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  4. A random variable ‘X’ has the following probability distribution: x 1 2 3 4 5 6 7 P(x) k-1 3k k 3k 3k2 k2 k2 + k Then the value of k is

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Q. A random variable ‘X’ has the following probability distribution:
x 1 2 3 4 5 6 7
P(x) k-1 3k k 3k $3k^2$ $k^2$ $k^2 + k$

Then the value of k is

KCETKCET 2019Probability

Solution:

$\sum P_{i}=1$
$k-1+3k+k+3k+3k^{2}+k^{2}+k^{2}+k=1$
$5k^{2}+9k-2=0$
$5k^{2}+10k-k-2=0$
$5k\left(k+2\right)-1\left(k+2\right)=0$
$\left(5k-1\right)\left(k+2\right)=0$
$k=\frac{1}{5},-2\left(k=-2\, is\, not\, possible\right).$