Probability

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} P(A) & =\frac{2}{3}, \quad P(B)=\frac{3}{4} \\ P(\text { not } A)=P\left(A^{\prime}\right) & =1-\frac{2}{3}=\frac{1}{3} \\ P\left(B^{\prime}\right)=1-\frac{3}{4} & =\frac{1}{4} \\ P(A \text { or } B) \text { or both }) & =1-P\left(A^{\prime}\right) P\left(B^{\prime}\right) \\ & =1-\frac{1}{3} \times \frac{1}{4}=\frac{11}{12} \end{aligned} \end{aligned} $

Given set $\{1,2,3,4,5,6,7\}$

Two number can be taken from set $=49$ ways.

Now, for $l x^2+m x+1>0$

$ m^2-4 l<0 \Rightarrow m^2<4 l $

If $l=1 \Rightarrow m^2<4 \Rightarrow m=1$

$ \begin{aligned} & l=2 \Rightarrow m^2<8 \Rightarrow m=1,2 \\ & l=3 \Rightarrow m^2<12 \Rightarrow m=1,2,3 \\ & l=4 \Rightarrow m^2<16 \Rightarrow m=1,2,3 \\ & l=5 \Rightarrow m^2<20 \Rightarrow m=1,2,3,4 \\ & l=6 \Rightarrow m^2<24 \Rightarrow m=1,2,3,4 \\ & l=7 \Rightarrow m^2<28 \Rightarrow m=1,2,3,4,5 \end{aligned} $

$\therefore$ Required probability $=\frac{22}{49}=\frac{22}{7^2}$

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} & \quad P(A \text { wins })=0.6 \\ & P(A \text { loses })=0.3 \\ & P(\text { game draw })=0.1 \\ & P(\text { A win at least two game }) \\ & \Rightarrow \quad P(W W W)+P(W W L)+P(W L W) \\ & \quad+P(L W W)+P(W W D) \\ & \quad+P(D W W)+P(W D W) \end{aligned} \end{aligned} $

$ \begin{aligned} & =(0.6)^3+3(0.6)^2(0.3)+3(0.6)^2(0.1) \\ & =(0.6)^2(0.6+0.9+0.3) \\ & =0.36(1.8)=0.648=\frac{648}{1000}=\frac{81}{125} \end{aligned} $

Given that

$U r n U_1$ contains 5 red, 3 white, 2 black balls

$U r n U_2$ contains 4 red, 4 white, 2 black balls

$U r n U_3$ contains 3 red, 4 white, 3 black balls

Total number of balls $=30$

Non-black balls = 23

Total probability of selecting one $\operatorname{Ur} n=\frac{1}{3}$

Probability of getting not a black balls from $U_1$ or $U_2$ and $U_3$.

$ =\frac{1}{3}\left(\frac{8}{10}+\frac{8}{10}+\frac{7}{10}\right) $

$\therefore$ Required probability $=\frac{1}{3} \times \frac{23}{10}=\frac{23}{30}$

$ \begin{aligned} &\text { We know, } \Sigma P_i=1\\ &\begin{aligned} & \therefore \quad 3 k+5 k+3 k^2+4 k^2+k+3 k^2=1 \\ & \Rightarrow \quad 10 k^2+9 k-1=0 \\ & \Rightarrow \quad(10 k-1)(k+1)=0 \\ & \Rightarrow \quad k=\frac{1}{10} \\ & \quad k \neq-1 \\ & P(X \leq 2)=P(X=0)+P(X=2)+P(X=3) \\ & =3 k+5 k+3 k^2=\frac{8}{10}+\frac{3}{100}=\frac{83}{100} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \text { Variance }=2=\lambda\\ &\sigma=2 \end{aligned} $

$ \begin{aligned} & P(X=k)=\frac{e^{-\lambda} \lambda^k}{k!} \\ & \begin{aligned} P(X \geq 3) & =1-P(X=0)-P(X=1) \\ -P(X & =2) \\ & =1-\left(\frac{e^{-2} 2^0}{0!}+\frac{e^{-2} 2^1}{1!}+\frac{e^{-2} 2^2}{2!}\right) \\ & =1-5 e^{-2}=\frac{e^2-5}{e^2} \end{aligned} \end{aligned} $

