Each of two urns contains green balls and red balls. Urn 1 contains 10 green balls and 14 red balls. Urn 2 contains 4 green balls and 11 red balls. If a ball is drawn from each urn, what is p (red and red)
There are two ways to approach this problem, depending on whether the draws are considered independent or dependent events.
Assuming the draws are independent:
The probability of drawing a red ball from Urn 1 is 14/24 (or 7/12), and the probability of drawing a red ball from Urn 2 is 11/15. Since the two draws are independent, we can use the product rule of probability:
p(red and red) = p(red from Urn 1) × p(red from Urn 2)
= (14/24) × (11/15)
= 77/180
≈ 0.4286
Therefore, the probability of drawing two red balls is approximately 0.4286.
Assuming the draws are dependent:
If we assume that a ball is drawn from Urn 1 first, and then without replacement, a ball is drawn from Urn 2, the probability of drawing two red balls depends on whether the first ball drawn is red or green.
If the first ball drawn is red, then there are 13 red balls left in Urn 1 out of a total of 23 balls, and 11 red balls left in Urn 2 out of a total of 14 balls. Therefore, the probability of drawing a red ball from each urn in this case is:
p(red and red | first ball is red) = (13/23) × (11/14)
= 143/322
≈ 0.444
If the first ball drawn is green, then there are 14 red balls left in Urn 1 out of a total of 23 balls, and 11 red balls left in Urn 2 out of a total of 14 balls. Therefore, the probability of drawing a red ball from each urn in this case is:
p(red and red | first ball is green) = (14/23) × (11/14)
= 77/161
≈ 0.478
To find the overall probability of drawing two red balls, we need to consider the probability of drawing a red ball first and multiply it by the respective conditional probabilities:
p(red and red) = p(first ball is red) × p(red and red | first ball is red) + p(first ball is green) × p(red and red | first ball is green)
= (14/24) × (143/322) + (10/24) × (77/161)
≈ 0.430
Again, the probability of drawing two red balls is approximately 0.430.