If the random variable X takes the values x1,x2,x3,....,x10 with probabilities p(X=xi)=ki, then the value of k is equal to

Question

If the random variable X takes the values x1,x2,x3,....,x10 with probabilities p(X=xi)=ki, then the value of k is equal to (A)  (1/10)  (B)  (1/4)  (C)  (1/55)  (D)  (7/12)

Tardigrade Admin · Accepted Answer

As we know, the sum of probability density function is one. ∴ p(X=x1)+p(X=x2)+...+p(X=x10)=1 ⇒ 1k+2k+3k+....+10k=1 ⇒ (10(10+1)/2)k=1 ⇒ k=(1/55)

Q. If the random variable $X$ takes the values $x_{1}, x_{2}, x_{3}, ...., x_{10}$ with probabilities $p (X = x_{i}) = ki,$ then the value of $k$ is equal to

$\frac{1}{10}$

$\frac{1}{4}$

$\frac{1}{55}$

$\frac{7}{12}$

$\frac{3}{4}$

As we know, the sum of probability density function is one.
$∴$ $p (X = x_{1}) + p (X = x_{2}) + ... + p (X = x_{10}) = 1$
$\Rightarrow$ $1 k + 2 k + 3 k + .... + 10 k = 1$
$\Rightarrow$ $\frac{10 ( 10 + 1 )}{2} k = 1$
$\Rightarrow$ $k = \frac{1}{55}$

Q. If the random variable X takes the values x1​,x2​,x3​,....,x10​ with probabilities p(X=xi​)=ki, then the value of k is equal to

101​

41​

551​

127​

43​

As we know, the sum of probability density function is one. ∴ p(X=x1​)+p(X=x2​)+...+p(X=x10​)=1 ⇒ 1k+2k+3k+....+10k=1 ⇒ 210(10+1)​k=1 ⇒ k=551​