Given, $P(A)=\frac{2}{5}, P(B)=\frac{3}{4}$

$P($ Problem is solved $)=1-P($ Problem is not solve)

$ \begin{aligned} P(A \cup B) & =1-P\left(A^{\prime} \cap B^{\prime}\right) \\ & =1-P(A) \cdot P(B) \\ & =1-\frac{3}{5} \times \frac{1}{4}=\frac{20-3}{20}=\frac{17}{20} \end{aligned} $

$A=$ getting a sum $>14$

$B=$ sum is multiple of 3

$ \begin{aligned} & n(A)=14 \quad P(A)=\frac{14}{216} \\ & \begin{array}{l} n(B)=78 \quad P(B)=\frac{78}{216} \\ n(A \cap B)=11 P(A \cap B)=\frac{11}{216} \\ \begin{aligned} \because P(A & \cap \bar{B})+P(\bar{A} \cap B) \\ & =P(A)+P(B)-2 P(A \cap B) \\ & =\frac{14}{216}+\frac{78}{216}-\frac{11 \times 2}{216} \\ & =\frac{92-22}{216}=\frac{70}{216}=\frac{35}{108} \end{aligned} \end{array} \end{aligned} $

Let $E_1$ mean the bulb was made by unit $A$.

Let $E_2$ mean the bulb was made by unit $B$.

Let $E_3$ mean the bulb was made by unit $C$.

Let $E$ mean the bulb is defective.

The chance a bulb comes from each unit is:
$P\left(E_1\right) = \frac{25}{100}$ (from $A$), $P\left(E_2\right) = \frac{35}{100}$ (from $B$), and $P\left(E_3\right) = \frac{40}{100}$ (from $C$).

The chance a bulb is defective from each unit is:
$P\left(\frac{E}{E_1}\right) = \frac{5}{100}$, $P\left(\frac{E}{E_2}\right) = \frac{4}{100}$, $P\left(\frac{E}{E_3}\right) = \frac{2}{100}$.

To find how likely it is a defective bulb came from unit $B$, use Bayes’ theorem:

$ P\left(\frac{E_2}{E}\right) = \frac{P\left(E_2\right) P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right) + P\left(E_2\right) P\left(\frac{E}{E_2}\right) + P\left(E_3\right) P\left(\frac{E}{E_3}\right) } $

Now plug in the numbers:

$ P\left(\frac{E_2}{E}\right) = \frac{\frac{35}{100} \times \frac{4}{100}}{\frac{25}{100} \times \frac{5}{100} + \frac{35}{100} \times \frac{4}{100} + \frac{40}{100} \times \frac{2}{100}} $

The top ($35 \times 4 = 140$) and the bottom adds up ($25 \times 5 = 125$, $35 \times 4 = 140$, $40 \times 2 = 80$) so:

$ P\left(\frac{E_2}{E}\right) = \frac{140}{125 + 140 + 80} $

$ P\left(\frac{E_2}{E}\right) = \frac{140}{345} = \frac{28}{69} $

$ \begin{array}{ccccc} \hline \boldsymbol{X}_{\boldsymbol{i}} & \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{X}_{\boldsymbol{i}} \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{X}_{\boldsymbol{i}}{ }^{\boldsymbol{2}} & \boldsymbol{P}_{\boldsymbol{i}} \boldsymbol{X}_{\boldsymbol{i}}{ }^{\boldsymbol{2}} \\ \hline 1 & \alpha & \alpha & 1 & \alpha \\ \hline 2 & \alpha & 2 \alpha & 4 & 4 \alpha \\ \hline 3 & \alpha & 3 \alpha & 9 & 9 \alpha \\ \hline 4 & \beta & 4 \beta & 16 & 16 \beta \\ \hline 5 & \beta & 5 \beta & 25 & 25 \beta \\ \hline 6 & 0.3 & 1.8 & 36 & 10.8 \\ \hline \end{array} $

