Total matches played = 4
Maximum points = \(2 \times 4 = 8\)
So, to get at least 7 points, India needs to win at least 3 matches and one can be draw.
India wins all 4 matches: There is only one possible outcome
Hence, the probability of 4 wins = \((0.5)\times(0.5)\times(0.5)\times(0.5) = 0.0625\)
India wins 3 matches and draws 1: probability in that case = \((0.5)\times(0.5)\times(0.5)\times(0.05) = 0.00625\)
Total probability of India wins 3 matches and draws 1 = \(4\times(0.00625) = 0.025\)
(since 4 cases, any one match can be draw)
\(P = 0.0625 + 0.025 = 0.0875\)
The correct option is (B): 0.0875
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 :
