# Probability-Conditional Probability (ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern)): Questions 1 - 4 of 6

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## Question number: 1

» Probability » Conditional Probability

Appeared in Year: 2011

### Describe in Detail

Consider the following bivariate p. m. f. of (X, Y):

p (0,10) = p (0,20) = 2/18;

p (l, 10) = p (l, 30) = 3/18;

p (1,20) = p (2,30) = 4/18;

Obtain the conditional mass functions p (y lx = 2), and p (y lx = 1).

### Explanation

The joint probability is given

First find the marginal probability of x at x = 0,1, 2

p (0) =p (0,10) +p (0,20) + p (0,30) = 2/9

p (1) =p (1,10) +p (1,20) +p (1,30) =5/9

p (2) =p (2,10) + p (2,20) + p (2,30) =2/9

The conditional distribution of the random variable X given Y = y is

… (42 more words) …

## Question number: 2

» Probability » Conditional Probability

Appeared in Year: 2009

### Describe in Detail

(i) Let X be a random variable such that P [X < 0] = 0 and E [x] exist. Show that P (X ≤2E [x] ) ≥l/2

(ii) Let E [X] = 0 and E [X ^{2}] be finite. Show that P (X ^{2} < 9E [X ^{2}] ) > 8/9

### Explanation

(i) Using Markov inequality, for any random variable and constant a > 0

Here a = 2E (X)

or

(ii) Using Chebyshev inequality, Let X have mean E (X) =µand Var (X) =σ ^{2}, then for any a > 0

… (107 more words) …

## Question number: 3

» Probability » Conditional Probability

Appeared in Year: 2011

### Describe in Detail

Let (X, Y) have the uniform distribution over the range 0 < y· < x < 1. Obtain the conditional mean and variance of X given Y = y.

### Explanation

The joint probability density function of (X, Y) is

The marginal distribution of X is

The conditional distribution of X given Y = y

The conditional mean is

… (25 more words) …

## Question number: 4

» Probability » Conditional Probability

Appeared in Year: 2012

### Describe in Detail

You arc given the following information:

(i) In random testing, you test positive for a disease.

(ii) In 5 % of cases, the test shows positive even when the subject does not have the disease.

(iii) In the population at large, one person in 1000 has the disease. What is the conditional probability that you have the disease given that you have been tested positive, assuming that if someone has the disease, he will test positive with probability 1?

### Explanation

Let X denotes the test is positive and Y denotes the person has disease. Given that 5 % of cases, the test shows positive even when the subject does not have the disease that is

where denotes the person has no disease. Also given the test positive with probability 1

Then the conditional probability that you ha

… (68 more words) …