Probability

To find the probability that the last ball drawn from the bag is red, consider the following:

The bag contains 2 white (W), 3 green (G), and 5 red (R) balls.

We need to calculate the probability of different combinations of drawing three balls where the last ball is red.

The possible successful sequences with a red ball at the end are:

Drawing a white, another white, and then a red (WWR)

Drawing two green balls and then a red (GGR)

Drawing a white, then a green, and then a red (WGR)

Drawing a green, then a white, and then a red (GWR)

Drawing one white, then two reds (WRR)

Drawing one green, then two reds (GRR)

Drawing a red first, then a white, then a red (RWR)

Drawing a red first, then a green, then a red (RGR)

All three drawn balls are red (RRR)

The probability calculation is as follows:

$ \begin{aligned} P(\text{last ball is red}) &= P(WWR) + P(GGR) + P(WGR) + P(GWR) \\ &\quad + P(WRR) + P(GRR) + P(RWR) + P(RGR) \\ &\quad + P(RRR) \\ &= \frac{2 \times 1 \times 5}{10 \times 9 \times 8} + \frac{3 \times 2 \times 5}{10 \times 9 \times 8} + \frac{2 \times 3 \times 5}{10 \times 9 \times 8} \\ &\quad + \frac{3 \times 2 \times 5}{10 \times 9 \times 8} + \frac{2 \times 5 \times 4}{10 \times 9 \times 8} + \frac{3 \times 5 \times 4}{10 \times 9 \times 8} \\ &\quad + \frac{5 \times 2 \times 4}{10 \times 9 \times 8} + \frac{5 \times 3 \times 4}{10 \times 9 \times 8} + \frac{5 \times 4 \times 3}{10 \times 9 \times 8} \\ &= \frac{10 + 30 + 30 + 30 + 40 + 60 + 40 + 60 + 60}{10 \times 9 \times 8} \\ &= \frac{360}{720} \\ &= \frac{1}{2} \end{aligned} $

Therefore, the probability that the last ball drawn is red is $\frac{1}{2}$.

To determine the probability that a black ball is from the first set of bags, let's define the following:

B: Probability of drawing a black ball.

$B_1$: Probability associated with drawing from the first set of bags, where each bag contains 3 white and 5 black balls.

$B_2$: Probability associated with drawing from the second set of bags, where each bag contains 6 white and 4 black balls.

Given:

Probability of choosing a bag from the first set ($P(B_1)$) is $ \frac{1}{2} $.

Probability of choosing a bag from the second set ($P(B_2)$) is also $ \frac{1}{2} $ since there are two bags in the first set and four bags in the second set but we simplify by considering overall probabilities.

Probabilities of drawing a black ball from each type of bag:

$P(B \mid B_1) = \frac{5}{8}$ for Bag 1.

$P(B \mid B_2) = \frac{4}{10} = \frac{2}{5}$ for Bag 2.

Calculation:

The total probability of drawing a black ball ($P(B)$) is calculated by adding the probabilities of drawing a black ball from each type of bag, weighted by the probability of choosing each type of bag:

$ \begin{aligned} P(B) & = P(B_1) \cdot P(B \mid B_1) + P(B_2) \cdot P(B \mid B_2) \\ & = \frac{1}{2} \times \frac{5}{8} + \frac{1}{2} \times \frac{2}{5} \\ & = \frac{5}{16} + \frac{1}{5} \\ & = \frac{41}{80} \end{aligned} $

Final Probability:

Using Bayes' Theorem, we find the probability that a black ball is from the first set of bags:

$ \begin{aligned} P(B_1 \mid B) & = \frac{P(B \mid B_1) \cdot P(B_1)}{P(B)} \\ & = \frac{\frac{5}{8} \times \frac{1}{2}}{\frac{41}{80}} \\ & = \frac{5}{16} \times \frac{80}{41} \\ & = \frac{25}{41} \end{aligned} $

Thus, the probability that the black ball is from the first set of bags is $ \frac{25}{41} $.

let $X$ be the random variable representing the number of kings drawn.

