Probability

$n-2, \sqrt{3 n}, n+2 \rightarrow$ G.P. $3 n=n^{2}-4$ $\Rightarrow n^{2}-3 n-4=0$ $\Rightarrow n=4,-1$ (rejected) $P(S=4)=\dfrac{3}{36}=\dfrac{1}{12}=\dfrac{4}{48}$ $\therefore k=4$

$x+y=66$

$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$ Therefore $S=\{11,12,13 \ldots 55\}$ $\Rightarrow n(S)=45$ Elements of $S$ which are multiple of 3 are

$\Rightarrow P(A)=\dfrac{15}{45}=\dfrac{1}{3}$

For unique solution $\Delta \neq 0$ i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$ $\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$ $\Rightarrow N^{2}-5 N+6 \neq 0$ $\therefore N \neq 2$ and $N \neq 3$ $\therefore$ Probability of not getting 2 or 3 in a throw of dice $=\dfrac{2}{3}$ As given $\dfrac{2}{3}=\dfrac{k}{6} \Rightarrow k=4$ $\therefore$ Required value $=1+4+5+6+4=20$

$\Omega=$ sample space $\mathrm{A}=$ be an event $\Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$ $\mathrm{A}=\left\{\dfrac{1}{2}\right\}$ = Selecting a point on the wire which is at $\left\{\dfrac{1} {2}\right\}$ or 0.5 As wire is an 1-D object so from geometrical probability $P(A)=\dfrac{\text{ Favourable Length }}{\text{ Total Length }}$ Here total length of wire = 1 unit and point has zero length so point A at $\left\{\dfrac{1} {2}\right\}$ or 0.5 has length = 0.

$\therefore$ Favorable length = 0 $\therefore$ $\mathrm{P}(\mathrm{A})=0 {\text{ but }} \mathrm{A} \neq \phi$ Now $\overline{\mathrm{A}}$ = $[0,1]$ - $\left\{\dfrac{1} {2}\right\}$ So, length of $\overline{\mathrm{A}}$ = Length of entire wire - Length of point A = 1 $\therefore$ $\mathrm{P}(\overline{\mathrm{A}})=1 {\text{ but }} \overline{\mathrm{A}} \neq \Omega$.

Then both statements are false.

Attention : According to NTA option A is correct.

Which is wrong.

That is proven here using geometrical probability.

Note : Geometrical probability : 1.

For 1-D object, $P(A)=\dfrac{\text{ Favourable Length }}{\text{ Total Length }}$ 2.

For 2-D object, $P(A)=\dfrac{\text{ Favourable Area }}{\text{ Total Area }}$ 3.

For 2-D object, $P(A)=\dfrac{\text{ Favourable volume }}{\text{ Total volume }}$

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 :

The probability of rolling any number from 1 to 6 on the die is equal, so $P(X=x) = \dfrac{1}{6}$ 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. $P(W|X=1) = {{{}^6{C_1}} \over {{}^{10}{C_1}}}= \dfrac{6}{10} = \dfrac{3}{5}$, since there are 6 white balls out of a total of 10 balls.

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

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

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

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

6. $P(W|X=6) = {{{}^6{C_6}} \over {{}^{10}{C_6}}} = \dfrac{1}{210}$, 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 :

To simplify this expression, find a common denominator :

Add the fractions :

A, E,G R D N

(when A & E are in order) Total case $=6$ ! Favourable case $={ }^6 \mathrm{C}_2 .4$ !

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

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

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 $P(H) = \dfrac{3}{4}$. - X = 2 : A tail is obtained on the first toss and a head on the second.

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

This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\dfrac{1}{4}\right)^2 \times \dfrac{3}{4} + \left(\dfrac{1}{4}\right)^3$.

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

Substituting the values from table :

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.

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 :

Here,

and

, and they are co-prime. The positive divisors of 35 are 1, 5, 7, and 35. The sum of these divisors is

.

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 $a, b, c, d \in\{0,1,2\}$

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 $a d=b c=1$

Case III : When $a d=b c=2$ There are two ways, when $a d=2$, i.e.

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 $a d=b c=4$

$\therefore$ Total number of ways, when