Box-I-has-3-red-and-5-white-balls-while-Box-II-contains-4-red-and-2-white-balls-A-ball-is-chosen-at-random-from-the-first-box-and-placed-in-the-second-box-without-observing-its-colour-Then-a-ball-

Question Number 3824 by Yozzii last updated on 21/Dec/15

B o x I h a s 3 r e d a n d 5 w h i t e b a l l s,

w h i l e B o x I I c o n t a i n s 4 r e d a n d 2

w h i t e b a l l s . A b a l l i s c h o s e n a t r a n d o m

f r o m t h e f i r s t b o x a n d p l a c e d i n t h e

s e c o n d b o x w i t h o u t o b s e r v i n g i t s c o l o u r .

T h e n a b a l l i s d r a w n f r o m t h e s e c o n d

b o x . F i n d t h e p r o b a b i l i t y t h a t i t i s w h i t e .

Commented by Rasheed Soomro last updated on 21/Dec/15

A c t - I : D r a w i n g a b a l l f r o m B o x I

T o t a l b a l l s = 8, o u t o f w h i c h a r e 5 w h i t e

b a l l s .

p r o b a b i l i t y o f b e i n g w h i t e b a l l i s \frac{5}{8}

A c t - I I : 4 r e d 2 w h i t e o n e \overset{\frac{5}{8}}{w h i t e} / \overset{\frac{3}{8}}{r e d}

O n e a d d e d b a l l w i t h r e s p e c t t o p r o b a b i l i t y

i s \frac{5}{8} p a r t s w h i t e a n d \frac{3}{8} p a r t s r e d

T o t a l b a l l s a r e 7

O u t o f w h i c h 2 + \frac{5}{8} = \frac{21}{8} w h i t e b a l l s

p r o b a b i l i t y b e i n g w h i t e = \frac{\frac{21}{8}}{7} = \frac{21}{8 \times 7} = \frac{3}{8}

Commented by prakash jain last updated on 21/Dec/15

E_{1} = White ball drawn from Box 1

E_{2} = Red ball drawn from Box 1

A = getting white ball from Box 2

P (A) = P (A ∣ E_{1}) P (E_{1}) + P (A ∣ E_{2}) P (E_{2})

P (E_{1}) = 5 / 8

P (E_{2}) = 3 / 8

P (A ∣ E_{1}) = 3 / 7

P (A ∣ E_{2}) = 2 / 7

= \frac{3}{7} \times \frac{5}{8} + \frac{2}{7} \times \frac{3}{8} = \frac{15 + 6}{56} = \frac{21}{56} = \frac{3}{8}

Commented by Yozzii last updated on 21/Dec/15

T h a n k s

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