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The probability distribution of a random...

The probability distribution of a random variable X is given as
`{:(X,-5,-4,-3,-2,-1,0,1,2,3,4,5),(P(X),p,2p,3p,4p,5p,7p,8p,9p,10p,11p,12p):}`
Then, the value of p is

A

`1/72`

B

`3/73`

C

`5/72`

D

`1/74`

Text Solution

AI Generated Solution

The correct Answer is:
To find the value of \( p \) in the given probability distribution of the random variable \( X \), we follow these steps: ### Step 1: Write down the probability distribution The random variable \( X \) takes the values: \[ X = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} \] with corresponding probabilities: \[ P(X) = \{p, 2p, 3p, 4p, 5p, 7p, 8p, 9p, 10p, 11p, 12p\} \] ### Step 2: Set up the equation for total probability The sum of all probabilities must equal 1: \[ p + 2p + 3p + 4p + 5p + 7p + 8p + 9p + 10p + 11p + 12p = 1 \] ### Step 3: Combine like terms Add all the coefficients of \( p \): \[ (1 + 2 + 3 + 4 + 5 + 7 + 8 + 9 + 10 + 11 + 12)p = 1 \] Calculating the sum of the coefficients: \[ 1 + 2 + 3 + 4 + 5 + 7 + 8 + 9 + 10 + 11 + 12 = 72 \] Thus, we have: \[ 72p = 1 \] ### Step 4: Solve for \( p \) To find \( p \), divide both sides by 72: \[ p = \frac{1}{72} \] ### Final Answer The value of \( p \) is: \[ \boxed{\frac{1}{72}} \]
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Knowledge Check

  • A random variable has the following probability dustribution. {:(x:,0,1,2,3,4,5,6,7),(p(x):,0,2p,2p,3p,p^2,2p^2,7p^2,2p):} The value of p, is

    A
    `1//10`
    B
    `-1`
    C
    `-1//10`
    D
    none of these

  • The probability distribution of random variable X with two missing probabilities p_(1) and p_(2) is given below {:(X, P(X)),(1, k), (2, p_(1)),(3, 4k),(4, p_(2)),(5, 2k):} It is further given that P(X le 2) = 0.25 and P(X ge 4) = 0.35 . Consider the following statements 1. p_(1) = p_(2) 2. p_(1) + p_(2) = P(X = 3) which of the statements given above is/are correct ?

    A
    1 only
    B
    2 only
    C
    Both 1 and 2
    D
    Neither 1 nor 2

  • For a binomial variable X if n=5 and P(X=1)=8P(X=3), then p=

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    `(4)/(5)`
    B
    `(1)/(5)`
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    `(1)/(3)`
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    `(2)/(3)`

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