Question

Calculate the probability mass function of Z = X + Y where X and Y are statistically independent and identically distributed binomial random variables with N = 2 and p = 0.4 . The probability mass functions for X and Y are
P ( X = j ) = P ( Y = j ) = ( 2 j ) ( 0.4 ) j ( 0.6 ) 2 − j = { 0.36 j = 0 0.48 j = 1 0.16 j = 2

Determine P(Z=0), P(Z=1), P(Z=2), and P(Z=3)

Answer 1

Answer:-

Given That:-

Calculate the probability mass function of Z = X + Y where X and Y are statistically independent and identically distributed binomial random variables with N = 2 and p = 0.4 . The probability mass functions for X and Y are
P ( X = j ) = P ( Y = j ) = ( 2 j ) ( 0.4 ) j ( 0.6 ) 2 − j = { 0.36 j = 0 0.48 j = 1 0.16 j = 2

Determine P(Z=0), P(Z=1), P(Z=2), and P(Z=3)

Given,

Given the probability meass function for X and Y are

$P[X=j]=P[Y=j]=\binom{2}{j}(0.4)^j(0.6)^{2-j}$

  

0.36, j = 0 0.48, j = 1 0.16, j = 2

The probability mass function of Z = X + Y

P[Z = 0] = P[X + Y = 0]

= P[X = 0 and Y = 0]

= P[X = 0] P[Y = 0]

= 0.36 * 0.36

=0.1296

P[Z =1] = P[X + Y = 1]

= P[X = 0 and Y = 1] + P[X = 1 and Y = 0]

= P[X = 0] P[Y = 1] + P[X = 1] P[Y = 0]

= 0.39 * 0.48 + 0.48 * 0.36

= 0.1728 + 0.1728

= 0.3456

P[Z = 2] = P[X + Y = 2]

= P[X = 0 and Y = 2] + P[X = 2 and Y = 0] + P[X = 1 and Y = 1]

= 0.36 * 0.16 + 0.16 * 0.36 + 0.48 * 0.48

= 0.0576 * 2 + 0.2304

= 0.3456

P[Z = 3] = P[X + Y = 3]

= P[X = 1 and Y = 2] + P[X = 2 and Y = 1]

= P[X = 1] * P[Y = 2] + P[X = 2] P[Y = 1]

= 0.48 * 0.16 + 0.16 * 0.48

= 0.0768 * 2

= 0.1536

P[Z = 4] = P[X + Y = 4]

= P[X = 2 and Y = 2]

= P[X = 2] P[Y = 2]

= 0.16 * 0.16

= 0.0256

Hence

$P[Z=k]=\left\{\begin{matrix} 0.1296, &k=0 \\ 0.3456, & k=1\\ 0.3456 ,&k=2 \\ 0.1536, &k=3 \\ 0.0256, &k=4 \end{matrix}\right.$

Thank you for your supporting. Please upvote my answer...