Question

# A bag contains 6 red balls and 7 white balls. Two balls are drawn without replacement....

A bag contains 6 red balls and 7 white balls. Two balls are drawn without replacement. (Enter your probabilities as fractions.)
(a) What is the probability that the second ball is white, given that the first ball is red?

(b) What is the probability that the second ball is red, given that the first ball is white?

(c) Answer part (a) if the first ball is replaced before the second is drawn.

a)

P(Second ball white | first ball red) = P(Second ball white and first ball red) / P(First balls is red)

= (6/13 * 7/12) / (6/13 * 7/12 + 6/13 * 5 / 12)

= 7 / 12

b)

P(Second ball red | first ball white) = P(Second ball red and first ball white) / P(First balls is white)

= (6/12 * 7/13) / (6/12 * 7/13 + 7/13 * 6 / 12)

= 1 / 2

c)

If first ball is replaced before second drawn,

P(Second ball white | first ball red) = P(Second ball white and first ball red) / P(First balls is red)

= ( 6/13 * 7/13) / (6/13 * 7 / 13 + 6/13 * 6/13)

= 7 / 13

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