bag 1 contains 5 red balls and 3 black balls. bag 2 contains 3 red balls and 1 black ball. bag 3 contains 4 red balls and 2 black balls. if a bag is selected at random and a ball is drawn, find the probability it will be red.
Answers
Answer:
49 / 72
Step-by-step explanation:
By the Law of Total Probability:
P(red)
= P(red | bag 1)×P(bag 1) + P(red | bag 2)×P(bag 2) + P(red | bag 3)×P(bag 3)
= (5/8)×(1/3) + (3/4)×(1/3) + (4/6)×(1/3)
= (1/3)×(5/8 + 3/4 + 2/3)
= (1/3)×(15 + 18 + 16)/24
= 49/72
Probability of drawing a red ball from any of the 3 bags = 49/72
Step-by-step explanation:
Given: A bag 1 contains 5 red balls and 3 black balls. A bag 2 contains 3 red balls and 1 black ball. A bag 3 contains 4 red balls and 2 black balls.
Find: If a bag is selected at random and a ball is drawn, find the probability it will be red.
Solution:
Probability of red ball in first bag = number of red balls / total balls in the bag = 5/8 = p1
Probability of red ball in second bag = number of red balls / total balls in the bag = 3/4 = p2
Probability of red ball in third bag = number of red balls / total balls in the bag = 4/6 = p3
According to the law of total probability, the probability that a random ball drawn from any of the 3 bags is red = (probability of drawing a bag) * (Sum of probability of drawing a red ball in the 3 bags
= (1/3)*(5/8 + 3/4 + 2/3)
= (1/3)*(15 + 18 + 16)/24
= 49/72