$ \begin{aligned} & P(X=0)=\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{188}{221} \\ & P(X=1)=\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4 \times 48 \times 2}{52 \times 51}=\frac{32}{221} \\ & P(X=2)=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221} \\ & E(X)=p_i x_i=0+\frac{32}{221}+\frac{2}{221}=\frac{34}{221}=\frac{2}{13} \end{aligned} $

P(defective articles)

$ \begin{aligned} &=\frac{3}{15}=\frac{1}{5}=p(let) \\ & q=1-p=1-\frac{1}{5}=\frac{4}{5} \end{aligned} $

Here, $n=5$

$ P(X=2)={ }^5 C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^{5-2}=\frac{10 \times 64}{3125}=\frac{128}{625} $

To determine the probability that at least one letter is placed in the wrong envelope, consider the following:

Total Arrangements: There are 5 letters, each meant for a specific envelope. The total number of ways to place the letters is calculated using permutations because each letter can go into any of the 5 envelopes. This number is given by $5! = 120$.

Correct Arrangement: There is only 1 way to place all letters in the correctly addressed envelopes, which means every letter is in its intended envelope.

At Least One Mistake: To find the probability that at least one letter is in the wrong envelope, we subtract the probability that all letters are correctly placed from 1. The probability that all letters are correctly placed is $\frac{1}{120}$.

Therefore, the probability that at least one letter is placed in the wrong envelope is:

$ 1 - \frac{1}{120} = \frac{119}{120} $

Here, $x=8, P=\frac{1}{2}$ and $q=\frac{1}{2}$

frobabllity that he will pass

$ \begin{aligned} & =F X=\theta+P(X=7)+P(X=8 \\ & ={ }^8 C_4\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2+{ }^8 C_7\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)+{ }^8 C_3\left(\frac{1}{2}\right)^2 \\ & =\left(\frac{1}{2}\right)^2 P^3 C_2+{ }^3 C_1+{ }^8 C_4 1=\frac{37}{256} \end{aligned} $

Probability that he will fail is the cxamination $ =1-\frac{37}{256}=\frac{219}{256} $

To find the probability that a person traveled to college by car, given that they arrived on time, let's denote the events as follows:

$ E_1 $: The event that a person travels by car.

$ E_2 $: The event that a person travels by bus.

$ E_3 $: The event that a person travels by train.

The probabilities for choosing each mode of transport are:

$ P(E_1) = \frac{1}{5}, \quad P(E_2) = \frac{2}{5}, \quad P(E_3) = \frac{3}{5} $

Let $ A $ be the event of reaching the college on time. The given probabilities of being late for each mode of transport are:

$ P(\bar{A} \mid E_1) = \frac{2}{7} $ leading to $ P(A \mid E_1) = 1 - \frac{2}{7} = \frac{5}{7} $

$ P(\bar{A} \mid E_2) = \frac{4}{7} $ leading to $ P(A \mid E_2) = 1 - \frac{4}{7} = \frac{3}{7} $

$ P(\bar{A} \mid E_3) = \frac{1}{7} $ leading to $ P(A \mid E_3) = 1 - \frac{1}{7} = \frac{6}{7} $

Using Bayes' Theorem, we calculate $ P(E_1 \mid A) $, the probability that the person traveled by car, given they arrived on time:

$ P(E_1 \mid A) = \frac{P(A \mid E_1) \cdot P(E_1)}{P(A \mid E_1) \cdot P(E_1) + P(A \mid E_2) \cdot P(E_2) + P(A \mid E_3) \cdot P(E_3)} $

Substituting the known values:

$ = \frac{\left(\frac{5}{7}\right) \times \left(\frac{1}{5}\right)}{\left(\frac{5}{7} \times \frac{1}{5}\right) + \left(\frac{3}{7} \times \frac{2}{5}\right) + \left(\frac{6}{7} \times \frac{3}{5}\right)} $

Calculate the numerator and the denominator:

$ = \frac{\frac{1}{5} \times \frac{5}{7}}{\frac{1}{5} \times \frac{5}{7} + \frac{2}{5} \times \frac{3}{7} + \frac{3}{5} \times \frac{6}{7}} $

Simplify:

$ = \frac{\frac{5}{35}}{\frac{5}{35} + \frac{6}{35} + \frac{18}{35}} = \frac{\frac{5}{35}}{\frac{29}{35}} = \frac{5}{29} $

Thus, the probability that the person traveled by car, given they arrived on time, is $\frac{5}{29}$.

