An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$ .

3 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A b e t h e e v e n t h a v i n g m w h i t e a n d n b l a c k b a l l s . \\ E_{1} = {f i r s t b a l l d r a w n o f w h i t e c o l o u r} \\ E_{2} = {f i r s t b a l l d r a w n o f b l a c k c o l o u r} \\ E_{3} = {second b a l l d r a w n o f w h i t e c o l o u r} \\ P (E_{1}) = \frac{m}{m + n}, P (E_{2}) = \frac{n}{m + n}, P (E_{3} / E_{1}) = \frac{m + k}{m + n + k}, P (E_{3} / E_{2}) = \frac{m}{m + n + k} \\ N o w, P (E_{3}) = P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2}) \\ = \frac{m}{m + n} * \frac{m + k}{m + n + k} + \frac{n}{m + n} * \frac{m}{m + n + k} \\ = \frac{m}{m + n + k} [\frac{m + k}{m + n} + \frac{n}{m + n}] \\ = \frac{m}{m + n + k} [\frac{m + n + k}{m + n}] = \frac{m}{m + n} \\ H e n c e, t h e p r o b a b i l i t y o f d r a w n i n g a w h i t e b a l l d o e s n o t d e p e n d u p o n k . \end{array}$

Upvote

Similar Questions for you

Can I get admission with compartment tag?

Sumit Joshi

It's difficult but in some colleges you may can get

Alok Kumar Singh

$\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$

⇒ $\vec{a} + 5 \vec{b}$ = $\vec{c}$ …(1)

$\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$

⇒ $\vec{b} + 6 \vec{c} = μ \vec{a}$ …(2)

From (1) and (2)

$\vec{b} + 6 \vec{c} = μ (λ \vec{c} - 5 \vec{b})$

-> $(1 + 5 μ) \vec{b} + (6 - λ μ) \vec{c} = 0$

$? \vec{b}$ and $\vec{c}$

$\int \frac{(s i n x - c o s x) s i n^{2} x}{t a n x (s i n^{3} x + c o s^{3} x)} d x$

$\int \frac{(s i n x - c o s x) s i n x c o s x}{s i n^{3} x + c o s^{3} x} d x$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3} \int \frac{d t}{t}$

$= \frac{l n t}{3} + c$

$= \frac{l n | s i n^{3} x + c o s^{3} x |}{3} + c$

Alok Kumar Singh

$f (x) = \frac{(2^{x} + 2^{- x}) t a n x \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)}}{{(7 x^{2} - 3 x + 1)}^{3}}$

$f (x) = (2^{x} + 2^{- x}) . t a n x . \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)} . {(7 x^{2} - 3 x + 1)}^{- 3}$

$f' (x) = (2^{x} + 2^{- x}) . s e c^{2} x . \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)} . {(7 x^{2} - 3 x + 1)}^{- 3} + t a n x . (Q (x))$

$f' (0) = 2.1 \sqrt[]{\frac{π}{4}} . 1$

$= \sqrt[]{π}$

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.8L

Reviews

1.8M

Answers

Learn more about...

Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering

An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k .