Probability

The expected value of a discrete random variable is given by the formula :

$ E(X)=\sum X_i \cdot P\left(X_i\right) $

Using the data from the table :

$ E(X)=\left(4 k \cdot \frac{2}{15}\right)+\left(\frac{30 k}{7} \cdot \frac{1}{15}\right)+\left(\frac{32 k}{7} \cdot \frac{2}{15}\right)+\left(\frac{34 k}{7} \cdot \frac{1}{5}\right)+\left(\frac{36 k}{7} \cdot \frac{1}{15}\right)+\left(\frac{40 k}{7} \cdot \frac{1}{5}\right)+\left(6 k \cdot \frac{1}{15}\right) $

Take $\frac{k}{15}$ factor common

$ \begin{gathered} E(X)=\frac{k}{15}\left[8+\frac{30}{7}+\frac{64}{7}+\frac{102}{7}+\frac{36}{7}+\frac{76}{7}+\frac{120}{7}+6\right] \\ E(X)=\frac{k}{15}\left[14+\frac{30+64+102+36+76+120}{7}\right] \\ E(X)=\frac{k}{15}\left[14+\frac{428}{7}\right] \end{gathered} $

$ E(X)=\frac{k}{15}\left[\frac{98+428}{7}\right]=\frac{k}{15} \cdot \frac{526}{7}=\frac{526 k}{105} $

Given that $E(X)=\frac{263}{15}$ :

$ \begin{gathered} \frac{263}{15}=\frac{526 k}{105} \\ 263=\frac{526 k}{7} \\ k=\frac{263 \times 7}{526}=\frac{1 \times 7}{2}=\frac{7}{2}=3.5 \end{gathered} $

Identify $X$ values where $X<20$

Substitute $k=\frac{7}{2}$ into each $X$ :

$ \begin{array}{ll} \text { } X_1=4 \times \frac{7}{2}=14(<20) & X_4=\frac{34}{7} \times \frac{7}{2}=17(<20) \\ \text { } X_2=\frac{30}{7} \times \frac{7}{2}=15(<20) & X_5=\frac{36}{7} \times \frac{7}{2}=18(<20) \\ \text { } X_3=\frac{32}{7} \times \frac{7}{2}=16(<20) & X_6=\frac{38}{7} \times \frac{7}{2}=19(<20) \\ \text { } & X_7=\frac{40}{7} \times \frac{7}{2}=20(\text { Not }<20) \\ \text { }& X_8=6 \times \frac{7}{2}=21(\text { Not }<20) \end{array} $

Calculate $P(X<20)$ :

Sum the probabilities for $X=14,15,16,17,18$, and 19:

$ P(X<20)=P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19) $

$\Rightarrow $ $ P(X<20)=\frac{2}{15}+\frac{1}{15}+\frac{2}{15}+\frac{3}{15}+\frac{1}{15}+\frac{2}{15} $

$\Rightarrow $ $ P(X<20)=\frac{11}{15} $

Given Terms

• Total Balls ( $n$ ) : 10

• Red Balls ( $k$ ) : Variable from 0 to 10 .

• Black Balls : $(10-k)$.

• Sample Drawn ( $r$ ) : 3 balls, without replacement.

• Observed Event (A) : All 3 balls drawn are black.

• Target Event : Find the probability that the bag contained exactly $\mathbf{1}$ red and $\mathbf{9}$ black balls (which means $k=1$ ).

Let $E_k$ be the event that the bag contains $k$ red balls. Since $k$ can be any integer from 0 to 10 , there are 11 possible compositions. Assuming they are equally likely:

$ P\left(E_k\right)=\frac{1}{11} $

The event $A$ is drawing 3 black balls. The probability of this happening depends on how many black balls $(10-k)$ were in the bag :

$ P\left(A \mid E_k\right)=\frac{{ }^{10-k} C_3}{{ }^{10} C_3} $

Note : For this to be possible, there must be at least 3 black balls, so $10-k \geq 3$, meaning $k \leq 7$.

Using Bayes' Theorem

We need $P\left(E_1 \mid A\right)$, which is the probability of having 1 red ball given that 3 black balls were drawn :

$ P\left(E_1 \mid A\right)=\frac{P\left(E_1\right) \cdot P\left(A \mid E_1\right)}{\sum\limits_{k=0}^7 P\left(E_k\right) \cdot P\left(A \mid E_k\right)} $

Since $P\left(E_k\right)$ is constant $\left(\frac{1}{11}\right)$ for all $k$, it cancels out :

$ P\left(E_1 \mid A\right)=\frac{{ }^{10-1} C_3}{{ }^{10} C_3+{ }^9 C_3+\cdots+{ }^3 C_3} $

denominator ${ }^{10} C_3+{ }^9 C_3+\cdots+{ }^3 C_3$ can be written as ${ }^3 C_3+{ }^4 C_3+{ }^5 C_3 \cdots{ }^{10} C_3$

using identity ${ }^k C_k+{ }^{k+1} C_k+{ }^{k+2} C_k+\ldots{ }^n C_k={ }^{n+1} C_{k+1}$

so, denominator become $={ }^{11} C_4=\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}=330$

Numerator : ${ }^{10-1} C_3={ }^9 C_3=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84$

$P\left(E_1 \mid A\right)=\frac{84}{330}=\frac{14}{55}$

total number of defective bulbs $n(D)=10$

total number of non defective bulbs $n(\bar{D})=90$

total bulbs $=n(D)+n(\bar{D})=100$

probability of picking a defective bulbs: $p=\frac{10}{100}=\frac{1}{10}$

probability of picking a non-defective bulbs $q=\frac{90}{100}=\frac{9}{10}$

number of trials $n=8$ bulbs are selected

let $X$ be the event of getting atleast 7 defective bulbs

$ \begin{aligned} & P(X)=P(X=7)+P(X=8) \\ & ={ }^8 C_7 p^7 \cdot q^1+{ }^8 C_8 p^8 q^0 \\ & =8 \cdot\left(\frac{1}{10}\right)^7 \cdot \frac{9}{10}+1 \cdot\left(\frac{1}{10}\right)^8 \end{aligned} $