To find the probability that the target is hit by $P$ or $Q$ but not by $R$, let's define the events:

$E_1$: $P$ hits the target.

$E_2$: $Q$ hits the target.

$E_3$: $R$ hits the target.

The given probabilities are:

$P(E_1) = \frac{2}{3}$

$P(E_2) = \frac{3}{5}$

$P(E_3) = \frac{5}{7}$

These events are independent, so:

The probability that $P$ hits and both $Q$ and $R$ miss is:

$ P(E_1 \cap \overline{E_2} \cap \overline{E_3}) = P(E_1) \cdot (1 - P(E_2)) \cdot (1 - P(E_3)) = \frac{2}{3} \cdot \frac{2}{5} \cdot \frac{2}{7} $

The probability that $Q$ hits and both $P$ and $R$ miss is:

$ P(\overline{E_1} \cap E_2 \cap \overline{E_3}) = (1 - P(E_1)) \cdot P(E_2) \cdot (1 - P(E_3)) = \frac{1}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} $

The probability that both $P$ and $Q$ hit but $R$ misses is:

$ P(E_1 \cap E_2 \cap \overline{E_3}) = P(E_1) \cdot P(E_2) \cdot (1 - P(E_3)) = \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} $

Now, summing these probabilities gives the required probability that the target is hit by $P$ or $Q$, but not by $R$:

$ \begin{align*} P(\text{hit by } P \text{ or } Q \text{ but not } R) &= P(E_1 \cap \overline{E_2} \cap \overline{E_3}) + P(\overline{E_1} \cap E_2 \cap \overline{E_3}) + P(E_1 \cap E_2 \cap \overline{E_3}) \\ &= \frac{2}{3} \cdot \frac{2}{5} \cdot \frac{2}{7} + \frac{1}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} + \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} \\ &= \frac{8}{105} + \frac{6}{105} + \frac{12}{105} \\ &= \frac{26}{105} \end{align*} $

Thus, the probability that the target is hit by $P$ or $Q$ but not by $R$ is $\frac{26}{105}$.

To solve for the probability that exactly 3 out of 5 bulbs chosen are defective, given that 20% of the bulbs are defective, we can use the binomial probability formula. Here's how it's done:

Define the variables:

Total number of trials, $ n = 5 $

Probability of a defective bulb, $ p = 20\% = \frac{1}{5} $

Probability of a non-defective bulb, $ q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} $

Calculate the probability:

We need to find the probability of exactly 3 defective bulbs, $ P(X=3) $.

Apply the binomial probability formula:

$ P(X = 3) = {^5C_3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2 $

Calculate the individual components:

The coefficient, $ ^5C_3 $, is the number of ways to choose 3 defective bulbs from 5. This is calculated as:

$ ^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = 10 $

Plug in the values:

$ \left(\frac{1}{5}\right)^3 = \frac{1}{125} $

$ \left(\frac{4}{5}\right)^2 = \frac{16}{25} $

Calculate the probability:

$ P(X = 3) = 10 \times \frac{1}{125} \times \frac{16}{25} = 10 \times \frac{16}{3125} = \frac{160}{3125} = \frac{32}{625} $

Therefore, the probability that exactly 3 out of the 5 bulbs chosen are defective is $\frac{32}{625}$.

To find the probability that a random variable $ X $, which follows a Poisson distribution with a mean ($\lambda$) of 5, is less than 3, we calculate $ P(X < 3) $.