$ \begin{array}{ll} \because & \Sigma P_i=1 \\ & 3 \alpha+2 \beta+0.3=1 \\ & 3 \alpha+2 \beta=0.7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \because & \mu=\Sigma X_i P_i \\ & 6 \alpha+9 \beta+1.8=4.2 \\ & 2 \alpha+3 \beta=0.8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Solving Eqs. (i) and (ii) we get,

$ \begin{aligned} \operatorname{var}(x) & =\sigma^2=\Sigma P_i X_i^2-(\mu)^2 \\ & \sigma^2+\mu^2=\Sigma P_i X_i^2 \\ & =14 \alpha+41 \beta+10.8 \\ & =1.4+8.2+10.8=20.4 \end{aligned} $

$X=$ Students get distinction

$ \begin{gathered} p=\frac{2}{3}, q=1-p=\frac{1}{3}, n=5 \\ P(X \geq 3)=P(X=3)+P(X=4)+P(X=5)\end{gathered} $

We know that, $P(X=r)={ }^n C_r p^r q^{n-r}$

$ \begin{aligned} & P(X=3)={ }^5 C_3\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=10 \times \frac{8}{3^5}=\frac{80}{3^5} \\ & P(X=4)={ }^5 C_4\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3}=\frac{5 \times 16}{3^5}=\frac{80}{3^5} \\ & P(X=5)={ }^5 C_5\left(\frac{2}{3}\right)^5=\frac{32}{3^5} \\ & P(X \geq 3)=\frac{192}{3^5}=\frac{64}{3^4}=\frac{64}{81} \end{aligned} $

Given, $P(A \cup B)=\frac{3}{4} P(A \cap B)=\frac{1}{4}$

$ P(\bar{A})=\frac{2}{3} \Rightarrow P(A)=\frac{1}{3} $

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

Now, $P(B)=\frac{2}{3}$

Now, $P(\bar{A} \cap B)=P(B)-P(A \cap B)$

$ =\frac{2}{3}-\frac{1}{4}=\frac{8-3}{12}=\frac{5}{12} $

$A=$ Both cards are black

$B=$ At least one card is a face card.

$ P(A)=\frac{26}{52} \times \frac{25}{51}=\frac{650}{2652} $

$\because 20$ Black card are not a face card.

$\because P$ (two black non-face card)

$ =\frac{20}{52} \times \frac{19}{51}=\frac{380}{2652} $

$P(B)=P($ two black card $)$

$-P$ (two non-face card)

$ =\frac{650}{2652}-\frac{380}{2652}=\frac{270}{2652} $

$\because P\left(\frac{B}{A}\right)=\frac{\frac{270}{2652}}{\frac{650}{2652}}=\frac{270}{650}=\frac{27}{65}$

The person tells the truth 3 out of every 4 times.

Let’s define:
- $A$: The die really shows six.
- $B$: The die does not show six.
- $E$: The person says the die shows six.

The chance of rolling a six when throwing a die, $P(A)$, is $\frac{1}{6}$.

The chance of not rolling a six, $P(B)$, is $\frac{5}{6}$.

If a six really comes up, the chance the person will say “six” (tells the truth) is $P(E|A) = \frac{3}{4}$.

If a six did NOT come up, the chance the person wrongly says “six” (tells a lie) is $P(E|B) = \frac{1}{4}$.

We want to find: What is the chance the die REALLY shows six if the person says it is a six? This is $P(A|E)$.