$P(X)=\left(\frac{1}{10}\right)^8[72+1]=\frac{73}{10^8}$

Bag A : 9 white, 8 black (Total $=17$ )

Bag B : 6 white, 4 black (Total $=10$ )

One ball is moved from Bag B to Bag A. There are two mutually exclusive events for this transfer:

Case 1 : The ball moved is white

The probability of choosing a white ball from Bag B is:

$ P\left(W_B\right)=\frac{6}{10} $

After adding this white ball to Bag A, Bag A now contains 10 white and 8 black balls ( 18 total). The probability of then drawing a white ball from Bag A is :

$ P\left(W_A \mid W_B\right)=\frac{10}{18} $

Case 2: The ball moved is black

The probability of choosing a black ball from Bag B is :

$ P\left(B_B\right)=\frac{4}{10} $

After adding this black ball to Bag A, Bag A now contains 9 white and 9 black balls ( 18 total). The probability of then drawing a white ball from Bag A is :

$ P\left(W_A \mid B_B\right)=\frac{9}{18} $

Using total probability theorem

$ P\left(W_A\right)=P\left(W_B\right) \cdot P\left(W_A \mid W_B\right)+P\left(B_B\right) \cdot P\left(W_A \mid B_B\right) $

$\Rightarrow $$P\left(W_A\right)=\left(\frac{6}{10} \times \frac{10}{18}\right)+\left(\frac{4}{10} \times \frac{9}{18}\right)$

$ P\left(W_A\right)=\frac{60}{180}+\frac{36}{180}=\frac{96}{180}=\frac{8}{15} $

It is given that $P\left(W_A\right)=\frac{p}{q}$.

$ \begin{aligned} & \operatorname{gcd}(8,15)=1 \\\\ & \text { So } p=8 \& q=15 \end{aligned} $

$p+q=8+15=23$

Total ways to choose two distinct numbers from $1,2,\dots,50$ is

$ \binom{50}{2}=1225. $

The product $ab$ is not divisible by $3$ only if neither $a$ nor $b$ is divisible by $3$.

Multiples of $3$ in $1$ to $50$: $\left\lfloor \frac{50}{3}\right\rfloor=16$.

So numbers not divisible by $3$: $50-16=34$.

Unfavorable pairs:

$ \binom{34}{2}=\frac{34\cdot 33}{2}=561. $

Therefore favorable pairs:

$ 1225-561=664, $

and the required probability is

$ \frac{664}{1225}. $

Answer: $\boxed{\frac{664}{1225}}$ (Option C).

$ \begin{aligned} & 2 K+K+3 K+2 K^2+2 K+K^2+K+7 K^2=1 \\ & (K+1)(10 K-1)=0 \\ & K=-1 \text { or } K=\frac{1}{10} \\ & P(3 < x \leq 6)=\frac{2}{100}+\frac{2}{10}+\frac{1}{100}+\frac{1}{10}=0.33 \end{aligned} $

$ \begin{aligned} & \text { Mean }=8 \\ & \frac{2+4+10+x+y+12+14}{7}=8 \\ & \Rightarrow x+y=14 \\ & \text { variance }=16 \\ & 16=\frac{2^2+4^2+10^2+x^2+y^2+12^2+14^2+y^2}{7}-8^2 \\ & \Rightarrow x^2+y^2=100 \\ & \therefore x=8, y=6 \\ & \text { New set }:\{1,2,3,4,5,6\} \\ & \text { Required probability }=1-P(\text { both number are } \geq 4) \\ & =1-\frac{{ }^3 C_2}{{ }^6 C_2} \\ & =1-\frac{3}{15} \\ & =\frac{4}{5} \end{aligned} $

$\begin{aligned} & \mathrm{P}(\mathrm{~A})=\frac{7}{10}, \mathrm{P}(\mathrm{~B})=\frac{4}{10} \\ & \mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})=\frac{5}{10} \\ & \mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~A} \cup \overline{\mathrm{~B}}}\right)=\frac{\mathrm{P}(\mathrm{~B} \cap(\mathrm{~A} \cup \overline{\mathrm{~B}}))}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})} \\ & =\frac{\mathrm{P}((\mathrm{~B} \cap \overline{\mathrm{~B}}) \cup(\mathrm{B} \cap \mathrm{~A}))}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})} \end{aligned}$

$\begin{aligned} & =\frac{\mathrm{P}(\mathrm{~A})-\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}})}{\mathrm{P}(\mathrm{~A})+\mathrm{P}(\overline{\mathrm{~B}})-\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}})}=\frac{\frac{7}{10}-\frac{5}{10}}{\frac{7}{10}+\left(1-\frac{4}{10}\right)-\frac{5}{10}} \\ & =\frac{2}{8}=\frac{1}{4} \end{aligned}$

Define the Events:

Let $A$ be the event that an unbiased coin is drawn.

Let $B$ be the event that the coin with heads on both sides is drawn.

Let $H$ be the event that a head turns up when the coin is tossed.

