# There are 4 cards

Clearly, x can take values 4, 6, 8, 10, 12

We have

P(x = 4) = P [getting pair (1, 3) or (3, 1)]

P(x = 6) = P [getting pair (1, 5) or (5, 1)]

P(x = 8) = P [getting pair (1, 7) or (7, 1) or (3, 5) or (5, 3)]

P(x = 10) = P [getting pair (3, 7) or (7, 3)]

P(x = 12) = P[getting pair (7, 5) or (5, 7)]

.

Mean = ∑ xi P(xi)

= 8

Variance = E(x2) - [E(x)]2

.

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