Question

Q1.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

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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,

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

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

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

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

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