Answer:
[tex]Probability = \frac{4}{15}[/tex]
Step-by-step explanation:
B1 = first bag
B2= second bag
B3 = third bag
Let A = ball drawn is red
Since, there are three bags.
Probability of choosing one bag= P(B1) = P(B2) = P(B3) = 1/3.
From B1: Total balls = 10
3 red + 7 black balls.
Probability of drawing 1 red ball from it , P(A) = 3/10.
From B2: Total balls = 10
8 red + 2 black
Probability of drawing 1 red ball is, P(A) = 8/10
From B3 : Total Balls = 10
4 red + 6 black
Probability of drawing 1 red ball, P(A) = 4/10 .
To find Probability given that the ball drawn is red, that the ball is drawn from the third bag by Bayes' rule.
That is , P(B3|A)
[tex]=\frac{\frac{1}{3} \times \frac{4}{10}} { \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times\frac{8}{10} + \frac{1}{3} \times \frac{4}{10}}[/tex]
[tex]=\frac{4}{30} \times \frac{30}{15}\\\\=\frac{4}{15}[/tex]
Therefore, the probability that it is drawn from the third bag is 4/15.
Answer:
4/15
Step-by-step explanation:
Solution of conditional probability problem:
Given:
Bags (3R,7B), (8R,2B), (4R,6B)
Let
P(R,i) = probability of drawing a red AND from bag i
P(R, 1) = 3/10 * (1/3) = 3/30
P(R, 2) = 8/10 * (1/3) = 8/30
P(R, 3) = 4/10 * (1/3) = 4/30
Let
Let P(R) = probability of drawing a red from any bag
P(R) = sum P(R,i) for i = 1 to 3 using the addition rule
= 3/30 + 8/30 + 4/30
= 15/30
= 1 / 2
Conditional Probability of drawing from the third bag GIVEN that it is a red
= P(3 | R)
= P(R, 3) / P(R)
= 4/30 / (1/2)
= 8/30
= 4 / 15
(Since all bags contain 10 balls, by intuition, 4 red from third / 15 total red = 4/15)