State the Given Probabilities:

$P(A) = \frac{19}{20}$ (since there are 19 unbiased coins out of 20 total coins)

$P(B) = \frac{1}{20}$ (since there is 1 coin with two heads out of 20 total coins)

$P(H|A) = \frac{1}{2}$ (probability of getting a head given an unbiased coin is drawn)

$P(H|B) = 1$ (probability of getting a head given the coin with two heads is drawn)

Use Bayes' Theorem:

We want to find $P(A|H)$, which is the probability that the drawn coin was unbiased given that a head turned up. Bayes' Theorem states:

$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)}$

Calculate $P(H)$:

We can find $P(H)$ using the law of total probability:

$P(H) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B)$

$P(H) = \left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right) + (1) \cdot \left(\frac{1}{20}\right) = \frac{19}{40} + \frac{1}{20} = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$

Apply Bayes' Theorem to find $P(A|H)$:

$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right)}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}$

Determine $m$ and $n$:

Since $P(A|H) = \frac{m}{n}$ and $\gcd(m, n) = 1$, we have $m = 19$ and $n = 21$.

Calculate $n^2 - m^2$:

$n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = (40)(2) = 80$

Therefore, $n^2 - m^2 = 80$.

So, the correct answer is:

Option B: 80

$\begin{aligned} & 2 p+2 q=\frac{1}{2} \\ & p+q \\ & E\left(x^2\right)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 . p+4 \cdot q+9 q \\ & =p+13 q \\ & E(x)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 . p+1 . p+2 q+3 q=p+5 q \\ & p+13 q=2(p+5 q) \\ & p=3 q \\ & \text { So, } q=\frac{1}{8} \& p=\frac{3}{8} \\ & \text { So, } 8 p-1=2 \end{aligned}$

$\begin{array}{lll} 3 E, & 1 D, & 8 P \\ 3 E, & 2 D, & 7 P \\ 4 E, & 1 D, & 7 P \\ 4 E, & 2 D, & 6 P \end{array}$

$\begin{aligned} & P=\frac{{ }^4 C_3 \cdot{ }^2 C_1 \cdot{ }^{10} C_8+{ }^4 C_3 \cdot{ }^2 C_2 \cdot{ }^{10} C_7+{ }^4 C_4 \cdot{ }^2 C_1 \cdot{ }^{10} C_7+{ }^4 C_4 \cdot{ }^2 C_2 \cdot{ }^{10} C_6}{{ }^{10} C_{12}} \\ & =\frac{129}{182} \end{aligned}$

$X$	$P(X)$	$XP(X)$	$\left(X_i-\mu\right)^2$	$P_i X\left(X_i-\mu\right)^2$
$X=0$	$\frac{{ }^7 C_2}{{ }^{10} C_2}$	0	$\left(0-\frac{3}{5}\right)^2$	$\frac{7}{15}\left(\frac{9}{25}\right)$
$X=1$	$\frac{{ }^7 C_1{ }^3 C_1}{{ }^{10} C_2}$	$\frac{7}{15}$	$\left(1-\frac{3}{5}\right)^2$	$\frac{7}{15}\left(\frac{4}{25}\right)$
$x=2$	$\frac{{ }^7 C_0{ }^3 C_2}{{ }^{10} C_2}$	$\frac{2}{15}$	$\left(2-\frac{3}{5}\right)^2$	$\frac{2}{15}\left(\frac{49}{25}\right)$

$\begin{aligned} & \mu=\sum X_i P\left(X_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text { Variance }(X)= \\ & \sum P_i\left(X_i-\mu\right)^2=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75} \end{aligned}$

To find $ P(X \geq 3) $, we first determine the constant $ k $ using the total probability for $ X $.

The probability $ P(X = x) $ is given by:

$ P(X = x) = k(x+1) \cdot 3^{-x}, \quad x = 0, 1, 2, 3, \ldots $

The total probability must equal 1:

$ s = \sum_{x = 0}^{\infty} k(x+1) \cdot 3^{-x} $

Calculating that series:

$ s = \frac{k}{3^0} + \frac{2k}{3} + \frac{3k}{3^2} + \ldots $

Therefore, dividing the series by 3:

$ \frac{s}{3} = \frac{k}{3} + \frac{2k}{3^2} + \ldots $

Subtracting these:

$ s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \ldots $

The resulting series is a geometric series:

$ \frac{2s}{3} = k \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots \right) $

The sum of the infinite geometric series is:

$ \frac{2s}{3} = k \cdot \frac{1}{1-\frac{1}{3}} = \frac{3k}{2} $

Equating:

$ s = \frac{9k}{4} = 1 $

Thus, solving for $ k $:

$ k = \frac{4}{9} $

Next, compute $ P(X \geq 3) $:

$ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) $

Calculating these:

$ P(X = 0) = k = \frac{4}{9} $

$ P(X = 1) = \frac{2k}{3} = \frac{8}{27} $

$ P(X = 2) = \frac{3k}{9} = \frac{4}{27} $

Adding these probabilities:

$ P(X = 0) + P(X = 1) + P(X = 2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9} $

Finally, calculate $ P(X \geq 3) $:

$ P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9} $

Probability that a Red ball comes from Bag I ($p$):

$ p(B_1 / R) = \frac{p(B_1) \cdot p(R / B_1)}{p(R)} $

Substituting the values,

$ p(B_1 / R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} $

Simplify to find $ p $:

$ = \frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4} $

So, $ p = \frac{1}{4} $.

Probability that a Green ball comes from Bag III ($q$):

$ p(B_3 / G) = \frac{p(B_3) \cdot p(G / B_3)}{p(G)} $

Substitute the values,

$ p(B_3 / G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}} $

Simplify to find $ q $:

$ = \frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3} $

So, $ q = \frac{1}{3} $.

