4d
A bag contains 2 white and 3 red balls, and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red.
Find the probability that it was drawn from the bag B
A) \(\frac{5}{52}\)
B) \(\frac{15}{52}\)
C) \(\frac{25}{52}\)
D) \(\frac{35}{52}\)
E) \(\frac{45}{52}\)
This is a conditional probability problem since it is asking: given that the ball is red, what is the probability it is from bag B?
We can use the following formula:
P(bag B, given it’s a red ball) = P(a red ball from bag B)/P(a red ball)
Assuming there is an equal chance of picking either bag, we have:
P(a red ball from bag B) = 5/9 x 1/2 = 5/18
P(a red ball) = P(a red ball from bag B) + P(a red ball from bag A) = 5/9 x 1/2 + 3/5 x 1/2 = 5/18 + 3/10
Therefore, the probability in question is:
(5/18)/(5/18 + 3/10)
Multiply the numerator and the denominator by 90, we have:
25/(25 + 27) = 25/52
Alternate Solution:
We notice that the probability of picking a red ball from bag A is 3/5 and the probability of picking a red ball from bag B is 5/9. If the two probabilities were equal, then the probability that the ball was drawn from bag A would have been equal to the probability that the ball was drawn from bag B. However, since 3/5 = 27/45 > 5/9 = 25/45, there is a greater probability that the ball was drawn from bag A. Thus, the probability that the ball was drawn from bag B should be less than 1/2. Furthermore, since the two probabilities 27/45 and 25/45 are very close to each other, probability that the ball was drawn from bag B should be very close to (but less than) 1/2. Going over the answer choices, 25/52 is the choice that fits best with this.
Answer: C