\[ P(A) = \frac{1}{10}, \quad P(B|A) = \frac{3}{5}, \quad P(A|B^c) = \frac{1}{5} \] Using the law of total probability: \[ P(B) = P(B|A) P(A) + P(B|A^c) P(A^c) \] Since, \[ P(A^c) = 1 - P(A) = 1 - \frac{1}{10} = \frac{9}{10} \] And given: \[ P(B|A^c) = 1 - P(A|B^c) = 1 - \frac{1}{5} = \frac{4}{5} \] Substituting the values: \[ P(B) = \frac{3}{5} \times \frac{1}{10} + \frac{4}{5} \times \frac{9}{10} \] \[ = \frac{3}{50} + \frac{36}{50} = \frac{39}{50} = 0.78 \]
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | kΒ² | 2kΒ² | 7kΒ² + k |
Then find: \( P(3 < x \leq 6) \)
The sum of the payoffs to the players in the Nash equilibrium of the following simultaneous game is ............
| Player Y | ||
|---|---|---|
| C | NC | |
| Player X | X: 50, Y: 50 | X: 40, Y: 30 |
| X: 30, Y: 40 | X: 20, Y: 20 | |