As box contains cards numbered as 1,1,2,2 and 3

∴ possible sums of card numbers are 2,3,4 and 5

Hence, X can take values 2,3,4 and 5

X=2 [ when drawn cards are (1,1)]

X=3 [when drawn cards are (1,2) or (2,1)]

X=4 [when drawn cards are (2,2) or (3,1) or (1,3)]

X=5 [when drawn cards are (2,3) or (3,2)]

As Y is a random variable representing maximum of the two numbers drawn

∴ Y can take values 1,2 and 3.

Y=1 [when drawn cards are 1 and 1]

Y=2 [when drawn cards are (1,2) or (2,2) or (2,1)]

Y=3 [when drawn cards are (1,3) or (3,1) or (2,3) or (3,2)]

Note : P(1) represents probability of drawing card numbered as 1, similarly P(2) and P(3)

∴ P(X=2) = P(1)P(1) =

[For drawing first card we had 2 favourable outcomes as 1,1 out of total 5 ,in second time of drawing ,as we drew a card numbered as 1 we are having 1 favourable outcome out of total remaining of 4]

Similarly,

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

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

P(X=5) = P(2)P(3)+P(3)P(2) =

Similarly,

P(Y=1) = P(1)P(1) =

P(Y=2) = P(1)P(2)+P(2)P(1)+P(2)P(2) =

P(Y=3) = P(2)P(3)+P(3)P(2)+ P(1)P(3)+P(3)P(1)

=

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean for(X) = 0.2+1.2+1.2+1 = 3.6

Variance for(X) = 0.4+3.6+4.8+5.0-3.6² = 13.8-3.6² = 0.84

Similarly probability distribution for Y is given below:

∴ Mean for(Y) = 0.1+1.0+1.2 = 2.3

Variance for(Y) = 0.1+2.0+3.6-2.3² = 5.7-2.3² = 0.41