Question

A bag contains 4 green marbles, 6 red marbles, 14 orange marbles, 5 brown marbles, and 8 blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
48/1369
14/1369
27/37
14/37

Answer 1

Each of two urns contains green balls and red balls. Urn I contains 10 green balls and 14 red balls. Urn II contains 4 green balls and 11 red balls. If a ball is drawn from each urn, what is P(red and red)?
1/9
25/39
79/60
77/180

Answer 2

The probability of choosing a red marble on the first draw is 6/37. Since the marble is replaced, the probability of choosing a blue marble on the second draw is 8/37. Therefore, the probability of P(red, then blue) is:

P(red, then blue) = P(red) * P(blue) = (6/37) * (8/37) = 48/1369

So the answer is 48/1369.

Answer 3

The probability of drawing a red ball from Urn I is 14/(10+14) = 7/12. The probability of drawing a red ball from Urn II is 11/(4+11) = 11/15.

To find the probability of both events happening together, we multiply the probabilities:

P(red and red) = P(red from Urn I) * P(red from Urn II)

P(red and red) = (7/12) * (11/15)

P(red and red) = 77/180

Therefore, the probability of drawing a red ball from both urns is 77/180.

Hence, the answer is 77/180.