Probability

\(
\begin{aligned}
& 2 \times 2 \rightarrow \text { Matrix } \\
& |A| \neq 0 \\
& n(s)=2 \times 2 \times 2 \times 2 \Rightarrow 16
\end{aligned}
\)

\(A\) can't be

\(
\begin{aligned}
& {\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right],\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right],\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]} \\
& {\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right],\left[\begin{array}{ll}
1 & 0 \\
1 & 0
\end{array}\right],\left[\begin{array}{ll}
0 & 0 \\
1 & 1
\end{array}\right]} \\ & {\left[\begin{array}{ll}
0 & 1 \\
0 & 1
\end{array}\right],\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right],\left[\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right]}
\end{aligned}
\)

$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$
So, total \(\Rightarrow 10\) types of matrix we can't take

\(
\begin{aligned}
& \therefore P(\bar{A})=\frac{10}{16}=\frac{5}{8} \\
& \therefore P(A)=1-P(\bar{A})=1-\frac{5}{8} \Rightarrow \frac{3}{8} \\
\end{aligned}
\)

We have

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

Use $\sum_{x=0}^{\infty}(x+1)r^x=\frac{1}{(1-r)^2}$ (for $|r|<1$) with $r=\frac13$:

$$\sum_{x=0}^{\infty}(x+1)\left(\frac13\right)^x = \frac{1}{\left(1-\frac13\right)^2} = \frac{1}{\left(\frac23\right)^2} = \frac94$$

Since total probability is 1:

$$k\cdot \frac94 = 1 \Rightarrow k = \frac49$$

Now

$$P(X\le2) = \sum_{x=0}^{2} k(x+1)3^{-x} = \frac49\left[1 + \frac{2}{3} + \frac{3}{9}\right] = \frac49 \cdot 2 = \frac89$$

So

$$P(X\ge 3)=1-P(X\le 2)=1-\frac{8}{9}=\frac{1}{9}\mathrm{.}$$

The probability that the first ball is black is $\frac{6}{10}$

After selecting a black ball, there are \(5\) black balls

left out of \(9\) total balls.

So, the probability that the second ball is black is $\frac{5}{9}$

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

Now, $P\left(B\right)=\frac{6}{10}\times \frac{5}{9}+\frac{4}{10}\times \frac{6}{9}=\frac{1}{3}+\frac{4}{15}=\frac{5}{15}+\frac{4}{15}$$=\frac{9}{15}=\frac{3}{5}$

By conditional probability,

$P\left(A\mid B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3}·\frac{5}{3}$$=\frac{5}{9}$

The probability $P(A\mid B)=\frac{5}{9}\text{.}$

$\text{Here, }m=5\text{ and }n=9\text{, and }gcd(5,9)=1\text{. }$

$m+n=5+9=14$

$\left|A\right|=\left|\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|$

$$\begin{gathered} =a_{11}{a}_{22}-a_{21}{a}_{12} \\ \text{possible values for X}=\{-1,0,1\} \end{gathered}$$

$\begin{array}{cccc}\mathrm{x}& {\mathrm{P}}_{\mathrm{i}}& {\mathrm{P}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{i}}& {\mathrm{P}}_{\mathrm{i}}{{\mathrm{X}}_{\mathrm{i}}}^2\\ -1& \frac{3}{16}& -\frac{3}{16}& \frac{3}{16}\\ 0& \frac{10}{16}& 0& 0\\ 1& \frac{3}{16}& \frac{3}{16}& \frac{3}{16}\\ & & \sum {\mathrm{P}}_{\mathrm{i}}{X}_{\mathrm{i}}=0& \sum {\mathrm{P}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{i}}^2=\frac{3}{8}\end{array}$

$$\begin{gathered} \therefore var\left(\mathrm{x}\right)=\sum {\mathrm{P}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{i}}^2-{\left(\sum {\mathrm{P}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{i}}\right)}^2 \\ =\frac{3}{8}-0=\frac{3}{8} \end{gathered}$$

Selection of any \(1\) out of \(2\) words \(=\frac{1}{2}\)

ANANTPURTgQPHd|[] (\(9\) letters):

Consecutive pairs are (A,N), (N,A), (A,N), (N,T), (T,P), (P,U), (U,R).

Total = \(7\) and "AN" appears \(2\) times.

P(AN|Anantpur) $=\frac{2}{7}$

KANPURTgQPHd|[] (\(6\) letters):

Consecutive pairs are (K,A), (A,N), (N,P), (P,U), (U,R).

Total = \(5\) and "AN" appears \(1\) time.

