---
title: "pdf definition; f(x)=Kx(1−x) probabilities"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-kx-1-minus-x-probabilities"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx-1-minus-x-probabilities
site: BCT Engineering
---

# pdf definition; f(x)=Kx(1−x) probabilities

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2075 Chaitra, 5 marks
- Video: [Find k if the function f(x) is defined by  `f(x)=kx(1-x),"for "0ltxlt1`  =0 , otherwise, is the](https://www.youtube.com/watch?v=cxN8HO9H2bk) — Doubtnut
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx-1-minus-x-probabilities
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

Define probability density function? A continuous probability distribution of a variable x is defined as f(x) = Kx(1−x) for 0 ≤ x ≤ 1. Compute (i) P(X ≥ 0.4) (ii) P(1/4 ≤ X ≤ 3/4).

## Short answer

A pdf is a nonnegative function integrating to 1. K=6 from ∫₀¹ Kx(1-x)dx=1. P(X≥0.4)=0.648; P(1/4≤X≤3/4)=0.6875.

## Answer

#### Probability density function
For a continuous random variable X, the **pdf** f(x) is a function such that
1. f(x) ≥ 0 for all x,
2. ∫ f(x) dx = 1 over the whole range, and
3. P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx.

#### Find K
f(x) = Kx(1 − x), 0 ≤ x ≤ 1.

∫₀¹ K(x − x²) dx = K(1/2 − 1/3) = K/6 = 1 → **K = 6**

So f(x) = 6x(1 − x), and the cumulative function is F(x) = 6(x²/2 − x³/3) = 3x² − 2x³.

#### (i) P(X ≥ 0.4)
P = 1 − F(0.4) = 1 − [3(0.16) − 2(0.064)] = 1 − (0.48 − 0.128) = 1 − 0.352 = **0.648**

#### (ii) P(1/4 ≤ X ≤ 3/4)
F(0.75) = 3(0.5625) − 2(0.421875) = 1.6875 − 0.84375 = 0.84375
F(0.25) = 3(0.0625) − 2(0.015625) = 0.1875 − 0.03125 = 0.15625

P = 0.84375 − 0.15625 = **0.6875**

**Answer: K = 6, (i) 0.648, (ii) 0.6875**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A pdf must satisfy:
   A. f(x) <= 1
   B. Sum of f = 1
   C. f(x) >= 0 and total area 1
   D. f(x) = P(X = x)
   Answer: C. Probabilities are areas, not heights. (Key point: pdf properties)

2. K is found from:
   A. Integral of Kx(1-x) over 0 to 1 = K/6 = 1, so K = 6
   B. K/2 = 1
   C. K/3 = 1
   D. K = 1/6
   Answer: A. 1/2 - 1/3 = 1/6. (Key point: K)

3. The cdf is:
   A. 6x^2 - 6x^3
   B. x^2 - x^3
   C. 3x - 2x^2
   D. F(x) = 3x^2 - 2x^3
   Answer: D. Integrate 6(x - x^2). (Key point: Cdf)

4. P(X >= 0.4) is:
   A. F(0.4) = 0.352
   B. 1 - F(0.4) = 0.648
   C. 0.4
   D. 1 - F(0.6)
   Answer: B. F(0.4) = 0.352. (Key point: P(X>=0.4))

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