Calculation of $\left(\frac{1}{p} + \frac{1}{q}\right)$:

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

Therefore,

$ \left(\frac{1}{p} + \frac{1}{q}\right) = 4 + 3 = \boxed{7} $

$\begin{aligned} & \text { Bag } 1=\{4 \mathrm{~W}, 5 \mathrm{~B}\} \\ & \text { Bag } \mathbf{2}=\{\mathbf{n W}, \mathbf{3 B}\} \\ & \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{Bag} 2}\right)=\frac{29}{45} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)+\mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)=\frac{29}{45} \\ & \frac{4}{9} \times \frac{\mathrm{n}+1}{\mathrm{n}+4}+\frac{5}{9} \times \frac{\mathrm{n}}{\mathrm{n}+4}=\frac{29}{45} \end{aligned}$

$\mathrm{n = 6}$

$E_1: B a g B_1$ is selected

$\begin{array}{lll} B_1 & B_2 & B_3 \\ 6 \mathrm{~W} 4 \mathrm{~B} & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B} \end{array}$

$E_2:$ bag $B_2$ is selected

$E_3: B a g B_3$ is selected

A : Drawn ball is white

We have to find $P\left(\frac{E_2}{A}\right)$

$\begin{aligned} & P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)} \\ & =\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \\ & =\frac{4}{15} \end{aligned}$

A, E,G R D N

$\text { Probabllity }(\mathrm{P})=\frac{\text { favourable case }}{\text { Total case }}$

(when A & E are in order)

Total case $=6$ !

Favourable case $={ }^6 \mathrm{C}_2 .4$ !

$P=\frac{(15) 4!}{(30) 4!}$

Probablity when not in order $=1-\frac{1}{2}=\frac{1}{2}$

$i^{k_1}+i^{k_2} \neq 0 \quad i^{k_1} \rightarrow 4$ option for $\mathrm{i},-1,-\mathrm{i}, 1$

Total cases $\Rightarrow 4 \times 4=16$

Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2}=0$

$\left\{\begin{array}{c} 1,-1 \\ -1,1 \\ i,-i \\ -i, i \end{array}\right\}$

4 Cases $\Rightarrow$ Probability $=\frac{16-4}{16}=\frac{3}{4}$

$\begin{aligned} &\text { Probability distribution }\\ &\begin{array}{c|c} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \\ \hline \mathrm{x}=0 & \frac{{ }^7 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{42}{90} \\ \hline \mathrm{x}=1 & \frac{{ }^7 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{10} \mathrm{C}_2}=\frac{42}{90} \\ \hline \mathrm{x}=2 & \frac{{ }^3 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{6}{90} \end{array} \end{aligned}$

$\begin{aligned} &\text { Now, }\\ &\begin{aligned} & \mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{42}{90}+\frac{12}{90}=\frac{54}{90} \\ & \sigma^2=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_1^2-\mu^2=\frac{42}{90}+\frac{24}{90}-\left(\frac{54}{90}\right)^2 \\ & \Rightarrow \frac{66}{90}-\left(\frac{54}{90}\right)^2 \\ & \sigma^2 \Rightarrow \frac{28}{75} \therefore(1) \end{aligned} \end{aligned}$

C-I $\left|\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right| \rightarrow 4$ ways

C-II $\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| \&\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right| \rightarrow 2$ ways

$\mathrm{P}=\frac{\text { favourable }}{\text { total }}=\frac{6}{16}=\frac{3}{8}$

$\begin{aligned} & \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{1}{9} \\ & \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{5}{36} \\ & \text { required prob }=\frac{1}{9}+\frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9}+\left(\frac{8}{9} \cdot \frac{31}{36}\right)^2 \cdot \frac{1}{9}+\ldots \infty \\ & =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19} \end{aligned}$

Total ways for selecting any two squares $={ }^{16} \mathrm{C}_2=120$

Total ways for selecting common side squares

$\begin{aligned} &=24\\ &\text { so required probability }\\ &\begin{aligned} & =1-\frac{24}{120} \\ & =\frac{4}{5} \end{aligned} \end{aligned}$

$\mathrm{a}=$ number or dice 1

$\mathrm{b}=$ number on dice 2

$(a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)$

Required probability

$\begin{aligned} & =\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6} \\ & =\frac{18}{36}=\frac{1}{2} \end{aligned}$

To solve this problem, start by considering the equation given for the probabilities:

$ 12x^2 - 7x + 1 = 0 $

The roots of this equation are:

$ x = \frac{1}{3}, \frac{1}{4} $

Assume $ P(A \mid B) = \frac{1}{3} $ and $ P(B \mid A) = \frac{1}{4} $.

From the definitions of conditional probability, we have:

$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{3} $

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

Given that $ P(A \cap B) = 0.1 $, we can use these equations to find $ P(B) $ and $ P(A) $:

From $ \frac{0.1}{P(B)} = \frac{1}{3} $, we find $ P(B) = 0.3 $.

From $ \frac{0.1}{P(A)} = \frac{1}{4} $, we find $ P(A) = 0.4 $.

Now, calculate $ P(A \cup B) $ using the formula for the union of two events:

$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.1 = 0.6 $

The task is to find the value of:

$ \frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} $

Using De Morgan's laws and the complements:

$ P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) = 1 - 0.1 = 0.9 $

$ P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4 $

Finally, compute the ratio:

$ \frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} = \frac{0.9}{0.4} = \frac{9}{4} $

Outcome	$x_i$	$p_i$
HHH	0	$\frac{1}{8}$
TTT	0	$\frac{1}{8}$
HHT	1	$\frac{1}{8}$
HTH	1	$\frac{1}{8}$
THH	0	$\frac{1}{8}$
TTH	0	$\frac{1}{8}$
THT	1	$\frac{1}{8}$
HTT	1	$\frac{1}{8}$

$\begin{aligned} & \mu=\sum x_i P_i=\frac{1}{2} \\ & \sigma^2=\sum x_i^2 P_i-\mu^2 \\ & =\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\ & 64\left(\mu+\sigma^2\right)=64\left[\frac{1}{2}+\frac{1}{4}\right] \\ & =64 \times \frac{3}{4}=48\end{aligned}$

In a bag containing 4 white and 6 black balls, two balls are selected one by one without replacement. We need to find the probability that the first ball is black, given that the second ball is black. Let’s define the events:

A: The first ball selected is black.

