Question

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red is:

Options

• `21/49`

• `27/49`

• `26/49`

• `32/49`

Answer 1

`32/49`

Explanation:

Step -1: Taking the events and framing an equation.

Let E₁ be an event in which a red ball is drawn and a green ball is placed in a bag.

E₂ represents the event of drawing a green ball and putting a red ball in the bag, while E represents the event of drawing a red ball in the second draw.

`P(E) = P(E_1) xx P (E/E_1) + P(E_2) xx P(E/E_2)`

Step -2: Calculating the required probability.

`P(E_1)P (E/E_1)` = drawing a red ball on the second draw when in the first draw we took a red ball and placed a green one 

`P(E_1)P(E/E_1) = 5/7 xx 4/7`  ......(As earlier there were 5 red balls out of total 7 and then we drew a red and placed a green so now there are 4 red balls out of 7.)

`P(E_2)P(E/E_2)` =  drawing a red ball on the second draw when in the first draw we took a green ball and placed a red one

`P(E_2) xx P(E/E_2) = 2/7 xx 6/7` ......(As there were 2 green balls out of 7 and then we put a red ball. So now we have to draw a red ball and there are 6 reds out of 7.)

P(E) = `5/7 xx 4/7 + 2/7 xx 6/7 = (20 + 12)/49 = 32/49`

Hence, the total probability is `32/49`.