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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

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

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alok kumar singh | Contributor-Level 10

4 months ago

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

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12

P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18

P(X = 12)

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

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12

P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18

P(X = 12) = P(6, 6) = 1 x 36

Therefore, the required probability distribution is as follows.

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	1/36	1/18	1/12	1/9	5/36	1/6	5/36	1/9	1/12	1/18	1/36

$\begin{array}{l} T h e n, E (X) = \sum_{}^{} X_{i} . P (X_{i}) \\ = 2 \times \frac{1}{36} + 3 \times \frac{1}{18} + 4 \times \frac{1}{12} + 5 \times \frac{1}{9} + 6 \times \frac{5}{36} + 7 \times \frac{1}{6} + 8 \times \frac{5}{36} + 9 \times \frac{1}{9} + 10 \times \frac{1}{12} + 11 \times \frac{1}{18} + 12 \times \frac{1}{36} \\ = \frac{1}{18} + \frac{1}{6} + \frac{1}{3} + \frac{5}{9} + \frac{5}{6} + \frac{7}{6} + \frac{10}{9} + 1 + \frac{5}{6} + \frac{11}{18} + \frac{1}{3} \\ = 7 \end{array}$

$\begin{array}{l} E (X^{2}) - \sum_{}^{} {X^{2}}_{i} . P (X_{i}) \\ = 4 \times \frac{1}{36} + 9 \times \frac{1}{18} + 16 \times \frac{1}{12} + 25 \times \frac{1}{9} + 36 \times \frac{5}{36} + 49 \times \frac{1}{6} + 64 \times \frac{5}{36} + 81 \times \frac{1}{9} + 100 \times \frac{1}{12} + 121 \times \frac{1}{18} + 144 \times \frac{1}{36} \\ = \frac{1}{9} + \frac{1}{2} + \frac{4}{3} + \frac{25}{9} + 5 + \frac{49}{6} + \frac{80}{9} + 9 + \frac{25}{3} + \frac{121}{18} + 4 \\ = \frac{987}{18} = \frac{329}{6} = 54.833 \end{array}$

$\begin{array}{l} T h e n, V a r (X) = E (X^{2}) - [E {(X)}^{2}] \\ = 54.833 - {(7)}^{2} \\ = 54.833 - 49 \\ = 5.833 \end{array}$

$∴$ Standard derivation $\begin{array}{l} =\sqrtvar(X) \\ =\sqrt5.833 \\ = 2.415 \end{array}$

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

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