B: The second ball selected is black.

The probability $ P(A|B) $ is given by the formula:

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

Let's compute each part separately:

Probability of both balls being black $ P(A \cap B) $:

The chance that the first ball is black is $\frac{6}{10}$. Given that one black ball is removed, the probability that the second is also black is $\frac{5}{9}$.

$ P(A \cap B) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} $

Probability that the second ball is black $ P(B) $:

This can occur if either the first ball was white or black:

First ball white, second black: $\frac{4}{10} \times \frac{6}{9}$

Both balls black: $\frac{6}{10} \times \frac{5}{9}$

$ P(B) = \frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{5}{9} = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} = \frac{3}{5} $

Now, using these probabilities:

$ P(A|B) = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} $

Here, $ \frac{5}{9} $ is in its simplest form ($\operatorname{gcd}(5, 9) = 1$), so $ m = 5$ and $ n = 9 $.

Therefore, $ m + n = 5 + 9 = 14 $.

Let's denote the outcomes of the three rolls as $X_1$, $X_2$, and $X_3$, where $X_i$ represents the number obtained in the $i^{\text{th}}$ roll. We are looking for the probability that:

$X_2 > X_1 \text{ and } X_3 > X_2$

The total number of outcomes when rolling a dice three times is:

$6^3 = 216$

Let's count the favorable outcomes. For each satisfying outcome, we must ensure that both inequalities are adhered to. Consider the possible sequences where every subsequent roll's number is higher than the previous roll's number. These sequences are:

• (1, 2, 3)
• (1, 2, 4)
• (1, 2, 5)
• (1, 2, 6)
• (1, 3, 4)
• (1, 3, 5)
• (1, 3, 6)
• (1, 4, 5)
• (1, 4, 6)
• (1, 5, 6)
• (2, 3, 4)
• (2, 3, 5)
• (2, 3, 6)
• (2, 4, 5)
• (2, 4, 6)
• (2, 5, 6)
• (3, 4, 5)
• (3, 4, 6)
• (3, 5, 6)
• (4, 5, 6)

Clearly, there are 20 such favorable outcomes. Hence, the probability is given by:

$ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{216} = \frac{5}{54} $

Therefore, the correct option is:

Option A: $\frac{5}{54}$

To solve this problem, we use Bayes' theorem. Let's define the events:

• $X$: Selecting bag $X$
• $Y$: Selecting bag $Y$
• $Z$: Selecting bag $Z$
• $A$: Drawing a one-rupee coin

We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:

$P(X) = P(Y) = P(Z) = \frac{1}{3}$

Next, we need the probability of drawing a one-rupee coin from each bag:

• From bag $X$: $P(A|X) = \frac{5}{5+4} = \frac{5}{9}$
• From bag $Y$: $P(A|Y) = \frac{4}{4+5} = \frac{4}{9}$
• From bag $Z$: $P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}$

We need to find the probability that the coin came from bag $Y$ given that a one-rupee coin was drawn, i.e., we need $P(Y|A)$. Using Bayes' theorem:

$P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}$

To find $P(A)$, the total probability of drawing a one-rupee coin can be calculated as follows:

$P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)$

Substituting the values:

$P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$

Calculating the above, we get:

$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}$

Note that $\frac{1}{9} = \frac{3}{27}$, so:

$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}$

Now, substituting back into Bayes' theorem:

$P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}$

Hence, the probability that the coin came from bag $Y$ is:

Option B: $\frac{1}{3}$

Take two numbers as $a$ and $b$

$a+b=24$

For product to be maximum

$\begin{aligned} & \frac{a+b}{2} \geq \sqrt{a b} \\ & 144>a b \end{aligned}$

Maximum product is 144

Now, $a b \geq \frac{3}{4} \cdot 144=108$

Sample space $=\{(23,1),(22,2), \ldots\}$

Integer points on line in shaded region

$\begin{aligned} & \{(6,18),(7,17),(8,16), \ldots(18,6)\} \\ & P(E)=\frac{n(E)}{n(S)}=\frac{13}{23}=\frac{m}{n} \Rightarrow n-m=10 \end{aligned}$

We have 3 letters and 5 addresses, where 3 letters are posted to exactly 2 addresses. First, we will select 2 addresses in ${ }^5 C_2$ ways.

Now, 3 letters can be posted to exactly 2 addresses in 6 ways.

$\begin{aligned} \therefore \text { Probability }=\frac{{ }^5 C_2 \times 6}{5^3} & \\ & =\frac{60}{125}=\frac{12}{25} \end{aligned}$

$\begin{aligned} & P(\text { standard automobile from } A)=\frac{6}{10} \times \frac{8}{10}=\frac{12}{25} \\ & P(\text { standard automobile from } B)=\frac{4}{10} \times \frac{9}{10}=\frac{9}{25} \end{aligned}$

$\begin{aligned} & \text { Required Probability } \frac{\frac{9}{25}}{\frac{12}{25}+\frac{9}{25}} \\ & P=\frac{9}{21}=\frac{3}{7} \end{aligned}$

So, $126 P=126 \times \frac{3}{7}=54$

Equation is $a x^2+b x+c=0$

$\mathrm{D}>0$ [for roots to be real & distinct]

$\Rightarrow b^2-4 a c>0$

For $b<2$ no value of $a$ & $c$ are possible

$\begin{aligned} & \text { For } b=3 \Rightarrow a c<\frac{9}{4} \\ & (a, c) \in\{(1,1),(1,2),(2,1)\} \Rightarrow 3 \text { cases } \end{aligned}$

