<p><strong>There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let <math><mrow><mi>X</mi></mrow></math> denote the sum of the numbers on the two cards drawn. Find the mean and variance of <math><mrow><mi>X</mi></mrow></math> .</strong></p>

This is a Long Answer Type Questions as classified in NCERT ExemplarSol:Here, Sample space S={(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4)(4,2),(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)}∴ n(S)=20Let X be the random variable denoting the sum of the numbers on two cards drawn.∴ X=3,4,5,6,7,8,9So, P(X=3)=220 P(X=4)=220 P(X=5)=420 P(X=6)=420 P(X=7)=420 P(X=8)=220 P(X=9)=220∴ the mean E(X)=∑i=1nXiPi =3×220+4×220+5×420+6×420+7×420+8×220+9×220 =620+820+2020+2420+2820+1620+1820=12020=6∴ E(X2)=∑i=1nPiXi2 =9×220+16×220+25×420+36×420+49×420+64×220+81×220 =1820+3220+10020+14420+19620+12820+16220=78020=39∴ Variance(X)=E(X2)−[E(X)]2 =39−(6)2=39−36=3

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There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two cards drawn. Find the mean and variance of $X$ .

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alok kumar singh | Contributor-Level 10

3 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, S a m p l e s p a c e S = {\begin{cases} (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (1, 4), (4, 1), (1, 5), (5, 1), (2, 4) \\ (4, 2), (2, 5), (5, 2), (3, 4), (4, 3), (3, 5), (5, 3), (5, 4), (4, 5) \end{cases}} \\ ∴ n (S) = 20 \\ L e t X b e t h e r a n d o m v a r i a b l e d e n o t i n g t h e s u m o f t h e n u m b e r s o n t w o c a r d s d r a w n . \\ ∴ X = 3, 4, 5, 6, 7, 8, 9 \\ S o, P (X = 3) = \frac{2}{20} P (X = 4) = \frac{2}{20} \\ P (X = 5) = \frac{4}{20} P (X = 6) = \frac{4}{20} \\ P (X = 7) = \frac{4}{20} P (X = 8) = \frac{2}{20} \\ P (X = 9) = \frac{2}{20} \\ ∴ t h e m e a n E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = 3 \times \frac{2}{20} + 4 \times \frac{2}{20} + 5 \times \frac{4}{20} + 6 \times \frac{4}{20} + 7 \times \frac{4}{20} + 8 \times \frac{2}{20} + 9 \times \frac{2}{20} \\ = \frac{6}{20} + \frac{8}{20} + \frac{20}{20} + \frac{24}{20} + \frac{28}{20} + \frac{16}{20} + \frac{18}{20} = \frac{120}{20} = 6 \\ ∴ E (X^{2}) = \sum_{i = 1}^{n} P_{i} X_{i}^{2} \\ = 9 \times \frac{2}{20} + 16 \times \frac{2}{20} + 25 \times \frac{4}{20} + 36 \times \frac{4}{20} + 49 \times \frac{4}{20} + 64 \times \frac{2}{20} + 81 \times \frac{2}{20} \\ = \frac{18}{20} + \frac{32}{20} + \frac{100}{20} + \frac{144}{20} + \frac{196}{20} + \frac{128}{20} + \frac{162}{20} = \frac{780}{20} = 39 \\ ∴ V a r i a n c e (X) = E (X^{2}) - {[E (X)]}^{2} \\ = 39 - {(6)}^{2} = 39 - 36 = 3 \end{array}$

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There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two cards drawn. Find the mean and variance of $X$ .

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There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X .

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There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two cards drawn. Find the mean and variance of $X$ .