This is equivalent to finding the sum of probabilities $ P(X = 0) $, $ P(X = 1) $, and $ P(X = 2) $:

$ \begin{aligned} P(X < 3) &= P(X = 0) + P(X = 1) + P(X = 2) \\ &= \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!} + \frac{\lambda^2 e^{-\lambda}}{2!}. \end{aligned} $

Substituting $\lambda = 5$, we have:

$ \begin{aligned} P(X < 3) &= e^{-5} \left( 1 + 5 + \frac{25}{2} \right) \\ &= e^{-5} \left( \frac{2}{2} + \frac{10}{2} + \frac{25}{2} \right) \\ &= e^{-5} \times \frac{37}{2} \\ &= \frac{37}{2} e^{-5}. \end{aligned} $

When two dice are rolled together, then total outcomes are 36 and sample space is $[(1,1),(1,2),(1,3),(1,4),(1,5)(1,6), (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)]$

So, pairs with sum 9 are $(3,6)(4,5),(5,4),(6,3)$ i.e total 4 pairs.

Total outcomes $=36$

Favourable outcomes $=4$

Probability of getting the sum of 9

$=\frac{4}{36}=\frac{1}{9}$

To determine $P(\bar{A} \mid \bar{B})$, we need to use the definition of conditional probability and the properties of complements.

By definition, the conditional probability of $\bar{A}$ given $\bar{B}$ is:

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

Now, using De Morgan's law, we know that:

$\bar{A} \cap \bar{B} = \overline{A \cup B}$

So, we can rewrite the probability as:

$P(\bar{A} \mid \bar{B}) = \frac{P(\overline{A \cup B})}{P(\bar{B})}$

From the properties of complements, we know:

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

Thus, the expression for $P(\bar{A} \mid \bar{B})$ becomes:

$P(\bar{A} \mid \bar{B}) = \frac{1 - P(A \cup B)}{P(\bar{B})}$

This matches Option C.

Therefore, the correct answer is:

Option C

$\frac{1-P(A \cup B)}{P(\bar{B})}$

To find the probability that both brothers $X$ and $Y$ pass the exam, we need to consider the probability of the intersection of events $A$ and $B$. Assuming the events are independent, the probability of both events occurring is given by the product of their individual probabilities.

Let:

$P(A) = \frac{1}{7}$

$P(B) = \frac{2}{9}$

Since events $A$ and $B$ are independent, the probability of both $A$ and $B$ happening is:

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

Substituting the given probabilities:

$P(A \cap B) = \frac{1}{7} \cdot \frac{2}{9} = \frac{1 \cdot 2}{7 \cdot 9} = \frac{2}{63}$

Therefore, the probability that both brothers $X$ and $Y$ pass the exam is $\frac{2}{63}$.

The correct answer is:

Option C

A red ball can be drawn in two mutually exclusive ways.

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it

Let $E_1, E_2$ and $A$ denote the events defined as follows

$\begin{aligned} & E_1=\text { Selecting bag I } \\ & E_2=\text { Selecting bag II } \end{aligned}$

Since, one of the two bags is selected randomly.

$\therefore P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$

Now, $P\left(\frac{A}{E_1}\right)=$ Probability of drawing a red ball when the first bag has been selected $=4 / 7$

$P\left(\frac{A}{E_2}\right)=$ Probability of drawing a red ball when the second bag has been selected $=2 / 5$

$\begin{aligned} P(\text { red ball }) & =P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right) \\ & =\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{2}{5}=\frac{2}{7}+\frac{1}{5}=\frac{17}{35} \end{aligned}$

Let $X$ be the binomial variate for which mean $=4$ and variance $=3$, then $n p=4$ and $n p q=3$

$\Rightarrow q=\frac{3}{4}$

$\therefore P=(1-q)=\left(1-\frac{3}{4}\right)=\frac{1}{4}$ and $n p=4$

$\Rightarrow n=\frac{4}{1} \times 4=16$

Thus, $n=16, p=\frac{1}{4}$ and $q=\frac{3}{4}$

Hence, the binomial distribution

$ \begin{aligned} & 2^{32} P\left(X=\frac{n}{2}\right)=2^{32} \cdot{ }^{16} C_{\frac{16}{2}} \cdot\left(\frac{1}{4}\right)^{\frac{16}{2}}\left(\frac{3}{4}\right)^{16-\frac{16}{2}} . \\ & =2^{32} \cdot{ }^{16} C_8\left(\frac{1}{4}\right)^8 \times\left(\frac{3}{4}\right)^8=2^{32} \cdot{ }^{16} C_8 \frac{(3)^8}{(4)^{16}} \\ & ={ }^{16} C_8(3)^8 \quad {\left[\because(4)^{16}=(2)^{32}]\right.} \end{aligned}$

