Find k if the function f x is defined by f x = k x 1 - x , for 0 < x < 1 = 0 , otherwise , is the probability density function (p.d.f.) of a random variable (r.v.) X . Also find P x < 1 2 .

Given that the function f x = k x 1 - x , for 0 < x < 1 = 0 , otherwise , is the p.d.f. of X . ⇒ ∫ - ∞ ∞ f ( x ) d x = 1 ⇒ ∫ - ∞ 0 f ( x ) d x + ∫ 0 1 f ( x ) d x + ∫ 1 ∞ f ( x ) d x = 1 ⇒ 0 + ∫ 0 1 k x ( 1 - x ) d x + 0 = 1 ⇒ k ∫ 0 1 x - x 2 d x = 1 ⇒ k x 2 2 - x 3 3 0 1 = 1 ⇒ k 1 2 - 1 3 - 0 = 1 ⇒ k 6 = 1 Hence, k = 6 Now, we need to find: P x < 1 2 We know that P x < 1 2 = ∫ - ∞ 1 2 f ( x ) d x = ∫ - ∞ 0 f ( x ) d x + ∫ 0 1 2 f ( x ) d x = 0 + ∫ 0 1 2 k x ( 1 - x ) d x = k ∫ 0 1 2 x - x 2 d x = k x 2 2 - x 3 3 0 1 2 = k 1 8 - 1 24 - 0 = k 2 24 = 6 × 2 24 Hence, P x < 1 2 = 1 2 .

For the following probability density function (p.d.f) of X , f x = x 2 18 , − 3 < x < 3 0 , otherwise , find P ( X < 1 ) .

Given, f x = x 2 18 , − 3 < x < 3 0 , otherwise is a p.d.f Thus, P X < 1 = ∫ − 3 1 f ( x ) d x ⇒ P X < 1 = ∫ − 3 1 x 2 18 d x ⇒ P X < 1 = 1 18 x 3 3 - 3 1 ⇒ P X < 1 = 1 18 1 3 3 - - 3 3 3 ⇒ P X < 1 = 1 18 1 3 - - 27 3 ⇒ P X < 1 = 1 18 1 3 + 9 ⇒ P X < 1 = 0 . 5185

For the following probability density function (p.d.f) of X , f x = x 2 18 , − 3 < x < 3 0 , otherwise , find P ( X < 1 ) .

Given, f x = x 2 18 , − 3 < x < 3 0 , otherwise is a p.d.f Now, P X < 1 = P - 1 < X < 1 = ∫ − 1 1 f ( x ) d x ⇒ P X < 1 = ∫ − 1 1 x 2 18 d x ⇒ P X < 1 = 1 18 x 3 3 - 1 1 ⇒ P X < 1 = 1 18 1 3 3 - - 1 3 3 ⇒ P X < 1 = 1 18 1 3 - - 1 3 ⇒ P X < 1 = 1 18 1 3 + 1 3 ⇒ P X < 1 = 1 18 2 3 ⇒ P X < 1 = 1 27 = 0 . 03704

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 8 - Probability Distribution>Maharastra Board-2019

Embibe Experts Solutions for Chapter: Probability Distribution, Exercise 1: Maharastra Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Probability Distribution, Exercise 1: Maharastra Board-2018

Attempt the free practice questions on Chapter 8: Probability Distribution, Exercise 1: Maharastra Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Probability Distribution, Exercise 1: Maharastra Board-2018 with Hints & Solutions

MEDIUM

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 8 - Probability Distribution>Maharastra Board-2018>Q 1

Obtain the probability distribution of the number of sixes in two tosses of a fair die.

MEDIUM

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 8 - Probability Distribution>Maharastra Board-2018>Q 2

Find $k$ if the function $f (x)$ is defined by $\{\begin{cases} f (x) = k x (1 - x), & for 0 < x < 1 \\ = 0, & otherwise, \end{cases}$ is the probability density function (p.d.f.) of a random variable (r.v.) $X$ . Also find $P (x < \frac{1}{2})$ .

MEDIUM

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 8 - Probability Distribution>Maharastra Board-2019>Q 3

The probability distribution of a random variable $X$ , the number of defects per $10 meters$ of a fabric is given by:

$x$	$0$	$1$	$P (X = 0) = \frac{n (x)}{n (S)} = \frac{25}{36} . . . (2)$	$3$	$4$
$P (X = x)$	$0.45$	$0.35$	$0.15$	$0.03$	$0.02$

Find the variance of $X$ .

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12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 8 - Probability Distribution>Maharastra Board-2019>Q 4

For the following probability density function (p.d.f) of $X$ , $f (x) = \{\begin{cases} \frac{x^{2}}{18}, & - 3 < x < 3 \\ 0, & otherwise \end{cases}$ , find $P (X < 1)$ .

MEDIUM

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 8 - Probability Distribution>Maharastra Board-2019>Q 5

For the following probability density function (p.d.f) of $X$ , $f (x) = \{\begin{cases} \frac{x^{2}}{18}, & - 3 < x < 3 \\ 0, & otherwise \end{cases}$ , find $P (|X| < 1)$ .