Probability

$P({w_1}) + {{P({w_1})} \over 2} + {{P({w_1})} \over {{2^2}}}\, + \,..... = 1$

$\therefore$ $P({w_1}) = {1 \over 2}$

Hence, $P({w_n}) = {1 \over {{2^n}}}$

Every number except 1, 2, 3, 4, 6 is representable in the form

$2k + 3l$ where $k,l \in N$.

$\therefore$ $P(B) = 1 - P({w_1}) - P({w_2}) - P({w_3}) - P({w_4}) - P({w_6})$

$ = {3 \over {64}}$

We are given a biased coin with the probability of getting heads (H) as $\frac{1}{3}$ and thus, the probability of getting tails (T) is $\frac{2}{3}$ (since the total probability of getting either heads or tails should sum to 1).

The experiment involves tossing this coin repeatedly until we obtain two identical outcomes in a row. We are interested in the case where the experiment ends with two consecutive heads.

Let's consider two different scenarios for the sequences of coin tosses :

1. Sequences that begin with a head and end with HH (like HH, HTHH, HTHTHH, and so on).
2. Sequences that begin with a tail and end with HH (like THH, THTHH, THTHTHH, and so on).

Both types of sequences will result in the experiment ending, so we need to consider both in our final probability calculation.

Scenario 1 : Sequences beginning with H

The simplest sequence is getting HH right away. The probability of this event is $(\frac{1}{3})^2 = \frac{1}{9}$.

The next sequence is getting a head, then a tail, and then HH (i.e., HTHH). The probability of this sequence is $\frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{81}$.

We can see a pattern in these sequences. Each sequence in this scenario can be described as getting one head, followed by some number of "tail, head" pairs, and ending with HH.

Hence, the sequences in this scenario form a geometric series, where the ratio between consecutive terms is $r = P(HT) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$.

The sum of an infinite geometric series is given by the formula $\frac{a}{1 - r}$, where $a$ is the first term of the series and $r$ is the common ratio.

In this scenario, the first term is the probability of getting HH immediately, which is $\frac{1}{9}$. So, the sum of this series is :

$S_1 = \frac{1/9}{1 - 2/9} = \frac{1/9}{7/9} = \frac{1}{7}.$

Scenario 2 : Sequences beginning with T

In this scenario, we first get a tail, and then we follow the same pattern as in the first scenario.

The simplest sequence is THH, with a probability of $\frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{27}$.

The next sequence is getting a tail, a head, another tail, and then HH (i.e., THTHH). The probability of this sequence is $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{4}{243}$.

As before, these sequences form a geometric series, with the ratio between consecutive terms is $r = P(TH) = \frac{2}{9}$.

In this scenario, the first term is the probability of getting THH, which is $\frac{2}{27}$. So, the sum of this series is :

$S_2 = \frac{2/27}{1 - 2/9} = \frac{2/27}{7/9} = \frac{2}{21}.$

Finally, to calculate the total probability of the experiment ending with two consecutive heads, we need to sum up the probabilities calculated in both scenarios:

$P(\text{HH}) = S_1 + S_2 = \frac{1}{7} + \frac{2}{21} = \frac{5}{21}.$

So, the answer to the problem is $\frac{5}{21}$ (Option B).

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.

Let $P$ be the probability of getting an even number. Then, by hypothesis the probability of getting an odd number be $3 P$.

Since, event of getting an odd number and an even number are mutually exclusive and exhaustive.

$\therefore \quad 3 P+P=1 \Rightarrow P=\frac{1}{4}$

Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$. If the die is thrown twice, then the sum of the numbers in two throws is odd if one number is odd and another is even.

$\therefore$ Required probability

$=\frac{1}{4} \times \frac{3}{4}+\frac{3}{4} \times \frac{1}{4}=\frac{6}{16}=\frac{3}{8}$

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}}}$

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