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Home » There are three urns containing     white and     black balls,     white and     black balls, and     white and     black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
There are three urns containing

    \[2\]

white and

    \[3\]

black balls,

    \[3\]

white and

    \[2\]

black balls, and

    \[4\]

white and

    \[1\]

black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Probability
There are three urns containing

    \[2\]

white and

    \[3\]

black balls,

    \[3\]

white and

    \[2\]

black balls, and

    \[4\]

white and

    \[1\]

black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Given, we have

    \[3\]

urns:

Urn

    \[1\]

=

    \[2\]

white and

    \[3\]

black balls

Urn

    \[2\]

=

    \[3\]

white and 2 black balls

Urn

    \[3\]

=

    \[4\]

white and

    \[1\]

black balls

Now, the probabilities of choosing either of the urns are

    \[P({{U}_{1}})\text{ }=\text{ }P({{U}_{2}})\text{ }=\text{ }P({{U}_{3}})\text{ }=\text{ }1/3\]

Let H be the event of drawing white ball from the chosen urn.

So,

    \[P(H/{{U}_{1}})\text{ }=\text{ }2/5,\text{ }P(H/{{U}_{2}})\text{ }=\text{ }3/5\]

and

    \[P(H/{{U}_{3}})\text{ }=\text{ }4/5\]

By using Baye’s Theorem, we have

Therefore, the required probability is

    \[1/3\]

.

i

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