P(AN|Kanpur) $=\frac{1}{5}$

\(\therefore\) Required Probability \(=\frac{\frac{1}{2} \times \frac{2}{7}}{\frac{1}{2} \times \frac{1}{5}+\frac{1}{2} \times \frac{2}{7}}=\frac{\frac{2}{7} \times 5}{\frac{17}{7}}=\frac{10}{17}\)

outcomes for one dice =$\left\{1,1,2,2,3,4\right\}$

outcomes for another dice =$\left\{1,2,2,3,3,4\right\}$

Identifying Favorable Outcomes for a Sum of \(4\):

\( (1, 3) \): Probability = $\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

\( (2, 2) \): Probability $=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

\( (3, 1) \): Probability $=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$

Identifying Favorable Outcomes for a Sum of \(5\):

\( (1, 4) \): Probability $=\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}$

\( (2, 3) \): Probability $= \frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

\( (3, 2) \): Probability $=\frac{1}{6}\times \frac{1}{3}=\frac{1}{18}$

\( (4, 1) \): Probability $=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$

Calculating the Total Probability:

$P(\text{Sum}=4 \text{or} \text{Sum}=5)=P(\text{Sum}=4)+P(\text{Sum}=5)$

$=\left(\frac{1}{9}+\frac{1}{9}+\frac{1}{36}\right)+ \left(\frac{1}{18}+\frac{1}{9}+\frac{1}{18}+\frac{1}{36}\right)$
$=\frac{1}{2}$

Probability of getting \(2\) = $\frac{1}{6}$

Probability of getting other than \(2\) = $\frac{5}{6}$

Required Probability $=\frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+\dots \dots \dots$

$=\frac{5}{36}\left[1+{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^4+\dots \dots \dots \right]=\frac{5}{36}\times \frac{1}{1-\frac{25}{36}}=\frac{5}{11}$

Probability of getting \(2\) = $\frac{1}{6}$

Probability of getting other than \(2\) = $\frac{5}{6}$

Required Probability $=\frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+\dots \dots \dots$

$=\frac{5}{36}\left[1+{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^4+\dots \dots \dots \right]=\frac{5}{36}\times \frac{1}{1-\frac{25}{36}}=\frac{5}{11}$

We have

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

Use $\sum_{x=0}^{\infty}(x+1)r^x=\frac{1}{(1-r)^2}$ (for $|r|<1$) with $r=\frac13$:

$$\sum_{x=0}^{\infty}(x+1)\left(\frac13\right)^x = \frac{1}{\left(1-\frac13\right)^2} = \frac{1}{\left(\frac23\right)^2} = \frac94$$

Since total probability is 1:

$$k\cdot \frac94 = 1 \Rightarrow k = \frac49$$

Now

$P(X\le 2)=\sum _{x=0}^{2}k(x+1)3^{-x}$$=\frac{4}{9}[1+\frac{2}{3}+\frac{3}{9}]=\frac{4}{9}\cdot 2=\frac{8}{9}$

So

$$P(X\ge 3)=1-P(X\le 2)=1-\frac{8}{9}=\frac{1}{9}\mathrm{.}$$

$$\begin{gathered} P\left(A\right)=\frac{4}{36}=\frac{1}{9},P\left(B\right)=\frac{5}{36} \\ P\left(\text{ A wins }\right)=\frac{4}{36}+\frac{32}{36}·\frac{31}{36}·\frac{4}{36}+\dots ...\infty \\ P\left(\text{ A wins }\right)=\frac{\frac{4}{36}}{1-\frac{32}{36}·\frac{31}{36}} \\ =\frac{\frac{1}{9}}{1-\frac{8}{9}·\frac{31}{36}} \\ =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19} \end{gathered}$$

\( \mathrm{x}+\mathrm{y}=24, \mathrm{x}, \mathrm{y} \in \mathrm{~N} \)
\( \mathrm{AM}>\mathrm{GM} \Rightarrow \mathrm{xy} \leq 144 \)
\( x y \geq 108\)

Favorable pairs of \((x, y)\) are

$$\begin{gathered} \left\{\right(13,11),(12,12),(14,10),(15,9),(16,8),(17,7), \\ (18,6),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13\left)\right\} \end{gathered}$$
i.e. \(13\) cases
Total choices for \(\mathrm{x}+\mathrm{y}=24\) is \(23\)
Probability \(\frac{13}{23}=\frac{m}{n}\)

\(n-m=10\)

\( \mathrm{x}+\mathrm{y}=24, \mathrm{x}, \mathrm{y} \in \mathrm{~N} \)
\( \mathrm{AM}>\mathrm{GM} \Rightarrow \mathrm{xy} \leq 144 \)
\( x y \geq 108\)

