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Question
A random variable has the following probability distribution:
| X = xi : | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P (X = xi) : | 0 | 2 p | 2 p | 3 p | p2 | 2 p2 | 7 p2 | 2 p |
The value of p is
Options
1/10
−1
−1/10
1/5
Solution
1/10
We know that the sum of probabilities in a probability distribution is always 1.
∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1
\[\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2 p^2 + 7 p^2 + 2p = 1\]
\[ \Rightarrow 10 p^2 + 9p - 1 = 0\]
\[ \Rightarrow \left( 10p - 1 \right)\left( p + 1 \right) = 0\]
\[ \Rightarrow p = \frac{1}{10} \text{ or } - 1 \left( \text{ Neglecting - 1 as the value of the probabilitiy cannot be negative } \right)\]
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