For $b=4 \Rightarrow a c<4$

$(a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3)\} \Rightarrow 5 \text { cases }$

For $b=5 \Rightarrow a c<\frac{25}{4}$

$\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1)\}=14 \text { cases } \end{aligned}$

For $b=6 \Rightarrow a c<9$

$\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1),(2,4),(4,2)\}=16 \text { cases } \end{aligned}$

Total cases $=3+5+14+16=38$ cases

$\Rightarrow$ Probability, $p=\frac{38}{216}$

$\Rightarrow 216 p=38$

Given quadratic equation is

$a x^2+b x+c=0 \text { where } a, b, c \in\{1,2,3, \ldots, 8\}$

For repeated roots,

$\begin{aligned} & b^2-4 a c=0 \\ & \Rightarrow b^2=4 a c \end{aligned}$

$\Rightarrow a c$ must be a perfect square

$(a, c) \in\{(1,1),(1,4),(2,2),(2,8),(3,3),(4,1),(4,4),(5,5),(6,6),(7,7),(8,2),(8,8)\}$

Corresponding $b$ must lie in set $\{1,2,3, \ldots 8\}$

$\begin{aligned} & (a, b, c) \in\{(1,2,1),(1,2,4),(2,4,2),(2,8,8) \\ & (3,6,3),(4,4,1),(4,8,4),(8,8,2)\} \\ & \therefore \text { probability }=\frac{8}{8^3} \\ & =\frac{1}{64} \\ & \end{aligned}$

$X$	0	2	4	6	8
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

$\begin{aligned} & \text { Mean }=\sum x_i P\left(x_i\right) \\ & \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\ & \frac{46}{9}=8 a+40 b \\ & \frac{23}{9}=4 a+20 b \\ & 36 a+180 b=23 \quad \text{.... (1)} \end{aligned}$

Sum of probability is 1

$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$

$\begin{aligned} & \text { Solving (1) and (2) } \\ & a=\frac{1}{12}, b=\frac{1}{9} \\ & \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\ & =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\ & =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\ & =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\ \end{aligned}$

Let's denote the events as follows :

Let $ A_1, A_2, $ and $ A_3 $ be the events that urns A, B, and C are chosen, respectively.

Let $ B $ be the event that a black ball is drawn.

We need to find the probability that the chosen urn is A given that a black ball is drawn, which is $ P(A_1|B) $.

Using Bayes' theorem, we have :

$ P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)} $

First, we calculate each term individually :

1. The probability of choosing any urn, since they are chosen at random :

$ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} $

2. The probability of drawing a black ball from each urn :

Urn A: $ P(B|A_1) = \frac{5}{12} $

Urn B: $ P(B|A_2) = \frac{7}{12} $

Urn C: $ P(B|A_3) = \frac{6}{12} = \frac{1}{2} $

3. The total probability of drawing a black ball, $ P(B) $ :

$ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) $

$ P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} $

$ P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} $

$ P(B) = \frac{18}{36} = \frac{1}{2} $

Now, substitute these into Bayes' theorem :

$ P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)} $

$ P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}} $

$ P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}} $

$ P(A_1|B) = \frac{5}{18} $

Therefore, the probability that the black ball drawn is from urn A is option C :

$ \frac{5}{18} $

We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\frac{2}{7}$, and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\frac{1}{5}$.

We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this probability as $\mathrm{r}$.

To find $\mathrm{r}$, we need to use the concept of complementary events. The probability that Ajay will appear in the exam is the complement of the probability that he will not appear. So,

$ P(\text{Ajay appears}) = 1 - P(\text{Ajay does not appear}) = 1 - \mathrm{p} = 1 - \frac{2}{7} = \frac{5}{7}. $

The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not). Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ($\mathrm{q}$) is included in the probability of Ajay appearing:

$ \mathrm{r} = P(\text{Ajay appears}) - P(\text{Both Ajay and Vijay appear}) = \frac{5}{7} - \frac{1}{5}. $

To subtract these two fractions, we need a common denominator, which would be $35$ in this case. So,

$ \mathrm{r} = \frac{5}{7} \cdot \frac{5}{5} - \frac{1}{5} \cdot \frac{7}{7} = \frac{25}{35} - \frac{7}{35} = \frac{25 - 7}{35} = \frac{18}{35}. $

Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\mathrm{r} = \frac{18}{35}$, which corresponds to Option D.

$\begin{aligned} & \mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})= \\\\ & \frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)} \\ & +\ldots \ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)\end{aligned}$

$\begin{aligned} & =\frac{\frac{1}{5} \times \frac{{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}}{\frac{1}{5} \times \frac{{ }^2 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\frac{1}{5} \times \frac{{ }^3 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\ldots+\frac{1}{5} \times \frac{{ }^6 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}} \\\\ & =\frac{2}{7}\end{aligned}$

To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).

Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $ P(T) = p $ and the probability of getting a head as $ P(H) = 2p $.

These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :

$ P(H) + P(T) = 1 $

$ 2p + p = 1 $

$ 3p = 1 $

$ p = \frac{1}{3} $

Therefore, the probability of getting a tail (T) is $ P(T) = \frac{1}{3} $ and the probability of getting a head (H) is $ P(H) = 2 \times \frac{1}{3} = \frac{2}{3} $.

Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur. There are three unique sequences: TTH, THT, and HTT.

The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:

$ P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27} $

$ P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} $

$ P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27} $

The overall probability of getting two tails and one head in any order is the sum of these individual probabilities :

$ P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27} $

Simplifying this expression gives us: $ P(2T1H) = \frac{6}{27} = \frac{2}{9} $

Therefore, the correct answer is :

Option B : $\frac{2}{9}$

3 bad apples, 15 good apples.

