1
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

Assertion (A) Variance of $4 x_1, 4 x_2, \ldots, 4 x_n$ is 16 times the variance of $x_1, x_2, x_3, \ldots, x_n$

Reason (R) If $y=a x+b$, then variance of $y$ is a $($ variance of $x)+b$

The correct option among the following is

A

(A) is true, (R) is true and (R) is the correct explanation for (A).

B

(A) is true, (R) is true but (R) is not the correct explanation for (A).

C

(A) is true but (R) is false.

D

(A) is false but (R) is true.

2
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $\alpha, \beta$ are respectively the mean deviation about the mean and variance of the first five prime numbers, then the ordered pair ( $\alpha, \beta$ )

A

$(2.27,10.42)$

B

$(2.27,10.24)$

C

$(2.72,10.24)$

D

$(2.72,10.42)$

3
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $A_1, A_2, \ldots, A_{15}$ are the events of a random experiment, then which one of the following is true?

A

$P\left(\bigcap_{i=1}^{15} A_i\right) \leq \sum_{i=1}^{15} P\left(A_i\right)-15$

B

$P\left(\bigcap_{i=1}^{15} A_i\right) \geq \sum_{i=1}^{15} P\left(A_i\right)-14$

C

$P\left(\bigcup_{i=1}^{15} A_i\right) \geq \sum_{i=1}^{15} P\left(A_i\right)$

D

$$ P\left(\bigcup_{i=1}^{15} A_i\right) < \sum_{i=1}^{15} P\left(A_i\right)-\sum_{1 \leq i < j<15} P\left(A_i \cap A_j\right) $$

4
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

In an examination there are four Yes/No type of questions. The probability that the answer by the student to a question without guess to be correct is $2 / 3$. The probability that a student guesses a correct answer is $1 / 2$. A student writes the examination either by without guessing answers to all the 4 questions or by guessing answers to all 4 questions. The probability that he attempt the exam by guessing answers to all questions is $3 / 7$. Given that a student answered at least 3 questions correctly, the probability that he answered all the questions without guessing is

A

$\frac{13}{15}$

B

$\frac{405}{1429}$

C

$\frac{1024}{1429}$

D

$\frac{2}{15}$

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