Bag1 = 3 back, 4 white balls
Bag2 = 4 black, 3 white balls
Probability of selecting bag 1 \(=\frac{2}{6}=\frac{1}{3}\)
Probability of selecting bag 2 \(=1-\frac{1}{3}=\frac{2}{3}\)
Probability of choosing a black ball from bag 1 \(=\frac{3}{7}\)
Probability of choosing a black ball from bag 2 \(=\frac{4}{7}\)
Required probability \(=\frac{1}{3}\times\frac{3}{7}+\frac{2}{3}\times\frac{4}{7}\)
\(=\frac{11}{21}\).So the correct option is (A)
List-I | List-II (Adverbs) |
(A) P(exactly 2 heads) | (I) \(\frac{1}{4}\) |
(B) P(at least 1 head) | (II) \(1\) |
(C) P(at most 2 heads) | (III) \(\frac{3}{4}\) |
(D) P(exactly 1 head) | (IV) \(\frac{1}{2}\) |
LIST-I(EVENT) | LIST-II(PROBABILITY) |
(A) The sum of the number is greater than 11 | (i) 0 |
(B) The sum of the number is 4 or less | (ii) 1/15 |
(C) The sum of the number is 4 | (iii) 2/15 |
(D) The sum of the number is 4 | (iv) 3/15 |
Choose the correct answer from the option given below