Use Bayes’ Theorem:
$ P(A|E) = \frac{P(A) \cdot P(E|A)}{P(A)\cdot P(E|A) + P(B)\cdot P(E|B)} $

Put in the values:
$ P(A|E) = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} $

Solve:
$ P(A|E) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8} $

Let $E_1=$ Employee be a man $E_2=$ Employee be a woman $E=$ Employee be a technical assistants

$ \begin{aligned} Given\,\,& \mathrm{n}\left(E_1\right)=\frac{70}{100}=\frac{7}{10}, P\left(E_2\right)=\frac{3}{10} \\ & P\left(\frac{E}{E_1}\right)=\frac{3}{10} \\ & P\left(\frac{E}{E_2}\right)=\frac{15}{100} \end{aligned} $

$ \begin{aligned} P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) P\left(\frac{E}{E_1}\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right)} \\ & =\frac{\frac{7}{10} \times \frac{3}{10}}{\frac{7}{10} \times \frac{3}{10}+\frac{3}{10} \times \frac{15}{100}} \\ & =\frac{210}{210+45}=\frac{210}{255} \\ P\left(\frac{E_1}{E}\right) & =\frac{42}{51}=\frac{14}{17} \end{aligned} $

Given, $P(X=x)=K \cdot \frac{2^{2 x+1}}{(2 x+1)!}$

$ \begin{aligned} & \because \quad \sum P(X=x)=1 \\ & \Rightarrow \sum_{x=0}^{\infty} k \cdot \frac{2^{2 x+1}}{(2 x+1)!}=1 \\ & \Rightarrow 2 k \sum_{x=0}^{\infty} \frac{4^x}{(2 x+1)!}=1 \end{aligned} $

We know that $\sin h x=\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1)!}$

$ \begin{aligned} \frac{\sin h x}{x} & =\sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n+1)!} \\ \because \quad \sum_{x=0}^{\infty} \frac{4^x}{(2 x+1)!} & =\frac{\sin h(2)}{2} \\ 2 k \cdot \frac{\sin h(2)}{2} & =1 \\ k & =\frac{1}{\sin h 2}=\operatorname{cosech} 2 \end{aligned} $

Given, mean - variance $=1$

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

and given $2 P(X=2)=3 P(X=1)$

$ \begin{aligned} 2 \cdot{ }^n C_2 p^2 q^{n-2} & =3 \cdot{ }^n C_1 p q^{n-1} \\ 2 \cdot \frac{n(n-1)}{2} p & =3 \cdot n \cdot q \\ (n-1) p & =3 q=3(1-p) \\ p(n+2) & =3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Solving Eqs. (i) and (ii), we get

$ p=\frac{1}{2}, q=\frac{1}{2} \text { and } n=4 $

Now, $P(X>1)=1-P(X=0)-P(X=1)$

$ \begin{aligned} & =1-4\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4 \\ & =1-\frac{5}{16}=\frac{11}{16} \end{aligned} $

$ \because \quad n^2 P(X>1)=16 \cdot \frac{11}{16}=11 $

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 $\{1,2,3, \ldots 8\}$

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

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

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.

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:

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

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:

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:

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

$\mathrm{E}_1: \mathrm{A}$ 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.

We can solve this problem using Bayes' theorem. Let's define the following events :

$A$: The student knows the answer of a randomly chosen question.

$B$: The student guesses the answer of a randomly chosen question.

$C$: The student gives the correct answer to a question.

We are asked to find the probability that the student knows the answer of a randomly chosen question, i.e., $P(A | C)$. According to Bayes' theorem,

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

We need to find the values of $P(C | A)$, $P(A)$, and $P(C)$. Let's proceed step by step :

1. $P(C | A)$ is the probability that the student gives the correct answer given that he knows the answer. Since he knows the answer, he will definitely give the correct answer. Therefore,

$ P(C | A) = 1 $

2. $P(A)$ is the probability that the student knows the answer of a randomly chosen question. We don't have this value directly, and we will denote it as $P(A)$. Likewise, the probability that the student guesses the answer is $P(B) = 1 - P(A)$.

3. $P(C)$ is the total probability that the student gives the correct answer. This can be calculated using the Law of Total Probability :

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

Where:

$ P(C | B) = \frac{1}{2} $

(The probability that the student gives the correct answer given that he guessed it), and

$ P(B | C) = \frac{1}{6} $

(The probability that the answer was guessed given the student's answer is correct).

