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If X is the a random variable with pro...

If X is the a random variable with probability mass function `P(x) = kx `, for x = 1,2,3
= 0 , otherwise
then k = ……..

A

`(1)/(5)`

B

`(1)/(4)`

C

`(1)/(6)`

D

`(2)/(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the value of \( k \) in the probability mass function \( P(x) = kx \) for \( x = 1, 2, 3 \), we need to ensure that the total probability sums up to 1. Here’s how we can solve it step by step: ### Step 1: Write down the probability mass function The probability mass function is given as: \[ P(x) = kx \quad \text{for } x = 1, 2, 3 \] \[ P(x) = 0 \quad \text{otherwise} \] ### Step 2: Write the probabilities for each value of \( x \) We can calculate the probabilities for \( x = 1, 2, 3 \): - For \( x = 1 \): \[ P(1) = k \cdot 1 = k \] - For \( x = 2 \): \[ P(2) = k \cdot 2 = 2k \] - For \( x = 3 \): \[ P(3) = k \cdot 3 = 3k \] ### Step 3: Set up the equation for total probability The total probability must equal 1: \[ P(1) + P(2) + P(3) = 1 \] Substituting the values we calculated: \[ k + 2k + 3k = 1 \] ### Step 4: Simplify the equation Combine like terms: \[ 6k = 1 \] ### Step 5: Solve for \( k \) Now, divide both sides by 6: \[ k = \frac{1}{6} \] ### Conclusion Thus, the value of \( k \) is: \[ \boxed{\frac{1}{6}} \]
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Knowledge Check

  • Let X is a continuous random variable with probability density function f(x)={{:(x/6+k,0lexle3),(0," otherwise"):} The value of k is equal to

    A
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    B
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  • X is continuous random variable with probability density function f(x)=(x^(2))/(8),0 le x le 1 . Then, the value of P(0.2leXle0.5) is

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