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# The Random Variable X Can Take Only the Values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = P and P(X = 2) = P(X = 3) Such that Sigmap_I X_I^2 = 2sigmap_Ix_I. Find the Value of P - CBSE (Science) Class 12 - Mathematics

#### Question

The random variable X can take only the values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that Sigmap_i x_i^2 = 2Sigmap_ix_i. Find the value of p

#### Solution

It is given that the random variable X can take only the values 0, 1, 2, 3.

Given:

P(X = 0) = P(X = 1) = p

P(X = 2) = P(X = 3)

Let P(X = 2) = P(X = 3) = q

Now,

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1

=> p + p + (q + q) = 1

=> q = (1- 2p)/2

Since, Sigmap_ix_i^2 = 2Sigmap_ix_i

=> 0 + p(1)^2 + ((1-2p)/2)(2^2 + 3^2) = 2[0 + p + ((1-2p)/2) (2 + 3)]

=> p + 13/2 (1- 2p) = 2[p + 5/2(1-2p)]

=> p + 13/2 - 13p = 2p + 5 - 10p

=> 13/2 - 12p = -8p + 5

=> 4p = 3/2

=> p = 3/8

Is there an error in this question or solution?

#### APPEARS IN

Solution The Random Variable X Can Take Only the Values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = P and P(X = 2) = P(X = 3) Such that Sigmap_I X_I^2 = 2sigmap_Ix_I. Find the Value of P Concept: Variance of a Random Variable.
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The text passage contains a mathematical question and its solution, where the question is about finding the value of a probability p for a random variable X that can take values 0, 1, 2, 3 under certain conditions.
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First, describe the content structure you found in 1-2 sentences.

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# The Random Variable X Can Take Only the Values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = P and P(X = 2) = P(X = 3) Such that Sigmap_I X_I^2 = 2sigmap_Ix_I. Find the Value of P - CBSE (Science) Class 12 - Mathematics

#### Question

The random variable X can take only the values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that Sigmap_i x_i^2 = 2Sigmap_ix_i. Find the value of p

#### Solution

It is given that the random variable X can take only the values 0, 1, 2, 3.

Given:

P(X = 0) = P(X = 1) = p

P(X = 2) = P(X = 3)

Let P(X = 2) = P(X = 3) = q

Now,

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1

=> p + p + (q + q) = 1

=> q = (1- 2p)/2

Since, Sigmap_ix_i^2 = 2Sigmap_ix_i

=> 0 + p(1)^2 + ((1-2p)/2)(2^2 + 3^2) = 2[0 + p + ((1-2p)/2) (2 + 3)]

=> p + 13/2 (1- 2p) = 2[p + 5/2(1-2p)]

=> p + 13/2 - 13p = 2p + 5 - 10p

=> 13/2 - 12p = -8p + 5

=> 4p = 3/2

=> p = 3/8

Is there an error in this question or solution?

#### APPEARS IN

Solution The Random Variable X Can Take Only the Values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = P and P(X = 2) = P(X = 3) Such that Sigmap_I X_I^2 = 2sigmap_Ix_I. Find the Value of P Concept: Variance of a Random Variable.
S
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The text passage contains a mathematical question and its solution, where the question is about finding the value of a probability p for a random variable X that can take values 0, 1, 2, 3 under certain conditions.

TYPE: SINGLE_QUESTION_WITH_ANSWER