NCERT solution Class 12 probability

NCERT Solutions

Class 12 Maths

Probability

NCERT Questions

Ex.13.1 Q.1
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Ex.Misc.Q.1
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Ex.10.1 Q.1
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Ex.13.1 Q.1
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Ex.13.4 Q.1
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Ex.Misc. Q.1
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Ex.13.4 Q.13

Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1)

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

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1)

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1)

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1)

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1)

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2)

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) =

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4)

P(X = 11) = P(5, 6) + P(6, 5)

P(X = 12) = P(6, 6)

Therefore, the required probability distribution is as follows:

X	2	3	4	5	6	7	8	9	10	11	12
P(X)

Then, E(X) = Σ X_i × P(X_i)

Now, E(X) = Σ X²_i × P(X_i)

Again, Var(X) = E[X²] – [E(X)]²

= 54.833 – 7²

= 54.833 – 49

= 5.833

So, Standard deviation = √[Var(X)]

= √(5.833)

= 2.415

Probability Class 12 Maths NCERT Chapter 13 NCERT Solutions

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Attempt to Solve Yourself: Before looking at the solution, try to solve or answer the question independently. This practice will help you test your knowledge and problem-solving skills.
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