Favorable pairs of \((x, y)\) are

$$\begin{gathered} \left\{\right(13,11),(12,12),(14,10),(15,9),(16,8),(17,7), \\ (18,6),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13\left)\right\} \end{gathered}$$
i.e. \(13\) cases
Total choices for \(\mathrm{x}+\mathrm{y}=24\) is \(23\)
Probability \(\frac{13}{23}=\frac{m}{n}\)

\(n-m=10\)

• $U$: Selecting an unbiased coin. $P(U) = \frac{19}{20}$
• $B$: Selecting the biased (two-headed) coin. $P(B) = \frac{1}{20}$
• $H$: Getting a head on the toss.
  • $P(H|U) = \frac{1}{2}$ (for an unbiased coin)
  • $P(H|B) = 1$ (for a two-headed coin)

$P(U|H) = \frac{P(U) \cdot P(H|U)}{P(U) \cdot P(H|U) + P(B) \cdot P(H|B)}$

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

• Given $P(U|H) = \frac{m}{n} = \frac{19}{21}$, where $\text{gcd}(19, 21) = 1$.
• So, $m = 19$ and $n = 21$.

$n^2 - m^2 = (n - m)(n + m)$

$n^2-m^2=(21-19)(21+19)=2\times 40=\mathbf{80}$

$P(A) = 0.7$ $P(B) = 0.4$ $P(A \cap \overline{B}) = 0.5$

$$\begin{array}{l} P(A) - P(A \cap B) = 0.5 \\ P(A \cap B) = 0.2 \\ \end{array}$$

$P\left(\frac{B}{A\cup {\displaystyle \stackrel{-}{B}}}\right)=\frac{P(B\cap (A\cup \stackrel{-}{B}\left)\right)}{P(A\cup \stackrel{-}{B})}=\frac{0.2}{0.8}=\frac{1}{4}$

WHERE

$$\begin{array}{l}P(A\cup \bar{B})=P(A)+P(\bar{B})-P(A\cap \bar{B})\\ =0.7+0.6-0.5\\ =0.8\end{array}$$

$$\begin{gathered} P\left(A\right)=\frac{4}{36}=\frac{1}{9},P\left(B\right)=\frac{5}{36} \\ P\left(\text{ A wins }\right)=\frac{4}{36}+\frac{32}{36}·\frac{31}{36}·\frac{4}{36}+\dots ...\infty \\ P\left(\text{ A wins }\right)=\frac{\frac{4}{36}}{1-\frac{32}{36}·\frac{31}{36}} \\ =\frac{\frac{1}{9}}{1-\frac{8}{9}·\frac{31}{36}} \\ =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19} \end{gathered}$$

\(\begin{aligned}
& P(A)=\frac{4}{36}=\frac{1}{9}, P(B)=\frac{5}{36} \\
& P(\text { Awins })=\frac{4}{36}+\frac{32}{36} \cdot \frac{31}{36} \cdot \frac{4}{36}+\ldots \ldots \text { to } \propto \\
& =\frac{\frac{4}{36}}{2-\frac{32}{36} \cdot \frac{31}{36}}=\frac{\frac{1}{9}}{1-\frac{8}{9} \cdot \frac{31}{36}}=\frac{\frac{1}{9}}{1-\frac{62}{81}} \\
& =\frac{9}{19}
\end{aligned}\)

\(P(A \cap B)=\frac{1}{10}\)

\(\begin{aligned}
& P(A / B)+P(B / A)=\frac{7}{12} \\
& \text { and } P(A / B) \cdot P(B / A)=\frac{1}{12} \\
& \frac{P(A \cup B)}{P(B)} \cdot \frac{P(A \cap B)}{P(A)}=\frac{1}{12} \\
& \Rightarrow P(A) \cdot P(B)=12\left(\frac{1}{10}\right)^2=\frac{12}{100} \\
& P(A \cap B)\left[\frac{1}{P(A)}+\frac{1}{P(B)}\right]=\frac{7}{12} \\
& P(A)+P(B)=\frac{7}{10} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}=\frac{P(\overline{A \cap B})}{P(\overline{A \cup B})}=\frac{1-P(A \cap B)}{1-P(A \cup B)} \\
& \frac{1-\frac{1}{10}}{1-\frac{6}{10}} =\frac{\frac{9}{10}}{\frac{4}{10}}=\frac{9}{4}
\end{aligned}\)