For Poisson distribution, mean = variance

$l=m=4$

$\begin{aligned} \text { So, } e^4[1-P & P(X>2)] \\ & =e^4[P \leq 2] \\ & =e^4[P(X=0)+P(X=1)+P(X=2)] \\ & =e^4\left[\frac{e^{-4} \times 4^0}{0!}+\frac{e^{-4} \times 4^1}{1!}+\frac{e^{-4} \times 4^2}{2!}\right] \\ & =e^4 \times e^{-4}(1+4+8)=13 \end{aligned}$

Total number of balls $=(8+7+6)=21$

Let $E=$ Event that the ball drawn is neither red nor green

$\begin{array}{l} =\text { Event that the ball drawn is blue } \\ \therefore n(E) =7 \\ \therefore p(E) =\frac{n(E)}{n(S)}=\frac{7}{21}=\frac{1}{3} \end{array}$

We have,

$\begin{aligned} & P(\bar{A} \cup \bar{B})=P(\bar{A} \cap \bar{B}) \\ & \quad=1-P(A \cap B)=1-P(A) P\left(\frac{B}{A}\right) \end{aligned}$

Thus, $P(\bar{A} \cup \bar{B}) \neq 1-P(A \cup B)$

and $P(\bar{A} \cup \bar{B}) \neq P(A \cup B) \neq P(\bar{A})+P(\bar{B})$

A number divisible by 4 formed by the digits $1,2,3,4$ and 5 should have the last two digits 12 or 24 or 32 or 52.

In each cases, the five digit number can be formed using the remaining 3 digits in $3!=6$ ways A number divisible by 4 can be formed in $6 \times 4=24$ ways Total number of 5 -digit numbers that can be formed using the digits $1,2,3,4$ and 5 without repetition $5!=120$

Thus, required probability $=\frac{24}{120}=\frac{1}{5}$

The bigger part could be any number from 10000 to 20000.

Now, if the bigger part is to be at least twice as much as the smaller part, we have

$\begin{aligned} X & \geq 2 y \text { or } X \ge 2(20000-X)\\ \text { or } \quad X & \geq \frac{40000}{3}\end{aligned}$

We know that $X$ lies between 10000 and 20000, the probability that $X$ lies between $\frac{40000}{3}$ and 20000 is $\frac{20000-\frac{40000}{3}}{20000-10000}=\frac{2}{3}$

"The probability of two outcomes can change from one trial to the next" which is not a property of Binomial Distribution.

Given the following information for a Binomial distribution $ B(n, p) $:

• The mean is 15.


• The variance is 10.

We need to find the value of the parameter $ n $.

From the mean formula of a Binomial distribution, we have:

$ n \cdot p = 15 $

From the variance formula of a Binomial distribution, we have:

$ n \cdot p \cdot (1 - p) = 10 $

Since $ n \cdot p = 15 $, we substitute $ 15 $ for $ n \cdot p $ in the variance formula:

$ 15 \cdot (1 - p) = 10 $

Solving for $ p $:

$ 1 - p = \frac{10}{15} = \frac{2}{3} $

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

Now, substituting $ p $ back into the mean formula:

$ n \cdot \left( \frac{1}{3} \right) = 15 $

$ n = 15 \times 3 = 45 $

Therefore, the value of $ n $ is 45.

Total number of balls $=100$

odd numbered balls $\rightarrow 50$

Even numbered balls $\rightarrow 50$

Sum of three number will be odd when

(i) two numbers are even and one odd.

(ii) all three numbers are odd.

