}^{\infty} \left(\frac{2}{3}\right)^x$

is a geometric series with first term

$a = \left(\frac{2}{3}\right)$

and common ratio

$r = \frac{2}{3}.$

The sum of a geometric series starting at $x=1$ is given by:

$\sum\limits_{x=1}^{\infty} a\,r^{x-1} = \frac{a}{1-r}.$

However, since our first term is already $\left(\frac{2}{3}\right)^1$, we have:

$\sum\limits_{x=1}^{\infty} \left(\frac{2}{3}\right)^x = \frac{\frac{2}{3}}{1 - \frac{2}{3}}.$

Simplify the series sum:

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

Substitute back into the equation:

$c \cdot 2 = 1.$

Solve for $c$:

$c = \frac{1}{2}.$

Thus, the value of $c$ is $\frac{1}{2}$, which corresponds to Option C.

In a non-leap year, we need to determine the probability of having either 53 Sundays, 53 Tuesdays, or 53 Thursdays.

Calculation

A non-leap year consists of 365 days.

Since each week contains 7 days, a year typically has $\frac{365}{7} = 52$ full weeks with 1 extra day.

The extra day determines whether one of the days of the week occurs 53 times. The extra day can be any day of the week:

Sample Space (S): {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Thus, the probability sample space $n(S)$ is 7.

Event A: {Sunday, Tuesday, Thursday}

Since Sunday, Tuesday, and Thursday can each be this extra day, we have:

Number of favorable outcomes $n(A) = 3$ (Sunday, Tuesday, Thursday)

Using the probability formula, the probability of Event A is:

$ P(A) = \frac{n(A)}{n(S)} = \frac{3}{7} $

Thus, the probability of having either 53 Sundays, 53 Tuesdays, or 53 Thursdays in a non-leap year is $\frac{3}{7}$.

Wrong question

Given the problem statement that $P(A) + P(B) = 2 P(A \cap B)$, we can deduce several points based on probability theory.

First, recall the formula for the probability of the union of two events:

$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $

From the given condition, substitute the information provided:

$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A \cap B) $

This simplifies to:

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

This equation implies that the probability of the intersection of events $A$ and $B$ is equal to the probability of their union. For this statement to hold true, there are specific conditions that must be satisfied:

The events $A$ and $B$ are such that the probability of their overlap (intersection) is the entire probability of the union, meaning that the events are not just overlapping — they are covering the entire probability space available to them for their union. This can only happen under specific circumstances, which include:

Both events might be equivalent, which implies $P(A) = P(B)$. This means both events have the same probability and overlap completely.

Thus, from the given condition, $P(A) = P(B)$ is a valid conclusion.

S = {1, 2, 3, .......... 2022}

HCF (n, 2022) = 1

$\Rightarrow$ n and 2022 have no common factor

Total elements = 2022

2022 = 2 $\times$ 3 $\times$ 337

M : numbers divisible by 2.

{2, 4, 6, ........, 2022}$\,\,\,\,$ n(M) = 1011

N : numbers divisible by 3.

{3, 6, 9, ........, 2022}$\,\,\,\,$ n(N) = 674

L : numbers divisible by 6.

{6, 12, 18, ........, 2022}$\,\,\,\,$ n(L) = 337

n(M $\cup$ N) = n(M) + n(N) $-$ n(L)

= 1011 + 674 $-$ 337

= 1348

0 = Number divisible by 337 but not in M $\cup$ N

{337, 1685}

Number divisible by 2, 3 or 337

= 1348 + 2 = 1350

Required probability $ = {{2022 - 1350} \over {2022}}$

$ = {{672} \over {2022}}$

$ = {{112} \over {337}}$

$P(A/B) = {1 \over 7} \Rightarrow {{P(A \cap B)} \over {P(B)}} = {1 \over 7}$

$ \Rightarrow P(B) = {7 \over 9}$

$P(B/A) = {2 \over 5} \Rightarrow {{P(A \cap B)} \over {P(A)}} = {2 \over 5}$

$P(A) = {5 \over 2}\,.\,{1 \over 9} = {5 \over {18}}$

$S2:P(A' \cap B') = {1 \over {18}}$

$S1:$ and $P(A' \cup B) = {1 \over 9} + {6 \over 9} + {1 \over {18}} = {5 \over 6}.$

P (Female) $ = {{60} \over {100}} = {3 \over 5}$

P (Male) $ = {2 \over 5}$

P (Female/Qualified) $ = {{40} \over {60}} = {2 \over 3}$

P (Male/qualified) $ = {{20} \over {60}} = {1 \over 3}$

Mean $ = np = 16$

Variance $ = npq = 8$

$ \Rightarrow q = p = {1 \over 2}$ and $n = 32$

$P(x > n - 3) = p(x = n - 2) + p(x = n - 1) + p(x = n)$

$ = \left( {{}^{32}{C_2} + {}^{32}{C_1} + {}^{32}{C_0}} \right)\,.\,{1 \over {{2^n}}}$