$$\begin{gathered} (i{)}^{k_1}+(i{)}^{k_2}\text{ is non zero } \\ \Rightarrow k_1:4{\lambda }_1+r_1, r_1\in \{0,1,2,3\} \\ {K}_2:4{\lambda }_2+r_2, r_2\in \{0,1,2,3\} \\ The pairs to get zero will be (1,-1),(i,-i) \\ \Rightarrow for (1, -1) pair \\ (r_1,r_2)\in \left\{\right(2,0),(0,2\left)\right\} \\ \Rightarrow Also for (i,-i) pair, \\ (r_1,r_2)\in \left\{\right(1,3),(3,1\left)\right\} \\ \text{ probablity }\left(i^{k_1}+i^{k_2}\ne 0\right) \\ =1-\text{ probablity }\left(i^{k_1}+i^{k_2}=0\right) \\ =1-\frac{4}{16}=\frac{12}{16}=\frac{3}{4} \end{gathered}$$

$$\begin{gathered} A=\left[\begin{array}{cc}\dot{a}& b\\ c& d\end{array}\right], \mathrm{where} a,b,c,d \in \{0,1\} \\ \mathrm{then} \mathrm{possible} \mathrm{value} \mathrm{of} \left|A\right|=0,1,-1 \\ n\left(A\right) = 16 \end{gathered}$$

$$\begin{gathered} \mathrm{for} x=-1 \Rightarrow A=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{ll}0& 1\\ 1& 1\end{array}\right],\left[\begin{array}{ll}1& 1\\ 1& 0\end{array}\right] \\ \& \mathrm{for} x=1 \Rightarrow A=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right],\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right] \end{gathered}$$

Remaining $10$ matrices will occur for $x=0$

Now, var(x) = $\frac{6}{16}-0=\frac{3}{8}$

Given, Bag 1: 6 black & 4 White balls

Bag II: 4 black and 6 white

Ball III: 5 black and 5 white.

$$\begin{gathered} P\left(B_1\right)=\frac{1}{3}, P\left(B_2\right)=\frac{1}{3}, P\left(B_3\right)=\frac{1}{3} \\ P\left(\frac{w}{B_1}\right)=\frac{4}{10}; P\left(\frac{w}{B_2}\right)=\frac{6}{10}; P\left(\frac{w}{B_3}\right)=\frac{5}{10} \\ Now by Baye\text{'}s Theorem, \\ P\left(\frac{B_2}{w}\right)=\frac{\frac{1}{3}\times \frac{6}{10}}{\frac{1}{3}\times \frac{4}{10}+\frac{1}{3}\times \frac{6}{10}+\frac{1}{3}\times \frac{5}{10}}=\frac{6}{15}=\frac{2}{5} \end{gathered}$$

There are 3 bad oranges and 7 good oranges.

∴ X = number of bad oranges drawn.

Here we need to draw 2 oranges, then the possible random variable are as follows:

Now,

$$\begin{gathered} Var\left(X\right)=\sum _{x=0}^2{X}^2.P\left(X\right)-{\left[\sum _{x=0}^{2}X.P\left(X\right)\right]}^2 \\ =\left[0^2·\frac{{}^{7}C_2}{{}^{10}C_2}+1^2·\left(\frac{3\times 7}{{}^{10}C_2}\right)+2^2\left(\frac{3}{{}^{10}C_2}\right)\right] \\ -\left[{\left(0+1·\frac{3\times 7}{{}^{10}C_2}+2·\frac{3}{{}^{10}C_2}\right)}^2\right] \\ =\frac{28}{75} \end{gathered}$$

Required probability \(=1-(\) product not divisible by 3\()\)

total numbers from the given set which are multiple of \(3\) are \(16\)

total numbers from the given set which are not multiple of \(3\) are \(50-16=34\)

Required probability

$$\begin{gathered} =1-\frac{{ }^{34}{C}_2}{{ }^{50}{C}_2} \\ =1-\frac{561}{1225} \\ =\frac{664}{1225} \end{gathered}$$

1st die
\(1 \rightarrow 2\)
\(2 \rightarrow 1\)
\(3 \rightarrow 2\)
\(4 \rightarrow 1\)

2nd die

\(
\begin{aligned}
& 1 \rightarrow 2 \\
& 2 \rightarrow 2 \\
& 3 \rightarrow 1 \\
& 4 \rightarrow 1
\end{aligned}
\)

\(
\begin{aligned}
& P(\text { Sum is } 4 \text { or } 5)=P(1,3)+P(2,2)+P(3,1) \\
&+P(1,4)+P(2,3)+P(3,2)+P(4,1) \\
&=\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}+ \\
&\left.\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}=\frac{17}{36}\right)
\end{aligned}
\)

Let \(K=\) student knows the answer,

\(G=\) student guesses,

\(C=\) student's answer is correct.

Let \(P(K)=p\), so \(P(G)=1-p\).