$\begin{aligned} & \text { Required probability }=\frac{{ }^{50} C_2 \cdot{ }^{50} C_1+{ }^{50} C_3}{{ }^{100} C_3} \\ &=\frac{\frac{50 \cdot 49}{2 \cdot 1} \cdot 50+\frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1}}{\frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1}} \\ &=\frac{50 \cdot 49 \cdot 50 \cdot 3+50 \cdot 49 \cdot 48}{100 \cdot 99 \cdot 98} \\ &=\frac{50 \cdot 49[50 \times 3+48]}{100 \cdot 99 \cdot 98} \\ &=\frac{6[25+8]}{2 \cdot 99 \cdot 2}=\frac{6 \cdot 33}{2 \cdot 99 \cdot 2}=\frac{1}{2} \end{aligned}$

Total number of tickets $=35$

Number of tickets bearing a prize $=10$

$\begin{gathered} P \text { (not getting a prize })=1-P \text { (getting a prize) } \\ =1-\frac{10}{35}=\frac{25}{35}=\frac{5}{7} \end{gathered}$

Number of green balls $=7$

Number of black balls $=5$

Total number of balls $=7+5=12$

While drawing, ball is not replaced.

Required probability

$\begin{aligned} & =\frac{{ }^7 C_1 \cdot{ }^6 C_1 \cdot{ }^5 C_1}{{ }^{12} C_1{ }^{11} C_1 \cdot{ }^{10} C_1} \\ & =\frac{7 \cdot 6 \cdot 5}{12 \cdot 11 \cdot 10}=\frac{7}{44} \end{aligned}$

There are 12 possible values of $x y$.

$x \in\{1,2,3,4\}, y \in\{5,6,7\}$

Values of $x y$ are :

$\begin{aligned} & 1 \times 5=5 \\ & 1 \times 6=6 \end{aligned}$

$\begin{aligned} & 1 \times 7=7 \\ & 2 \times 5=10 \\ & 2 \times 6=12 \\ & 2 \times 7=14 \\ & 3 \times 5=15 \\ & 3 \times 6=18 \\ & 3 \times 7=21 \\ & 4 \times 5=20 \\ & 4 \times 6=24 \\ & 4 \times 7=28 \end{aligned}$

$P(x y \text { is even })=\frac{8}{12}=\frac{2}{3}$

Comparing with the standard binomial $(n, p)$

distribution, variance $=n p(1-p)$

$\begin{aligned} & =16 \times 0.25(1-0.29) \\ & =4 \times 0.75=3 \end{aligned}$

Comparing with the standard poisson ($\lambda$) distribution, variance $=\lambda=2$

$\therefore$ Sum of the variance of $X$ and $Y$

$=\operatorname{Var}(X)+\operatorname{Var}(Y)=3+2=5$

Probability distribution: $P(X=x)=\frac{e^{-\lambda} x^x}{x!}$

$\lambda=$ mean = 6

$\begin{aligned} & p(X \geq 3) =1-[P(X=0)+p(X=1)+p(X=2)] \\ & =1-\left[\frac{e^{-6} \cdot 6^0}{0!}+\frac{e^{-6} \cdot 6^1}{1!}+\frac{e^{-6} \cdot 6^2}{2!}\right] \\ & =1-\frac{1}{e^6}[1+6+, 18] \\ & =1-\frac{25}{e^6} \end{aligned}$

$ \begin{aligned} & \quad S=\{H, T H, T T H, T T T\} \\ & P(H)=\frac{1}{2}, P(T H)=\frac{1}{4} \\ & P(T T H)=\frac{1}{8}, P(T T T)=\frac{1}{8} \end{aligned}$

Let $E$ be the event "no head on first toss"

$\begin{aligned} & \therefore E=\{T H, T T H, T T T\} \\ & \therefore P(E)=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2} \end{aligned}$

Let $E^{\prime}$ be the event coin is tossed 3 times

$\begin{aligned} & \therefore E^{\prime}=\{T T H, T T T\} \\ & \therefore P\left(E^{\prime}\right)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} \\ & \therefore P\left(E^{\prime} / E\right)=\frac{P\left(E^{\prime} \cap E\right)}{P(E)}=\frac{P\left(E^{\prime}\right)}{P(E)}=\frac{1 / 4}{1 / 2}=\frac{1}{2} \end{aligned}$

$\Rightarrow$ No option is correct.