Let $\mathrm{X}$ be no of bad apples

$\begin{aligned} & \text { Then } \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{15} \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{105}{153} \\ & \mathrm{P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153} \\ & \mathrm{P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153} \\ & \mathrm{E}(\mathrm{X})=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153} \\ & =\frac{1}{3} \\ & \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-(\mathrm{E}(\mathrm{X}))^2 \\ & =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2 \\ & =\frac{57}{153}-\frac{1}{9}=\frac{40}{153} \end{aligned}$

To solve this problem, we need to calculate the probability of two independent events occurring in succession: the first marble drawn is red, and the second marble drawn is white. Since the drawing is with replacement, the number of marbles of each color remains the same for both draws.

The total number of marbles in the box is the sum of red, white, blue, and orange marbles:

$ \text{Total marbles} = 10 (\text{red}) + 30 (\text{white}) + 20 (\text{blue}) + 15 (\text{orange}) = 75. $

The probability of drawing a red marble in the first draw is the number of red marbles divided by the total number of marbles:

$ P(\text{First is red}) = \frac{10}{75}. $

Since the marble is replaced, the probability of drawing a white marble in the second draw remains as the number of white marbles divided by the total number of marbles:

$ P(\text{Second is white}) = \frac{30}{75}. $

The probability of both independent events occurring in succession (drawing a red marble first and then a white marble) is the product of their individual probabilities:

$ P(\text{First is red and second is white}) = P(\text{First is red}) \times P(\text{Second is white}) = \frac{10}{75} \times \frac{30}{75}. $

Now, let's calculate this probability:

$ P(\text{First is red and second is white}) = \frac{10 \times 30}{75 \times 75} = \frac{300}{5625}= \frac{4}{75}. $

Therefore, the correct answer is

Option D $\frac{4}{75}.$

$\mathrm{E}_1: \mathrm{A}$ is selected

$\mathrm{E}_2: \mathrm{B}$ is selected

$\mathrm{E}$ : white ball is drawn

$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right)=$

$\begin{aligned} & \frac{P(E) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}=\frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10}+\frac{1}{2} \times \frac{3}{5}} \\ & =\frac{3}{3+6}=\frac{1}{3} \end{aligned}$

If $x=0, y=6,7,8,9,10$

If $x=1, y=7,8,9,10$

If $x=2, y=8,9,10$

If $x=3, y=9,10$

If $x=4, y=10$

If $x=5, y=$ no possible value

Total possible ways $=(5+4+3+2+1) \times 2$

$=30$

Required probability $=\frac{30}{11 \times 11}=\frac{30}{121}$

Given set $=\{1,2,3, \ldots \ldots . .50\}$

$\mathrm{P}(\mathrm{A})=$ Probability that number is multiple of 4

$\mathrm{P(B)}=$ Probability that number is multiple of 6

$\mathrm{P}(\mathrm{C})=$ Probability that number is multiple of 7

Now,

$\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}$

again

$\begin{aligned} & \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \\ & \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0 \end{aligned}$

Thus

$\begin{aligned} P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\ & =\frac{21}{50} \end{aligned}$

Required probability $=$

$\begin{aligned} & \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\ & =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11} \end{aligned}$

$\frac{{ }^6 \mathrm{C}_4}{{ }^{15} \mathrm{C}_4} \times \frac{{ }^9 \mathrm{C}_4}{{ }^{11} \mathrm{C}_4}=\frac{3}{715}$

Hence option (3) is correct.

Let $X$ be the number rolled on the die, and let $W$ be the event that all balls drawn are white. We want to find the probability $P(W)$, which can be calculated using the law of total probability as follows :

$P(W) = \sum\limits_{x=1}^{6} P(W|X=x)P(X=x)$

The probability of rolling any number from 1 to 6 on the die is equal, so $P(X=x) = \frac{1}{6}$ for all $x \in \{1, 2, 3, 4, 5, 6\}$.

Now let's calculate the conditional probabilities $P(W|X=x)$ for each possible value of $x$ :

1. $P(W|X=1) = {{{}^6{C_1}} \over {{}^{10}{C_1}}}= \frac{6}{10} = \frac{3}{5}$, since there are 6 white balls out of a total of 10 balls.

2. $P(W|X=2) = {{{}^6{C_2}} \over {{}^{10}{C_2}}} = \frac{15}{45} = \frac{1}{3}$, since there are 15 ways to choose 2 white balls out of 6, and 45 ways to choose 2 balls out of 10.

3. $P(W|X=3) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = \frac{20}{120} = \frac{1}{6}$, since there are 20 ways to choose 3 white balls out of 6, and 120 ways to choose 3 balls out of 10.

4. $P(W|X=4) = {{{}^6{C_4}} \over {{}^{10}{C_4}}} = \frac{15}{210} = \frac{1}{14}$, since there are 15 ways to choose 4 white balls out of 6, and 210 ways to choose 4 balls out of 10.

5. $P(W|X=5) = {{{}^6{C_5}} \over {{}^{10}{C_5}}} = \frac{6}{252} = \frac{1}{42}$, since there are 6 ways to choose 5 white balls out of 6, and 252 ways to choose 5 balls out of 10.

6. $P(W|X=6) = {{{}^6{C_6}} \over {{}^{10}{C_6}}} = \frac{1}{210}$, since there are 1 ways to choose 6 white balls out of 6, and 210 ways to choose 6 balls out of 10.