We also have the relation :

$ P(B | C) = \frac{P(C | B) \cdot P(B)}{P(C)} $

Substitute the known values and simplify :

$ \frac{1}{6} = \frac{\frac{1}{2} \cdot P(B)}{P(C)} $

Solving for $P(C)$ gives :

$ P(C) = 3 \cdot P(B) $

From $P(B) = 1 - P(A)$, we get :

$ P(C) = 3 \cdot (1 - P(A)) $

We also know from our Law of Total Probability calculation :

$ P(C) = P(A) + \frac{1}{2} \cdot (1 - P(A)) $

Equate the two expressions for $P(C)$ :

$ P(A) + \frac{1}{2} \cdot (1 - P(A)) = 3 \cdot (1 - P(A)) $

Simplify this equation:

$ P(A) + \frac{1}{2} - \frac{1}{2} \cdot P(A) = 3 - 3 \cdot P(A) $

$ P(A) - \frac{1}{2} \cdot P(A) + \frac{1}{2} = 3 - 3 \cdot P(A) $

$ \frac{1}{2} \cdot P(A) + \frac{1}{2} = 3 - 3 \cdot P(A) $

$ \frac{7}{2} \cdot P(A) = \frac{5}{2} $

$ P(A) = \frac{5}{7} $

Thus, the probability that the student knows the answer of a randomly chosen question is :

$ \boxed{\frac{5}{7}} $

We have the Numbers $ 2,3,5,7,11 $ and 13.

3 Number are chosen

$ \therefore \quad n(s)={ }^{6} C_{3}+20 $

$ E= $ sum of number is divisible by 3

$ E=\{(2,37)(2,3,13)(3,5,7)(3,5,13) $

$ (3,11,13)\}(3,7,11)(2,5,11) $

$ n(E)=7 $

$ P(E)=\frac{7}{20} $

To determine the probability of getting a multiple of 3 as the sum of the numbers on the top faces of two dice, given that their sum is an odd number, we define the following:

Let $ A $ be the event that the sum is a multiple of 3.

Let $ B $ be the event that the sum is an odd number.

First, we calculate the total probabilities:

The probability of event $ A $, $ P(A) $, is the number of favorable outcomes divided by the total outcomes, i.e., $ \frac{12}{36} $.

The probability of event $ B $, $ P(B) $, is similarly $ \frac{18}{36} $.

Next, we need to determine the probability of both events occurring simultaneously, $ P(A \cap B) $.

The probability $ P(A \cap B) = \frac{6}{36} $, which represents the outcomes where the sum is both a multiple of 3 and an odd number.

Finally, we calculate the conditional probability of $ A $ given $ B $, denoted as $ P(A|B) $:

$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{36}}{\frac{18}{36}} = \frac{6}{18} = \frac{1}{3} $

Therefore, the probability of getting a multiple of 3 as the sum, given that the sum is an odd number, is $\frac{1}{3}$.

$ \Sigma p_{i}=1 $

$ \begin{array}{l} 4 k^{2}+3 k=1 \\\\ 4 k^{2}+4 k-k-1=0 \\\\ 4 k(k+1)-1(k+1)=0 \\\\ (k+1)(4 k-1)=0 \\\\ k=\frac{1}{4} \end{array} $

$ \operatorname{Var}(X)=\frac{15}{16} $

Mean $ =n p=\frac{16}{5} $

Variance $ =n p q=\frac{48}{25} $

$ \begin{array}{l} q=\frac{48}{25} \times \frac{5}{16}=\frac{3}{5} \\\\ p=\frac{2}{5} \end{array} $

$ \begin{aligned} & n \times \frac{2}{5}=\frac{16}{5} \\\\ n & =8 \\\\ P(x > 1) & =1-P(x=0)-P(x=1) \\\\ = & 1-{ }^{8} C_{0}\left(\frac{3}{5}\right)^{8}-{ }^{8} C_{1}\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)^{7} \\\\ = & 1-\left(\frac{3}{5}\right)^{7}\left\{\frac{3}{5}+\frac{16}{5}\right\} \\\\ = & 1-\left(\frac{3}{5}\right)^{7}\left(\frac{19}{5}\right) \\\\ \because \quad k & =\frac{19}{5} \\\\ \quad 5 k & =19 \end{aligned} $