Define a random variable. For the following bi-variants probability distribution of X and Y , find

  1. marginal probability mass function of X and Y ,
  2. P(x≤1, Y=2),
  3. P(X≤1)
X/Y 1 2 3 4 5 6
0 0 0 1/32 2/32 2/32 3/32
1 1/16 1/16 1/8 1/8 1/8 1/8
2 1/32 1/32 1/64 1/64 1/64 1/64

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A variable whose value id determined by the outcome of a random experiment is called a random variable.

Given,

xy 1 2 3 4 5 6 P(X=x)
0 0 0 1/32 2/32 2/32 3/32 8/32
1 1/16 1/16 1/8 1/8 1/8 1/8 5/8
2 1/32 1/32 1/64 1/64 1/64 1/64 1/8
P(Y=y) 3/32 3/32 1/64 13/64 13/64 13/64 sum = 1

Solution i):

Marginal pmf of x = p(X = xi) = \(\sum_{j=1}^n p(x_i y_i)\)

∴P(X = 0) = P(X=0, Y=1) + P(X = 0, Y = 2) + P(X = 0, Y = 3) + P(X = 0, Y = 4) + P(X = 0, Y = 5) + P(X = 0, Y = 6)

= 8/32

∴P(X = 1) = P(X=1, Y=1) + P(X = 1, Y = 2) + P(X = 1, Y = 3) + P(X = 1, Y = 4) + P(X = 1, Y = 5) + P(X = 1, Y = 6)

= 5/8

∴P(X = 2) = P(X=2, Y=1) + P(X = 2, Y = 2) + P(X = 2, Y = 3) + P(X = 2, Y = 4) + P(X = 2, Y = 5) + P(X = 2, Y = 6)

= 1/8

Also, Marginal pmf of Y = P(Y=yi) = \(\sum_{i=1}^n p(x_i  y_i)\)

∴P(Y = 1) = P(X=0, Y=1) + P(X = 1, Y = 1) + P(X = 1, Y = 2)

= 3/32

Similarly,

P(Y = 2) = 3/32

P(Y = 3) = 11/64

P(Y = 4) = 13/64

P(Y = 5) = 13/64

P(Y = 6) = 15/64

Solution ii):

P(X ≤ 1, Y = 2)

= P(X = 0, Y= 2) + P(X = 1, Y = 2)

= 0 + 1/16

= 1/16

Solution ii)

P(X ≤ 1)

= P(X = 0) + P(X = 1)

= 8/32 + 5/8

= 7/8

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