$ = {{529} \over {{2^n}}}$

Let P(a prime number) = $\alpha$

P(a composite number) = $\beta$

and P(1) = $\gamma$

$\because$ $3\alpha = 6\beta = 2\gamma = k$ (say)

and $3\alpha + 2\beta + \gamma = 1$

$ \Rightarrow k + {k \over 3} + {k \over 2} = 1 \Rightarrow k = {6 \over {11}}$

Mean = np where n = 2

and p = probability of getting perfect square

$ = P(1) + P(4) = {k \over 2} + {k \over 6} = {4 \over {11}}$

So, mean $ = 2\,.\,\left( {{4 \over {11}}} \right) = {8 \over {11}}$

Among the 5 digit numbers,

First number divisible by 7 is 10003 and last is 99995.

$\Rightarrow$ Number of numbers divisible by 7.

$ = {{99995 - 10003} \over 7} + 1$

$ = 12857$

First number divisible by 35 is 10010 and last is 99995.

$\Rightarrow$ Number of numbers divisible by 35

$ = {{99995 - 10010} \over {35}} + 1$

$ = 2572$

Hence number of number divisible by 7 but not by 5

$ = 12857 - 2572$

$ = 10285$

$9P. = {{10285} \over {90000}} \times 9$

$ = 1.0285$

Mean $ = 4 = \mu = np$

Variance $ = {\sigma ^2} = np(1 - P) = {4 \over 3}$

$4(1 - P) = {4 \over 3}$

$P = {2 \over 3}$

$n \times {2 \over 3} = 4$

$n = 6$

$P(X = k) = {}^n{C_k}\,{P^k}{(1 - P)^{n - k}}$

$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$ = {}^6{C_0}{P^0}{(1 - P)^6} + {}^6{C_1}{P^1}{(1 - P)^5} + {}^6{C_2}{P^2}{(1 - P)^4}$

$ = {}^6{C_0}{\left( {{1 \over 3}} \right)^6} + {}^6{C_1}\left( {{2 \over 3}} \right){\left( {{1 \over 3}} \right)^5} + {}^6{C_2}{\left( {{2 \over 3}} \right)^2}{\left( {{1 \over 3}} \right)^4}$

$ = {\left( {{1 \over 3}} \right)^6}[1 + 12 + 60] = {{73} \over {{3^6}}}$

$54P\,(X \le 2) = {{73} \over {{3^6}}} \times 54 = {{146} \over {27}}$

Given, mean $ = np = \alpha $.

and variance $ = npq = {\alpha \over 3}$

$ \Rightarrow q = {1 \over 3}$ and $p = {2 \over 3}$

$P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}$

$ \Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}$

$ \Rightarrow n = 6$

$P(X = 4\,\mathrm{or}\,5) = {}^6{C_4}\,.\,{\left( {{2 \over 3}} \right)^4}\,.\,{\left( {{1 \over 3}} \right)^2} + {}^6{C_5}\,.\,{\left( {{2 \over 5}} \right)^5}\,.\,{1 \over 3}$

$ = {{16} \over {27}}$

$0 \le {{2 + 3P} \over 6} \le 1 \Rightarrow P \in \left[ { - {2 \over 3},{4 \over 3}} \right]$

$0 \le {{2 - P} \over 8} \le 1 \Rightarrow P \in [ - 6,2]$

$0 \le {{1 - P} \over 2} \le 1 \Rightarrow P \in [ - 1,1]$

$0 < P({E_1}) + P({E_2}) + P({E_3}) \le 1$

$0 < {{13} \over {12}} - {P \over 8} \le 1$

$P \in \left[ {{2 \over 3},{{26} \over 3}} \right]$

Taking intersection of all

$P \in \left[ {{2 \over 3},1} \right)$

${P_1} + {P_2} = {5 \over 3}$

$P(A) = {1 \over 3},\,P(B) = {1 \over 5}$ and $P\left( {A \cup B} \right) = {1 \over 2}$

$\therefore$ $P\left( {A \cap B} \right) = {1 \over 3} + {1 \over 5} - {1 \over 2} = {1 \over {30}}$

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

$ = {{{9 \over {30}}} \over {{4 \over 5}}} + {{{5 \over {30}}} \over {{2 \over 3}}} = {5 \over 8}$

