Q. 9

# <span lang="EN-US

Given : There are 5 cards, numbers 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.

To find : mean (𝓊) and variance (σ2) of X

Formula used :

Mean = E(X) =

Variance = E(X2) -

There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement.

X denote the sum of the numbers on two cards drawn

The minimum value of X will be 3 as the two cards drawn are 1 and 2

The maximum value of X will be 9 as the two cards drawn are 4 and 5

For X = 3 the two cards can be (1,2) and (2,1)

For X = 4 the two cards can be (1,3) and (3,1)

For X = 5 the two cards can be (1,4) , (4,1) , (2,3) and (3,2)

For X = 6 the two cards can be (1,5) , (5,1) , (2,4) and (4,2)

For X = 7 the two cards can be (3,4) , (4,3) , (2,5) and (5,2)

For X = 8 the two cards can be (5,3) and (3,5)

For X = 9 the two cards can be (4,5) and (4,5)

Total outcomes = 20

P(3) = =

P(4) = =

P(5) = =

P(6) = =

P(7) = =

P(8) = =

P(9) = =

The probability distribution table is as follows,

Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5) + x6P(x6) + x7P(x7)

Mean = E(X) = 3() + 4() + 5() + 6() + 7() + 8() + 9()

Mean = E(X) = + + + + + + = = = 6

Mean = E(X) = 6

= = 36

E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5) + P(x6) + P(x7)

E(X2) = () + () + () + () + () + () + ()

E(X2) = + + + + + + = = = 39

E(X2) = 39

Variance = E(X2) - = 39 – 36 = 3

Variance = E(X2) - = 3

Mean = 6

Variance = 3

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