$x_1+x_2+x_3$ is odd if

(i) all three are odd

(ii) two even and one odd Now, total number of outcomes $=3 \times 5 \times 7=105$

Possible outcomes-

(i) (odd odd odd), then $P(O+O+O)$

$\text { Possibility }=\frac{2}{3} \times \frac{3}{5} \times \frac{4}{7}=\frac{24}{105}$

(ii) (odd even even), $P(O+E+E)$

$\text { Possibility }=\frac{2}{3} \times \frac{2}{5} \times \frac{3}{7}=\frac{12}{105}$

(iii) (even odd even), $p(E+O+E)$

$\text { Possibility }=\frac{1}{3} \times \frac{3}{5} \times \frac{3}{7}=\frac{9}{105}$

(iv) (even even odd), $P(E+E+O$ )$

Possibility $=\frac{1}{2} \times \frac{2}{5} \times \frac{4}{7}=\frac{8}{105}$

Probability that $x_1+x_2+x_3$ is odd is given as

$\frac{24}{105}+\frac{12}{105}+\frac{9}{105}+\frac{8}{105}=\frac{53}{105}$

We know $\Sigma p(x)=1$

$\Rightarrow \quad \frac{C}{1 !}+\frac{C^2}{2 !}+\frac{C^3}{3 !}+\ldots+=1$

Add 1 on both side

$\begin{aligned} 1+C+\frac{C^2}{2 !}+\frac{C^3}{3 !}+\ldots & =2 \\ \Rightarrow \quad e^C =2 \Rightarrow C=\log 2 \end{aligned}$

Let probability of getting head $=x$

Then, probability of getting tail $=1-x$

Tom wins if he gets head

H, TTH, TTTTH, ...(alternative through)

Probability of Tom winning game $=62.5 \%=\frac{625}{1000}$

$\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{TTH})+\mathrm{P}(\mathrm{TTTTH})+\ldots=\frac{625}{1000}$

$\begin{aligned} x+(1-x)(1-x) x+(1-x)^4 x+\ldots & =\frac{625}{1000} \\ x\left[1+(1-x)^2+(1-x)^4+\ldots\right] & =\frac{625}{1000} \\ x \cdot \frac{1}{1-(1-x)^2} & =\frac{625}{100} \\ \frac{x}{2 x-x^2} & =\frac{625}{1000} \\ \frac{1}{2-x} & =\frac{625}{1000} \end{aligned}$

or $\frac{1}{2-x}=\frac{625}{1000}$

or $x=\frac{2}{5}(\text { probability of head })$

or $1-x=\frac{3}{5} \text { (probability of tails) }$

$\therefore$ Probability of getting exactly 3 heads on tossing a coin or 5 times

$\begin{aligned} & ={ }^5 C_3 x^3(1-x)^2 \\ & =10 \times\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^2=\frac{144}{625} \end{aligned}$

Even or divisible by 5

$\begin{aligned} & A:\{2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26\} \\ & n(A)=16 \\ & \therefore \text { Probability }=\frac{n(A)}{n(S)}=\frac{16}{27} \end{aligned}$

$\begin{aligned} \text { Probability } & =\frac{{ }^5 C_3 \times{ }^7 C_6}{{ }^{12} C_9} \\ & =\frac{7 \times 10}{{ }^{12} C_9}=\frac{70}{\frac{12 \cdot 11 \cdot 10}{6}}=\frac{7}{22}\end{aligned}$

They will contradict each other in two case

Case I $A \rightarrow$ True, $B \rightarrow$ False

Case II $A \rightarrow$ False, $B \rightarrow$ True

$\text { Probability }=\frac{4}{5} \times \frac{1}{4}+\frac{1}{5} \times \frac{3}{4}=\frac{7}{20}$

