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

Question

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 (A) (2/7)  (B) (1/5)  (C) (1/10)  (D) -2

Tardigrade Admin · Accepted Answer

∑ Pi=1 k-1+3k+k+3k+3k2+k2+k2+k=1 5k2+9k-2=0 5k2+10k-k-2=0 5k(k+2)-1(k+2)=0 (5k-1)(k+2)=0 k=(1/5),-2(k=-2 is not possible).

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 $3 k^{2}$ $k^{2}$ $k^{2} + k$

Then the value of k is

$\frac{2}{7}$

$\frac{1}{5}$

$\frac{1}{10}$

-2

$\sum P_{i} = 1$
$k - 1 + 3 k + k + 3 k + 3 k^{2} + k^{2} + k^{2} + k = 1$
$5 k^{2} + 9 k - 2 = 0$
$5 k^{2} + 10 k - k - 2 = 0$
$5 k (k + 2) - 1 (k + 2) = 0$
$(5 k - 1) (k + 2) = 0$
$k = \frac{1}{5}, - 2 (k = - 2 i s n o t p oss ib l e) .$

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 3k2 k2 k2+k Then the value of k is

72​

51​

101​

-2

∑Pi​=1 k−1+3k+k+3k+3k2+k2+k2+k=1 5k2+9k−2=0 5k2+10k−k−2=0 5k(k+2)−1(k+2)=0 (5k−1)(k+2)=0 k=51​,−2(k=−2isnotpossible).

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 $3 k^{2}$ $k^{2}$ $k^{2} + k$

Then the value of k is

$\frac{2}{7}$

$\frac{1}{5}$

$\frac{1}{10}$

$\sum P_{i} = 1$
$k - 1 + 3 k + k + 3 k + 3 k^{2} + k^{2} + k^{2} + k = 1$
$5 k^{2} + 9 k - 2 = 0$
$5 k^{2} + 10 k - k - 2 = 0$
$5 k (k + 2) - 1 (k + 2) = 0$
$(5 k - 1) (k + 2) = 0$
$k = \frac{1}{5}, - 2 (k = - 2 i s n o t p oss ib l e) .$