A random variable has the following probability distribution:
$X = x_i$
$0$
$1$
$2$
$3$
$4$
$5$
$6$
$7$
$P(X = X_i)$
$0$
$2p$
$2p$
$3p$
$p^2$
$2p^2$
$7p^2$
$2p$
A$\frac{1}{10}$
B$-1$
C$-\frac{1}{10}$
D$\frac{1}{5}$
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A$\frac{1}{10}$
We know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
$\Rightarrow 10p^2+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1 ($Negleting $-1$ as the value of the probability cannot be negative$)$
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