Question

Two balls are drawn in succession out of a box containing 3 red and 3 white balls. Find the probability that at least 1 ball was​ red, given that the first ball was replaced before the second draw.

Answer 1

Since, the first ball was replaced before the second draw, the probability of getting a red ball on each draw is same and is given by:

P(red) = P(red ball on first draw) = P(red ball on second draw) = 3/6 [Since, there are a total of 6 balls out of which 3 are red]

=> P(red) = 1/2

Similarly, the probability of getting a white ball on each draw is same and is given by:

P(white) = P(white ball on first draw) = P(white ball on second draw) = 3/6 [Since, there are a total of 6 balls out of which 3 are white]

=> P(white) = 1/2

Now, the probability that on the draw of the two balls, at least 1 ball was red is given by:
P(at least one red ball in two draws) = 1 - P(no red ball on two draws)

= 1 - P(no red ball on first draw)*P(no red ball on second draw)

= 1 - P(white ball on first draw)*P(white ball on second draw)

= 1 - (1/2)*(1/2)

= 1 - 1/4

= 3/4 [ANSWER]

= 0.75 [ANSWER]

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