Given:

\(P(C \mid K)=1\),

\(P(C \mid G)=\frac{1}{2}\),

\(P(G \mid C)=\frac{1}{6}\)

Compute:

\(P(C)=P(C \mid K) P(K)+P(C \mid G) P(G)\)

\(=p+\frac{1}{2}(1-p)=\frac{1+p}{2}\)

\(P(G \cap C)=P(C \mid G) P(G)=\frac{1}{2}(1-p)\)

\(P(G \mid C)=\frac{P(G \cap C)}{P(C)}\)

\(=\frac{\frac{1}{2}(1-p)}{\frac{1+p}{2}}=\frac{1-p}{1+p}=\frac{1}{6}\)

\(6(1-p)=1+p \Rightarrow 6-6 p=1+p\)

\(\Rightarrow 5=7 p \Rightarrow p=\frac{5}{7}\)

\( (i)^{k_1}+(i)^{k_2} \) is non zero.
\( \Rightarrow k_1: 4 \lambda_1+r_1, r_1 \in\{0,1,2,3\} \)
\( \Rightarrow k_2: 4 \lambda_2+r_2, r_2 \in\{0,1,2,3\} \)
The pairs to get zero will be \((1,-1),(i,-i) \)
\( \Rightarrow \) for \((1,-1) \) pair.
\( \left(r_1, r_2\right) \in\{(2,0),(0,2)\} \)
\( \Rightarrow \) Also for \((i,-i) \) pair,
\( \left(r_1, r_2\right) \in\{(1,3),(3,1)\} \)
probability \(\left(i^{k_1}+i^{k_2} \neq 0\right) \)
\(=1-\text{probability }\left(i^{k_1}+i^{k_2}=0\right) \)
\(=1-\frac{4}{16}=\frac{12}{16}=\frac{3}{4}\)

Probability that the electronic component fails when first used is
$P\left(F\right)=0.10$.

$$\begin{gathered} \text{ Therefore, }P\left(F^{\text{'}}\right)=1-P\left(F\right) \\ =0.90 \end{gathered}$$

Let E be the event that a new component will last for one year.

$\text{ Then, }P\left(E\right)=P\left(F\right)P\left(\frac{E}{F}\right)+P\left(F^{\text{'}}\right)P\left(\frac{E}{F^{\text{'}}}\right)$

$$\begin{gathered} =0.10\times 0+0.90\times 0.99 \\ =0.891. \end{gathered}$$

\(P(A \cap B)=\frac{1}{10}\)

\(\begin{aligned}
& P(A / B)+P(B / A)=\frac{7}{12} \\
& \text { and } P(A / B) \cdot P(B / A)=\frac{1}{12} \\
& \frac{P(A \cup B)}{P(B)} \cdot \frac{P(A \cap B)}{P(A)}=\frac{1}{12} \\
& \Rightarrow P(A) \cdot P(B)=12\left(\frac{1}{10}\right)^2=\frac{12}{100} \\
& P(A \cap B)\left[\frac{1}{P(A)}+\frac{1}{P(B)}\right]=\frac{7}{12} \\
& P(A)+P(B)=\frac{7}{10} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}=\frac{P(\overline{A \cap B})}{P(\overline{A \cup B})}=\frac{1-P(A \cap B)}{1-P(A \cup B)} \\
& \frac{1-\frac{1}{10}}{1-\frac{6}{10}} =\frac{\frac{9}{10}}{\frac{4}{10}}=\frac{9}{4}
\end{aligned}\)

$$\begin{gathered} \text{Given,} \\ (i{)}^{k_1}+(i{)}^{k_2}\text{ is non zero } \\ \Rightarrow k_1:4{\lambda }_1+r_1, r_1\in \{0,1,2,3\} \\ {k}_2:4{\lambda }_2+r_2, r_2\in \{0,1,2,3\} \\ \text{ The pairs to get zero will be } \\ (1,-1),(i,-i), \\ \text{Now, for} (1,-1)\text{ pair}, \\ \Rightarrow \left(r_1,r_2\right)\in \left\{\right(2,0),(0,2\left)\right\} \\ \text{for} \left(i, -i\right) \text{pair} \\ \Rightarrow \left(r_1,r_2\right)\in \left\{\right(1,3),(3,1\left)\right\} \\ \Rightarrow \text{ Probablity }\left(i^{k_1}+i^{k_2}\ne 0\right) \\ =1-\text{ Probablity }\left(i^{k_1}+i^{k_2}\ne 0\right) \\ \Rightarrow \text{Probablity}=1-\frac{4}{16}=\frac{12}{16}=\frac{3}{4} \end{gathered}$$

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 combinations. Assuming they are equally likely:

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 :

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

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:

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

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

$=\frac{{ }^9{C}_3}{{ }^{10}{C}_3+{ }^9{C}_3+{ }^8{C}_3+\dots +{ }^3{C}_3}$

$=\frac{{ }^9{C}_3}{{ }^{11}{C}_4}=\frac{14}{55}$

Here matrix of order $2\times 2$ is given, and $\left|A\right|\ne 0$

Then $n\left(S\right)=2\times 2\times 2\times 2=16$

\(E=A\) is an invertible matrix

Now

$$\begin{gathered} \left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right],\left[\begin{array}{ll}1& 1\\ 1& 1\end{array}\right],\left[\begin{array}{ll}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{ll}1& 1\\ 0& 0\end{array}\right],\left[\begin{array}{ll}1& 0\\ 1& 0\end{array}\right],\left[\begin{array}{ll}0& 0\\ 1& 1\end{array}\right] \\ \left[\begin{array}{ll}0& 1\\ 0& 1\end{array}\right],\left[\begin{array}{ll}0& 0\\ 1& 0\end{array}\right],\left[\begin{array}{ll}0& 0\\ 0& 1\end{array}\right],\left[\begin{array}{ll}0& 1\\ 0& 0\end{array}\right] \end{gathered}$$

So these total \(10\) type of matrix we can't take, as these are singular matrices

$$\begin{gathered} \therefore P\left(\overline{E}\right)=\frac{10}{16}=\frac{5}{8} \\ \therefore P\left(E\right)=1-P\left(\overline{E}\right)=1-\frac{5}{8}=\frac{3}{8} \end{gathered}$$

Probability distribution

$$\begin{gathered} \mu =\sum {\mathrm{x}}_{\mathrm{i}}{\mathrm{p}}_{\mathrm{i}}=\frac{1}{2} \\ {\sigma }^2=\sum x_i^2{p}_i-{\mu }^2 \\ =\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\ 64\left(\mu +{\sigma }^2\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48 \end{gathered}$$

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 combinations. Assuming they are equally likely:

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 :

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

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:

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

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

$=\frac{{ }^9{C}_3}{{ }^{10}{C}_3+{ }^9{C}_3+{ }^8{C}_3+\dots +{ }^3{C}_3}$

$=\frac{{ }^9{C}_3}{{ }^{11}{C}_4}=\frac{14}{55}$

Bag \(1\) contains \(4\) white and \(5\) black balls

Bag \(2\) contains \(n\) white and \(3\) black balls

Given \(\mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{Bag} 2}\right)=\frac{29}{45}\)

so now we get

$$\begin{gathered} \Rightarrow \mathrm{P}\left(\frac{ \mathrm{W}}{{ \mathrm{B}}_1}\right)\times \mathrm{P}\left(\frac{ \mathrm{W}}{{ \mathrm{B}}_2}\right)+\mathrm{P}\left(\frac{ \mathrm{B}}{{ \mathrm{B}}_1}\right)\times \mathrm{P}\left(\frac{ \mathrm{W}}{{ \mathrm{B}}_2}\right)=\frac{29}{45} \\ \Rightarrow \frac{4}{9}\times \frac{n+1}{n+4}+\frac{5}{9}\times \frac{n}{n+4}=\frac{29}{45} \\ \Rightarrow \mathrm{n}=6 \end{gathered}$$

Let \(E_1\) be the event of selecting a fair coin

Let \(E_2\) be the event of selecting the two-headed coin

Let \(A\) be the event of getting a 'Head' when the selected coin is tossed

\(P\left(E_1\right)=\frac{N}{N+1}\)

\(P\left(E_2\right)=\frac{1}{N+1}\)

\(P\left(A \mid E_1\right)=\frac{1}{2}\)

\(P\left(A \mid E_2\right)=1\)

According to the Law of Total Probability,

\(P(A)=P\left(E_1\right) \cdot P\left(A \mid E_1\right)+P\left(E_2\right) \cdot P\left(A \mid E_2\right)\)

\(\frac{9}{16}=\left(\frac{N}{N+1} \cdot \frac{1}{2}\right)+\left(\frac{1}{N+1} \cdot 1\right)\)

\(\frac{9}{16}=\frac{N+2}{2 N+2}\)

\(N=7\)

$$\begin{gathered} A=\left[\begin{array}{cc}\dot{a}& b\\ c& d\end{array}\right], \mathrm{where} a,b,c,d \in \{0,1\} \\ \mathrm{then} \mathrm{possible} \mathrm{value} \mathrm{of} \left|A\right|=0,1,-1 \\ n\left(A\right) = 16 \end{gathered}$$

$$\begin{gathered} \mathrm{for} x=-1 \Rightarrow A=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{ll}0& 1\\ 1& 1\end{array}\right],\left[\begin{array}{ll}1& 1\\ 1& 0\end{array}\right] \\ \& \mathrm{for} x=1 \Rightarrow A=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right],\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right] \end{gathered}$$

