Algebra PYQ — Page 41

The probability distribution of a random variable $X$ is | x | 0 | 1 | 2 | 3 | | --- | --- | --- | --- | --- | | P(X = x) | $\frac{1}{4}$ | $\frac{1}{8}$ | $\frac{1}{8}$ | $\frac{1}{2}$ | The variance of $X$ is

In Binomial distribution with parameters $n = 100$ and p, Variance of distribution is maximum when p is equal to :

The number of all possible matrices of order 3 x 3 with each entry belonging to the set {0, 1} is:

If $n(A) = 3$, $n(B) = 2$, then number of all possible surjective function from set A to set B are :

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows : For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive. Based on the above informations, answer the question : The probability that the selected person will be diagonosed as COVID positive is :

If $A=\begin{bmatrix}2 & 0 & 0 \\ -1 & 2 & 3 \\ 3 & 3 & 5\end{bmatrix}$, then $A(\text{adj } A)$ is equal to :

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases. The probability of a randomly chosen seed to germinate is :

Three urns contain 6 red, 4 black; 4 red, 6 black and 5 red, 5 black marbles respectively. One of the urns is selected at random and a marble is drawn from it. If the marble drawn is red, then the probability that it is drawn from the first urn is

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases. The probability that seed will not germinate, given that the seed is of type $A_3$.

The value of $k$ is

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases. The probability that seed is not of type $A_1$, given that seed germinates.

Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. The probability distribution of the number of kings will be:

Consider the linear programming problem : Minimize $z = 50x + 70y$ Subject to $2x+y \geq 8$, $x+2y \geq 10$, $x \geq 0$, $y \geq 0$ The minimum value of objective function is :

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows : For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive. Based on the above informations, answer the question : The probability that randomly selected person from a population, not having COVID is :

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases. The probability that seed is of type $A_1$ given that seed doesn't germinate.

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are two non zero vectors inclined at an angle $\theta$, then identify the correct option out of the given options. (a) $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot |\vec{b}|}$ (b) $\vec{a}$ and $\vec{b}$ are perpendicular, if $a_1 b_1 + a_2 b_2 + a_3 b_3 = 0$ (c) $\vec{a}$ and $\vec{b}$ are perpendicular, if $\frac{a_1}{b_1} = \frac{a_2}{b_2} \neq \frac{c_1}{c_2}$ (d) for $\theta = \pi$, $\vec{a} \times \vec{b} = 0$ (e) $\cos\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| \cdot |\vec{b}|}$ Choose the most appropriate answer from the options given below

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows : For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive. Based on the above informations, answer the question : The probability that the person is actually having COVID given that he is tested as COVID positive is :

If $R$ is a relation on $Z$ (set of all integers) defined by $xRy$, iff $|x - y| \leq 1$, then (a) $R$ is reflexive (b) $R$ is symmetric (c) $R$ is transitive (d) $R$ is not symmetric (e) $R$ is not transitive Choose the most appropriate answer from the options given below

Let $A$ and $B$ be two non-singular, square matrices of same order, and A. $(AB)^{-1} = B^{-1} \cdot A^{-1}$ B. $(A+B)^{-1} = B^{-1} + A^{-1}$ C. $adj. A = |A| \cdot A^{-1}$ D. $det(A^{-1}) = [det A]^{-1}$ Choose the correct answer from the options given below

A feasible solution is :

The maximum value of $z = 4x + 3y$, if the feasible region for an LPP is as shown below is:

Area of the rectangular plot is:

The probability that the task is completed on time by none of them is

Match List - I with List - II. | | List - I (Two given vector) | | List - II (Projection of $\vec{a}$ on $\vec{b}$) | |---|---|---|---| | (A) | $\vec{a} = \hat{i} - \hat{j}$, $\vec{b} = \hat{i} + \hat{j}$ | (I) | $\frac{2}{\sqrt{5}}$ | | (B) | $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = 2\hat{i} - \hat{k}$ | (II) | 0 | | (C) | $\vec{a} = \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + \hat{k}$ | (III) | $\sqrt{2}$ | | (D) | $\vec{a} = 2\hat{i} + 3\hat{j}$, $\vec{b} = \hat{i} - \hat{k}$ | (IV) | $\frac{1}{\sqrt{2}}$ | Choose the correct answer from the options given below :