Question

The first of three urns contains 7 White and 10 Black balls, the second contains 5 White and 12 Black balls and the third contains 17 White balls and no Black ball. A person chooses an urn at random and draws a ball from it. And the ball is found to be White. Find the probabilities that the ball comes from

1. the first urn
2. the second urn
3. the third urn

Answer 1

	White	Black	Total
Urn I	7	10	17
Urn II	5	12	17
Urn III	17	0	17

Let E₁, E₂, and E₃ be the event of choosing the first, second, and third urn.

Let A be the event of drawing a white ball.

Then P(E₁) = P(E₂) = P(E₃) = `1/3`

P`("A"/"E"_1)` = P(drawing a white ball from the first urn) = `7/17`

P`("A"/"E"_2)` = P(drawing a white ball from the second urn) = `5/17`
P`("A"/"E"_3)` = P(drawing a white ball from the third urn) = `17/17`

(i) The probability of drawing a ball from the first urn, being given that it is white, is `"P"("E"_1/"A")`,

`"P"("E"_1/"A") = ("P"("E"_1) xx "P"("A"/"E"_1))/("P"("E"_1) xx "P"("A"/"E"_1) + "P"("E"_2) xx "P"("A"/"E"_2) + "P"("E"_3) xx "P"("A"/"E"_3))`

= `(1/3 xx 7/17)/(1/3 xx 7/17 + 1/3 xx 5/17 + 1/3 xx 17/17)`

= `(1/3 xx 7/17)/(1/3 [7/17 + 5/17 + 17/17])`

= `(7/17)/(29/17)`

= `7/29`

(ii) the second urn

`"P"("E"_2/"A") = ("P"("E"_2) xx "P"("A"/"E"_2))/("P"("E"_1) xx "P"("A"/"E"_1) + "P"("E"_2) xx "P"("A"/"E"_2) + "P"("E"_3) xx "P"("A"/"E"_3))`

= `(1/3 xx 5/17)/(1/3 xx 7/17 + 1/3 xx 5/17 + 1/3 xx 17/17)`

= `(5/17)/(7/17 + 5/17 + 17/17)`

= `(5/17)/(29/17)`

= `5/29`

(iii) the third urn

`"P"("E"_3/"A") = ("P"("E"_3) xx "P"("A"/"E"_3))/("P"("E"_1) xx "P"("A"/"E"_1) + "P"("E"_2) xx "P"("A"/"E"_2) + "P"("E"_3) xx "P"("A"/"E"_3))`

= `(1/3 xx 17/17)/(1/3 xx 7/17 + 1/3 xx 5/17 + 1/3 xx 17/17)`

= `(17/17)/(7/17 + 5/17 + 17/17)`

= `(17/17)/(29/17)`

= `17/29`