1
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A diagnostic test has the probability 0.95 of giving a positive result when applied to a person suffering from a certain disease and a probability 0.10 of giving a positive result when given to a non-sufferer. It is estimated that $0.5 \%$ of the population are suffering from the disease. If this test is now administered to a person from this population about whom there is no information relating to the incidence of this disease and the test gives a positive result, then the probability that he is a sufferer, is

A

0.9545

B

0.2194

C

0.0455

D

0.9499

2
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

Consider the following statements

Assertion (A) If $P_1, P_2, P_3$ are probability of happening of three independent events, then probability of happening of atleast one of them is $1-\left[\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)\right]$

Reason (R) For any three independent events $A, B$ and $C$

$$ \begin{array}{r} P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A) P(B)-P(A) P(C) -P(B) P(C)+P(A) P(B) P(C) \end{array} $$

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

3
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If probability function of a discrete random variable $X$ is $P(X=r)=r / k, r=1,2,3,4,5$, then $P\left(X=2\right.$ or $\left.X=\frac{k}{3}\right)$, is

A

$P(X=1$ or $X=6)$

B

$P\left(X=4\right.$ or $\left.X=\frac{k}{5}\right)$

C

$P\left(X=\frac{k}{5}\right.$ or $\left.X=5\right)$

D

$P\left(X=\frac{k}{3}\right.$ or $\left.X=0\right)$

4
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If the probability that an individual will suffer a reaction from an injection of a drug is 0.001 , then the probability that out of 2000 individuals having that injection, more than 2 individuals will suffer a reaction, is

A

$\frac{5}{e^2}$

B

$1-\frac{5}{e^2}$

C

$1-\frac{4}{e^2}$

D

$\frac{4}{e^2}$

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