Given,

$n = 5$

and $P(X = 0) = P(X = 1)$

We know, $p(x = r) = {}^n{C_r}\,.\,{p^r}\,.\,{q^{n - r}}$

where $p + q = 1$

$\therefore$ $P(X = 0) = P(X = 1)$

$ \Rightarrow {}^5{C_0}\,.\,{p^0}\,.\,{q^5} = {}^5{C_1}\,.\,{p^1}\,.\,{q^4}$

$ \Rightarrow 1\,.\,1\,.\,{(1 - p)^5} = 5\,.\,p\,.\,{(1 - p)^4}$

$ \Rightarrow 1 - p = 5p$

$ \Rightarrow 6p = 1$

$ \Rightarrow p = {1 \over 6}$

$\therefore$ $q = 1 - p = 1 - {1 \over 6} = {5 \over 6}$

Now, ${{P(X = 2)} \over {P(X = 3)}} = {{{}^5{C_2}\,.\,{p^2}\,.\,{q^3}} \over {{}^5{C_3}\,.\,{p^3}\,.\,{q^2}}}$

$ = {{10\,.\,q} \over {10\,.\,p}}$

$ = {5 \over 6} \times {6 \over 1}$

$ = 5$

First 10 prime numbers are

={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Let A is a 2 $\times$ 2 matrix,

$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$

Given that matrix A is singular.

$\therefore$ | A | = 0

$ \Rightarrow \left| {\matrix{ a & b \cr c & d \cr } } \right| = 0$

$ \Rightarrow ad = bc$

Case I :

ad = bc condition satisfy when a = b = c = d.

For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.

Now there are 10 prime numbers.

We can choose any one of the 10 prime number in ${}^{10}C_{1}$ = 10 ways and put them in the four positions of the matrix and matrix will be singular.

$\therefore$ In this case, total favorable case = 10

Case 2 :

ad = bc condition satisfies when

(1)

a = 2, d - 3 then

(a) b = 2, c = 3

(b) b = 3, c = 2

or

a = 3, d = 2 then

(a) b = 2, c = 3

(b) b = 3, c = 2

So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.

Two different values of a and d can be chosen from 10 prime numbers = ${}^{10}C_{2}$ ways

And for each combination of a and d there are 4 possible values of b and c.

$\therefore$ Total possible values = ${}^{10}C_{2}$ $\times$ 4

From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition

= 10 + ${}^{10}C_{2}$ $\times$ 4

For sample space,

Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = ${}^{10}C_{1}$ ways

Similarly,

For element b of matrix A = ${}^{10}C_{1}$ ways

For element c of matrix A = ${}^{10}C_{1}$ ways

For element d of matrix A = ${}^{10}C_{1}$ ways

$\therefore$ Sample space = ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ = 10⁴

$\therefore$ Probability $ = {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}$

$ = {{10 + 180} \over {{{10}^4}}}$

$ = {{190} \over {{{10}^4}}}$

$ = {{19} \over {{{10}^3}}}$

The required probability

$ = {\text{Area of Region PQCAP} \over \text{Area of Region ABCA}}$

$ = {{{1 \over 2} \times 8 \times 6 - {1 \over 2} \times 2 \times 4} \over {{1 \over 2} \times 8 \times 6}}$

$ = {5 \over 6}$

Given P(X = 3) = 5P(X = 4) and n = 7

$ \Rightarrow {}^7{C_3}{p^3}{q^4} = 5\,.\, \Rightarrow {}^7{C_4}{p^4}{q^3}$

$ \Rightarrow q = 5p$ and also $p + q = 1$

$ \Rightarrow p = {1 \over 6}$ and $q = {5 \over 6}$

Mean $ = {7 \over 6}$ and variance $ = {{35} \over {36}}$

Mean + Variance $ = {7 \over 6} + {{35} \over {36}} = {{77} \over {36}}$

Coin is tossed 5 times, so n = 5

Let, p = probability of getting heads

q = probability of getting tails.

$\therefore$ p + q = 1 ...... (1)

$\therefore$ Probability of getting 4 heads

= ⁵C₄ . p⁴ . q

And probability of getting 5 heads

= ⁵C₅ . p⁵

Given, ⁵C₄ . p⁴ . q = ⁵C₅ . p⁵

$\Rightarrow$ 5q = p ....... (2)

From equation (1) and (2), we get,

5q + q = 1

$\Rightarrow$ 6q = 1

$\Rightarrow$ q = ${1 \over 6}$

$\therefore$ p = 1 $-$ ${1 \over 6}$ = ${5 \over 6}$

Now, probability of getting atmost two heads

= p (x = 0) + p (x = 1) + p (x = 2)

p (x = 0) = Getting zero head in 5 trials

= ⁵C₀ . p⁰ . q⁵

p (x = 1) = Getting one head in 5 trials

= ⁵C₁ . p¹ . q⁴

p (x = 2) = Getting two heads in 5 trials

= ⁵C₂ . p² . q³

= ⁵C₀ . q⁵ + ⁵C₁ . pq⁴ + ⁵C₂ . p²q³

$ = {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}$

$ = {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}$ = ${{46} \over {{6^4}}}$

There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)

So, required probability

$ = 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)$

$ = {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}$

$ = {{13} \over {{2^{12}}}}$