Using the law of total probability, we have :

$P(W) = \frac{1}{6} \left(P(W|X=1) + P(W|X=2) + P(W|X=3) + P(W|X=4) + P(W|X=5) + P(W|X=6)\right)$

$P(W) = \frac{1}{6} \left(\frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}\right)$

To simplify this expression, find a common denominator :

$P(W) = \frac{1}{6} \left(\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210}\right)$

Add the fractions :

$P(W) = \frac{1}{6}\left(\frac{126+70+35+15+5+1}{210}\right)=\frac{42}{210}=\frac{1}{5}$

$ \begin{aligned} & n p-n p q=1 \\\\ \Rightarrow & n p(1-q)=1 \\\\ \Rightarrow & n p^2=1 \end{aligned} $

$ \begin{aligned} & 2 P(X=2)=3 P(X=1) \\\\ & 2 \cdot{ }^n C_2 p^2 q{ }^{n-2}=3 \cdot{ }^n C_1 p \cdot q^{n-1} \\\\ \Rightarrow & 2 \cdot \frac{n \cdot(n-1)}{2} \cdot p=3 \cdot n \cdot q \\\\ \Rightarrow & (n-1) p=3(1-p) \\\\ \Rightarrow & \left(\frac{1}{p^2}-1\right) p=3(1-p) \\\\ \Rightarrow & \frac{(1-p)(1+p)}{p}=3(1-p) \\\\ \Rightarrow & 1+p=3 p \\\\ \Rightarrow & p=\frac{1}{2} \\\\ \therefore & n=4 \end{aligned} $

$ \begin{aligned} n^2 P(x>1) & =n^2(1-P(x=1)-P(x=0)) \\\\ & =16\left(1-{ }^4 C_1 \cdot\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4\right)=11 \end{aligned} $

The given probabilities for getting a head (H) and a tail (T) are as follows:

$ P(H) = \frac{3}{4}, \quad P(T) = \frac{1}{4} $

The random variable X can take the values 1, 2, or 3. These correspond to the following events:

- X = 1 : A head is obtained on the first toss. This happens with probability $P(H) = \frac{3}{4}$.

- X = 2 : A tail is obtained on the first toss and a head on the second. This happens with probability $P(T)P(H) = \frac{1}{4} \times \frac{3}{4}$.

- X = 3 : Either two tails and then a head are obtained, or three tails are obtained.
This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3$.

Now, we calculate the mean (expected value) of X:

$ \begin{aligned} E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\ & = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\ & = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\ & = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\ & = 3 \cdot \left(\frac{7}{16}\right) \\\\ & = \frac{21}{16}. \end{aligned} $

$ \begin{array}{|c|l|c|} \hline \alpha-\beta & {\text { Case }} & \mathbf{P} \\ \hline 5 & (6,1) & 1 / 36 \\ \hline 4 & (6,2)(5,1) & 2 / 36 \\ \hline 3 & (6,3)(5,2)(4,1) & 3 / 36 \\ \hline 2 & (6,4)(5,3)(4,3)(3,1) & 4 / 36 \\ \hline 1 & (6,5)(5,4)(4,3)(3,2)(2,1) & 5 / 36 \\ \hline 0 & (6,6)(5,5) \ldots \ldots(1,1) & 6 / 36 \\ \hline-1 & ----- & 5 / 36 \\ \hline-2 & -----& 4 / 36 \\ \hline-3 & ----- & 3 / 36 \\ \hline-4 & (2,6)(1,5) & 2 / 36 \\ \hline-5 & (1,6) & 1 / 36 \\ \hline \end{array} $

$E[X^2] = \sum\limits_{i=-5}^{5} (x_i)^2 P(x_i)$

Substituting the values from table :

$E[X^2] = 2\left[\frac{25}{36}+\frac{32}{36}+\frac{27}{36}+\frac{16}{36}+\frac{5}{36}\right] = \frac{105}{18} = \frac{35}{6}$

Next, we calculate the expected value of the differences. The expected value is calculated as the sum of the products of each outcome and its corresponding probability. Given that the table is symmetric around 0, the expected value is 0.

$E[X] = \sum\limits_{i=-5}^{5} x_i P(x_i) = 0$

Now, we can calculate the variance, which is the expected value of the squared differences minus the square of the expected value of the differences :

$Var[X] = E[X^2] - (E[X])^2 = \frac{35}{6} - 0^2 = \frac{35}{6}$

Here, $p = 35$ and $q = 6$, and they are co-prime.

The positive divisors of 35 are 1, 5, 7, and 35. The sum of these divisors is $1 + 5 + 7 + 35 = 48$.

We have, $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$

Let $M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, where $a, b, c, d \in\{0,1,2\}$

$ n(s)=3^4=81 $

If $A$ is invertible, then $|A| \neq 0$

Now, if $|A|=0$, then $|M|=0$

$\therefore a d-b c=0$ or $a d=b c$

Case I : When $a d=b c=0$, then

There are five ways when $a d=0$ i.e.,

$(a, d)=(0,0),(0,1),(0,2),(1,0),(2,0)$

Similarly, there are again five ways, when $b c=0$

$\therefore$ There are total $5 \times 5=25$ ways, when $a d=b c=0$

Case II : When $a d=b c=1$

There is only one way, when $a d=b c=1$

$ \text { i.e. } \quad a=b=c=d=1 $

Case III : When $a d=b c=2$

There are two ways, when $a d=2$, i.e.

$ (a, d)=(1,2) \text { or }(2,1) $

Similarly, there are two ways

when $b c=2$ i.e., $(b, c)=(1,2)$ or $(2,1)$

Case IV : When $a d-b c=4$

There is only way, when $a d=b c=4$

$ \text { i.e., } a=b=c=d=2 $

$\therefore$ Total number of ways, when

$ \begin{aligned} & (\bar{A})=\frac{31}{81}|A|=0 \text { is } 25+1+4+1=31 \\\\ & \text { Hence, } P(A)=1-P(\bar{A})=1-\frac{31}{81}=\frac{50}{81} \end{aligned} $