Remaining $10$ matrices will occur for $x=0$

Now, var(x) = $\frac{6}{16}-0=\frac{3}{8}$

$\begin{array}{|lcc|}\hline & \text{ A }& \text{ B }\\ \text{ Manufactured }& 60\%& 40\%\\ \text{ Standard quality }& 80\%& 90\%\\ \hline\end{array}$

\(A\) : Found Standard quality
$E_1$ : Manufactured at \(B\)
$E_2$ : Manufactured at \(A\)
$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{40}{100}$
$\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{60}{100}$
$\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_1\right)=\frac{90}{100}$
$\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_2\right)=\frac{80}{100}$
$\because \mathrm{P}\left({\mathrm{E}}_1|\mathrm{A}\right)=\frac{\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_1\right)\mathrm{P}\left({\mathrm{E}}_1\right)}{\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_1\right)\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_2\right)\mathrm{P}\left({\mathrm{E}}_2\right)}=p$

\(p=\frac{\frac{4}{10} \times \frac{9}{10}}{\frac{6}{10} \times \frac{8}{10}+\frac{4}{10} \times \frac{9}{10}}\)
$\therefore 126p=54$

$\begin{array}{|lcc|}\hline & \text{ A }& \text{ B }\\ \text{ Manufactured }& 60\%& 40\%\\ \text{ Standard quality }& 80\%& 90\%\\ \hline\end{array}$

\(A\) : Found Standard quality
$E_1$ : Manufactured at \(B\)
$E_2$ : Manufactured at \(A\)
$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{40}{100}$
$\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{60}{100}$
$\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_1\right)=\frac{90}{100}$
$\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_2\right)=\frac{80}{100}$
$\because \mathrm{P}\left({\mathrm{E}}_1|\mathrm{A}\right)=\frac{\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_1\right)\mathrm{P}\left({\mathrm{E}}_1\right)}{\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_1\right)\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left( \mathrm{A}|{\mathrm{E}}_2\right)\mathrm{P}\left({\mathrm{E}}_2\right)}=p$

\(p=\frac{\frac{4}{10} \times \frac{9}{10}}{\frac{6}{10} \times \frac{8}{10}+\frac{4}{10} \times \frac{9}{10}}\)
$\therefore 126p=54$

Required probability \(=1-(\) product not divisible by 3\()\)

total numbers from the given set which are multiple of \(3\) are \(16\)

total numbers from the given set which are not multiple of \(3\) are \(50-16=34\)

Required probability

$$\begin{gathered} =1-\frac{{ }^{34}{C}_2}{{ }^{50}{C}_2} \\ =1-\frac{561}{1225} \\ =\frac{664}{1225} \end{gathered}$$

$U$: Selecting an unbiased coin. $P(U) = \frac{19}{20}$

$B$: Selecting the biased (two-headed) coin. $P(B) = \frac{1}{20}$

$H$: Getting a head on the toss.

$P(H|U) = \frac{1}{2}$

$P(H|B) = 1$

$P(U|H) = \frac{P(U) \cdot P(H|U)}{P(U) \cdot P(H|U) + P(B) \cdot P(H|B)}$

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

Given $P(U|H) = \frac{m}{n} = \frac{19}{21}$, where $\text{gcd}(19, 21) = 1$.

So, $m = 19$ and $n = 21$.

$n^2 - m^2 = (n - m)(n + m)$

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

Total number of ways to divide \(12\) distinct balls into \(6\) unordered pairs is \(\frac{12!}{(2!)^6 6!}\)

For every pair to contain one blue and one green ball, each blue ball must be paired with one green ball.

Number of favourable pairings \(=6!\)

Required probability \(=\frac{6!}{\frac{12!}{(2!)^6 6!}}=\frac{6!(2!)^6 6!}{12!}\)

Required probability \(=\frac{720 \times 64 \times 720}{479001600}=\frac{16}{231}\)

$$\begin{gathered} P\left(A\right)=\frac{4}{36}=\frac{1}{9},P\left(B\right)=\frac{5}{36} \\ P\left(\text{ A wins }\right)=\frac{4}{36}+\frac{32}{36}·\frac{31}{36}·\frac{4}{36}+\dots ...\infty \\ P\left(\text{ A wins }\right)=\frac{\frac{4}{36}}{1-\frac{32}{36}·\frac{31}{36}} \\ =\frac{\frac{1}{9}}{1-\frac{8}{9}·\frac{31}{36}} \\ =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19} \end{gathered}$$

Let $P(X=0) = P(X=1) = p$. Let $P(X=2) = P(X=3) = q$.

