Question

In: Statistics and Probability

Two balls are drawn, without replacement, from a bag containing 11 red balls numbered 1-11 and...

Two balls are drawn, without replacement, from a bag containing 11 red balls numbered 1-11 and 5 white balls numbered 12-16. (Enter your probabilities as fractions)

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

(b) What is the probability that both balls are even numbered?

(c) What is the probability that the first ball is red and even-numbered and the second ball is even numbered?

Expert Solution

(a)

Total number of balls in the bag is

11 + 5 = 16

Out of 16 balls, 5 are white so

P(first white) = 5 /16

After first draw, out of remaining 15 balls, 11 are red so

P(Second red | first white) = 11 /15

Answer: 11 /15 = 0.7333

(b)

Even number on red balls are : 2, 4, 6, 8, 10

Even number on white balls are : 12, 14, 16

That is out of 16 balls, 8 are even numbered so the probability that both are even numbered is

Answer: 0.2333

(c)

Out of 16 balls, 5 are red and even-numbered so

P(first ball red and even-numbered) = 5 / 16

After first draw, out of 15 remaining balls 7 are even numbered so

P(second even numbered | first ball red and even-numbered) = 7 / 15

Therefore the probability that the first ball is red and even-numbered and the second ball is even numbered is

P(second even numbered and first ball red and even-numbered) =P(second even numbered | first ball red and even-numbered)* P(first ball red and even-numbered) = (7/15) * (5/16) = 0.1458

Answer: 0.1458

orchestra answered 1 year ago

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