Question

A bag A contains a white and 3 red balls and
a bag B contains 4 white and 5 red balls. One
ball is drawn at random from one of the bags
and
it is found to be red. Find the probability that the
red ball drawn is from bag B.​

Answer 1

Answer:

Probability = 20/47

Step-by-step explanation:

Given:

• Bag A contains a white ball and 3 red balls
• Bag B contains 4 white balls and 5 red balls

To Find:

• Probability that the red ball drawn is from bag B

Solution:

Let E be the event that the ball is red.

Probability of choosing the bag A or B is given by,

$\sf P(A)=P(B)=\dfrac{1}{2}$

Given that bag A contains 3 red balls and 1 white ball.

Hence the probability of drawing a red ball from bag A is given by,

$\sf P(E/A)=\dfrac{3}{4}$

Also by given bag B contains 5 red balls and 4 white balls.

Therefore probability of drawing a red ball from bag B is given by,

$\sf P(E/B)=\dfrac{5}{9}$

Now we have to find the probability that the red ball was drawn from the bag B, that is P(B/E)

By Bayes theorem,

$\sf P(B/E)=\dfrac{P(E/B)\times P(B)}{P(E/B)\times P(B)+P(E/A)\times P(A)}$

Substituting the values we get,

$\sf P(B/E)=\dfrac{5/9 \times 1/2}{5/9\times 1/2+3/4\times 1/2}$

$\sf \implies \dfrac{5/18}{5/18+3/8}$

$\sf \implies \dfrac{5/18}{47/72}$

$\sf \implies \dfrac{20}{47}$

Hence the probability that the ball drawn is from bag B is 20/47.

Answer 2

Given : A bag A contains one white and 3 red balls and a bag B contains 4 white and 5 red balls.

Need To Find : The probability that the red ball drawn is from bag B.

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider the A be the Probability of choosing Bag A , B be the Probability of choosing Bag B & R be the Probability of the ball is red .

Now ,

• PROBABILITY of choosing the Bag A or B is given by :

$\qquad \quad \leadsto \sf P(A) = \dfrac{1}{2}$

⠀⠀⠀⠀⠀AND ,

$\qquad \quad \leadsto \sf P(B) = \dfrac{1}{2}$

Now ,

• PROBABILITY of choosing a red ball from Bag A :

⠀⠀⠀⠀━━ A bag A contains one white and 3 red balls .

$\qquad \quad \leadsto \sf P \bigg( \dfrac{R}{A} \bigg) \:\:=\:\: \dfrac{3}{4}$

• PROBABILITY of choosing red ball from Bag B .

⠀⠀⠀⠀━━ A bag B contains 4 white and 5 red balls.

$\qquad \quad \leadsto \sf P \bigg( \dfrac{R}{B} \bigg) \:\:=\:\: \dfrac{5}{9}$

Now ,

• By BAYES'S THEOREM :

⠀⠀⠀⠀⠀⠀The probability that the red ball drawn is from bag B.

$\qquad \dag \qquad \boxed {\pmb{\mathrm{ \pink{ \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{P (R/ B ) \times P(B)}{P(R/B)\:\:\times P(B)\:\:+\: P(R/A)\:\times P(A) }}}}}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{P (R/ B ) \times P(B)}{P(R/B)\:\:\times P(B)\:\:+\: P(R/A)\:\times P(A)}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{P (R/ B ) \times P(B)}{P(R/B)\:\:\times P(B)\:\:+\: P(R/A)\:\times P(A)}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ (5/ 9 ) \times (1 /2 )}{(5/9)\:\:\times (1/2)\:\:+\: (3/4)\:\times (1/2)}$

$\qquad \dashrightarrow \sf \:\:\: \bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ (5/ 18 )}{(5/9)\:\:\times (1/2)\:\:+\: (3/4)\:\times (1/2)}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ (5/ 18 )}{(5/18)\:\:\:\:+\: (3/8)\:}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ (5/ 18 )}{(47/72)\:\:\:\:\:}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ (5/ \cancel {18} )}{(47/\cancel {72})\:\:\:\:\:}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ (5 )}{(47/4)\:\:\:\:\:}$

$\qquad \dashrightarrow \sf \:\:\: P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ 20}{47\:\:\:\:\:}$

$\qquad \therefore \pmb{\underline{\purple{\:P\bigg( \dfrac{B}{R} \bigg) \:\:=\:\dfrac{ 20}{47\:\:\:\:\:} }} }\:\bigstar$

$\qquad \therefore \underline{\sf Hence, \:\:The \:Probability \:of \:choosing \: red\;ball\:from \:Bag \: B \:is \:\bf 20/47 }$