Respuesta :
Answer:
a) 0.5664 = 56.64% probability that the urn selected was the first one.
b) 0.4336 = 43.36% probability that the urn selected was the second one
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
If both balls are white, what are the following probabilities?
This means that event A is both balls being white.
Probability of both balls being white:
0.8 probability of 4/10 = 0.4 squared(first urn, which each time the ball is drawn will have 4 white balls and 6 red)
1 - 0.8 = 0.2 probability of 7/10 = 0.07 squared(second urn). So
[tex]P(A) = 0.8*(0.4)^2 + 0.2*(0.7)^2 = 0.226[/tex]
a. the probability that the urn selected was the first one
Event B: First urn selected.
The intersection of events A and B is given by both balls being white and selected from the first urn, which has the following probability:
[tex]P(A \cap B) = 0.8*(0.4)^2 = 0.128[/tex]
The desired probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.128}{0.226} = 0.5664[/tex]
0.5664 = 56.64% probability that the urn selected was the first one
b. the probability that the urn selected was the second one
The intersection of events A and B is given by both balls being white and selected from the second urn, which has the following probability:
[tex]P(A \cap B) = 0.2*(0.7)^2 = 0.098[/tex]
The desired probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.098}{0.226} = 0.4336[/tex]
0.4336 = 43.36% probability that the urn selected was the second one