Since the sum of all probabilities must be \(1\):

$p + p + q + q = 1$

$\Rightarrow 2p+2q=1$

$\Rightarrow p+q=\frac{1}{2}⟹q=\frac{1}{2}-p$

$E(X) = \sum x \cdot P(X=x)$

$\Rightarrow E\left(X\right)=0\left(p\right)+1\left(p\right)+2\left(q\right)+3\left(q\right)$

$\Rightarrow E\left(X\right)=p+5q$

Substitute $q = \frac{1}{2} - p$:

$E\left(X\right)=p+5(\frac{1}{2}-p)=p+\frac{5}{2}-5p$$=\frac{5}{2}-4p$

$E(X^2) = \sum x^2 \cdot P(X=x)$

$\Rightarrow E\left(X^2\right)=0^2\left(p\right)+1^2\left(p\right)+2^2\left(q\right)+3^2\left(q\right)$

$\Rightarrow E\left(X^2\right)=p+4q+9q=p+13q$

Substitute $q = \frac{1}{2} - p$:

$E\left(X^2\right)=p+13(\frac{1}{2}-p)=p+\frac{13}{2}-13p$$=\frac{13}{2}-12p$

Since given that $E(X^2) = 2E(X)$:

So, $\frac{13}{2} - 12p = 2\left(\frac{5}{2} - 4p\right)$ $\Rightarrow \frac{13}{2}-12p=5-8p$

$\Rightarrow \frac{13}{2}-5=12p-8p$ $\Rightarrow \frac{13}{2}-\frac{10}{2}=4p$ $\Rightarrow \frac{3}{2}=4p$ $\Rightarrow p=\frac{3}{8}$

Hence, $8p - 1 = 8\left(\frac{3}{8}\right) - 1$ $= 3 - 1$ $=2$

Given set \(=\{1,2,3, \ldots \ldots . .50\}\)
\(\mathrm{P}(\mathrm{A})=\) Probability that number is multiple of 4
\(\mathrm{P}(\mathrm{B})=\) 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

\( \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\frac{4}{50}, \mathrm{P}(\mathrm{~B} \cap \mathrm{C})=\frac{1}{50} \)
\( \mathrm{P}(\mathrm{~A} \cap \mathrm{C})=\frac{1}{50} \)
\( \mathrm{P}(\mathrm{~A} \cap \mathrm{~B} \cap \mathrm{C})=0\)

Thus

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

Given set \(=\{1,2,3, \ldots \ldots . .50\}\)
\(\mathrm{P}(\mathrm{A})=\) Probability that number is multiple of 4
\(\mathrm{P}(\mathrm{B})=\) 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

\( \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\frac{4}{50}, \mathrm{P}(\mathrm{~B} \cap \mathrm{C})=\frac{1}{50} \)
\( \mathrm{P}(\mathrm{~A} \cap \mathrm{C})=\frac{1}{50} \)
\( \mathrm{P}(\mathrm{~A} \cap \mathrm{~B} \cap \mathrm{C})=0\)

Thus

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

Let \(B, S, C\) be the events that the candidate travels by bus, scooter and car, respectively.

\(P(B)=\frac{2}{5}, P(S)=\frac{1}{5}, P(C)=\frac{2}{5}\)

Let \(L\) be the event that the candidate reaches late.

\(P(L\mid B)=\frac{1}{5}, P(L\mid S)=\frac{1}{3}, P(L\mid C)=\frac{1}{4}\)

By Bayes’ theorem,

\(P(B\mid L)=\frac{P(B)P(L\mid B)}{P(B)P(L\mid B)+P(S)P(L\mid S)+P(C)P(L\mid C)}\)

\(P(B\mid L)=\frac{\frac{2}{5}\cdot\frac{1}{5}}{\frac{2}{5}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{3}+\frac{2}{5}\cdot\frac{1}{4}}\)

\(P(B\mid L)=\frac{\frac{2}{25}}{\frac{2}{25}+\frac{1}{15}+\frac{1}{10}}\)

\(P(B\mid L)=\frac{\frac{2}{25}}{\frac{12+10+15}{150}}\)

\(P(B\mid L)=\frac{\frac{2}{25}}{\frac{37}{150}}=\frac{2}{25}\cdot\frac{150}{37}=\frac{12}{37}\)

Given, $P\left(A\right)=\frac{1}{5},P\left(A\cup B\right)=\frac{7}{10}$

Also, A and B are independent events,

$\therefore P\left(A\cap B\right)=P\left(A\right)\cdot P\left(B\right)$

$\Rightarrow P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)$

$=P\left(A\right) ·P\left(B\right)$

$\Rightarrow \frac{1}{5}+P\left(B\right)-\frac{7}{10}=\frac{1}{5}\times P\left(B\right)$