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)
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :
