Question: Suppose there is a room filled with urns of two types. Type I urns contain 5 blue balls and 5 red balls. Type II urns contain 2 red balls and 8 blue balls. There are 700 Type I urns and 300 Type II urns in the room. They are distributed randomly and look alike. An urn is selected at random from the room and a ball is drawn from it.
A) What is the probability that the urn is Type I?
So that will be: total type 1 urns/total urns $= 700/1000 = 0.7$
B) What is the probability that the ball drawn is red?
I'm confused with this part of the question. My answer is:
(type 1 red balls) $\times$ (type 2 red balls)
(5 red balls/10 total balls) $\times$ (2 red balls/10 total balls) $= 1/10$
Is $1/10$ the probability to draw a red ball correct?
2 Answers
Type 1 Urn:
700 Total Urns (70%) 50% Chance Red
Type 2 Urn:
300 Total Urns (30%) 20% Chance Red
The Urns are indistinguishable, as are the balls. Therefore, when selecting a ball, there is a 70% chance the urn is Type 1 with a 50% chance of leading to a Red. There is a 30% chance the urn is Type 2 with a 20% chance of being a Red.
To find the total probability of drawing a red ball, you can take 70% $\times$ 50% $+$ 30% $\times$ 20%
$0.35+.06=0.41$ or 41%
700 Urns x 5 Red balls per urn = 3500 Red 300 Urns x 2 Red balls per urn = 600 Red
4100 Red 10,000 Total
$\frac{4100}{10000} = 0.41$ or 41%
Think of this as a weighted average.
The probability of a type I urn is $0.7$
The probability of then selecting a red ball is $\frac{1}{2}$
The probability of a type II urn is $0.3$
The probability of then selecting a red ball is $\frac{1}{5}$
Thus, the desired probability is $$\left(0.7\cdot\frac{1}{2}\right)+\left(0.3\cdot\frac{1}{5}\right)=0.41$$
