Question

An urn contains 6 red balls and 3 blue balls. One ball is selected at random and is replaced by a ball of the other color. A second ball is than chosen. What is conditional probability that the first ball selected is red, given that the second ball was red?
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Answer 1

Answer:

ans: there is a probability of getting red ball is more because in red colour there are 6 ball where as blue colour is 3

Answer 2

The conditional probability that the first ball selected is red, given that the second ball is red, is 10 / 17.

• Given,

Number of red balls in the urn = 6

Number of blue balls in the urn = 3

(i) Case 1 : Drawing a red ball in the first turn

• Probability that the first ball drawn is red = 6 / 9 = 2 / 3

• According to the question, if a red ball is drawn from the urn, then a blue ball is added to the urn.

∴  Number of blue balls in the urn now = 3 + 1 = 4

• Number of red ball in the urn now = 6 - 1 = 5

• Total number of balls in the urn now = 4 + 5 = 9

• Probability of drawing a red ball in the second turn if the first ball drawn is red = (2 / 3) × (5 / 9)

= 10 / 27 = Event A

(ii) Case 2 : Drawing a blue ball in the first turn

• Probability of drawing a blue ball from the urn in first turn = 3 / 9 = 1 / 3

• According to the question, if a blue ball is drawn in the first turn, then a red ball is added to the urn.

∴  Number of red balls in the urn now = 6 + 1 = 7

• Number of blue balls in the urn now = 3 - 1 = 2

• Total number of balls in the urn now = 7 + 2 = 9

• Probability of drawing a red ball in the second turn if the first ball drawn is blue = (1 / 3) × (7 / 9)

= 7 / 27 = Event B

• Probability of either event A or event B taking place = (10 / 27) + (7 / 27)

= (10 + 7) / 27

= 17 / 27

• Now, conditional probability of drawing a red ball in the first turn, provided the second ball drawn is red [P(A)] = Probability of event A taking place / Probability of either event A or event B taking place

=> P(A) = (10 / 27) / (17 / 27)

=> P(A) = (10 / 27) / (17 / 27)

=> P(A) = (10 / 27) × (27 / 17)

=> P(A) = (10 × 27) / (27 / 17)

=> P(A) = 10 / 17 (Answer)