$\begin{aligned} & \text { Mean }=n p=2 \\ & \text { Variance }=n p q=1 \\ & \Rightarrow \quad q=\frac{1}{2} \text { and } p=\frac{1}{2} \Rightarrow n\left(\frac{1}{2}\right)=2 \\ & \Rightarrow \quad n=4 \\ & P(x>1)=P(x=2)+P(x=3)+P(x=4) \\ & =\left({ }^4 C_2+{ }^4 C_3+{ }^4 C_4\right) \frac{1}{2^4}=\frac{11}{16}\end{aligned}$

$P$ speaks truth $=70 \%=7 / 10$

$Q$ speaks truth $=80 \%=8 / 10$

They agree when both speaks truth or both does not speaks truth

$\begin{aligned} & =\frac{7}{10} \times \frac{8}{10}+\left(1-\frac{7}{10}\right)\left(1-\frac{8}{10}\right) \\ & =\frac{56}{100}+\left(\frac{3}{10}\right)\left(\frac{2}{10}\right)=\frac{56+6}{100}=\frac{62}{100}=62 \% \end{aligned}$

Given,

$\begin{array}{ll} & P(A \cap B)=\frac{1}{3}, P(A \cup B)=\frac{5}{6}, P\left(A^c\right)=\frac{1}{2} \\ \because & P(A)+P(B)=P(A \cup B)+P(A \cap B) \\ \because & P(A)+P(B)=P(A \cup B)+P(A \cap B) \\ \text { or } & 1-P\left(A^c\right)+1-P\left(B^c\right)=\frac{5}{6}+\frac{1}{3}=\frac{7}{6} \\ \text { or } & P\left(B^c\right)=2-\frac{7}{6}-\frac{1}{2}=\frac{1}{3} \end{array}$

Every time we toss, probability of getting head is $1 / 2$.

So, on tossing $n$ times, probability of getting head at $r$th toss $=1 / 2$

$\therefore$ Probability of getting head on 1947th toss $=1 / 2$

Expected value of $X=E(X)=\Sigma px$

$=10\times0.3+20\times0.3+30\times0.2+40\times0.2$

$=3+6+6+8=23$

$\therefore \Sigma P=1$

$\Rightarrow K+8K+27K+64K=1$

$\Rightarrow K=\frac{1}{100}$

$\begin{aligned} P\left(\frac{1}{2} < X < \frac{5}{2} X>1\right) & =\frac{P\left(\left(\frac{1}{2} < X < \frac{5}{2}\right) \cap(X >1)\right)}{P(X > 1)} \\ & =\frac{P\left(1 < X < \frac{5}{2}\right)}{1-P(X \leq 1)}=\frac{P(X=2)}{1-P(X=1)} \\ & =\frac{\frac{8}{100}}{1-\frac{1}{100}}=\frac{8}{99}\end{aligned}$

Total distribution = 3$^{12}$

Number of distribution when first box contains three balls $ = {}^{12}{C_3} \times {2^9}$

Required probability $ = {{{}^{12}{C_3} \times {2^9}} \over {{3^{12}}}}$

$\begin{aligned} & \text { } P(E)=P(X=2)+P(X=3)+P(X=5) \\ & +P(X=7) \\ & =0.23+0.12+0.20+0.07=0.62 \\ & P(F)=P(X=1)+P(X=2)+P(X=3) \\ & =0.15+0.23+0.12=0.50 \\ & P(E \cap F)=P(X=2)+P(X=3) \\ & =0.23+0.12=0.35 \\ & P(E \cup F)=P(E)+P(F)-P(E \cap F) \\ & =0.62+0.50-0.35 \\ & =0.77 \\ & \end{aligned}$

Sample space of a dice $S=\{1,2,3,4,5,6\}$

Let $E=$ getting even number $\Rightarrow E=\{2,4,6\}$

$P(E)=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}$

If dice is thrown three times

Let $X=$ success, then

Probability of getting at least 2 successes

$\begin{aligned} & =P(X \geq 2)=P(X=2)+P(X=3) \\ & ={ }^3 C_2\left(\frac{1}{2}\right)^3+{ }^3 C_3\left(\frac{1}{2}\right)^3=\frac{3}{8}+\frac{1}{8}=\frac{1}{2} \end{aligned} $