3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses. If the coin is unbalanced and a head has a 40% chance of occurring, find (a) the joint probability distribution of W and Z; (b) the marginal distribution of W; (c) the marginal distribution of Z

Respuesta :

Answer:

a)  The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

b)  P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

  P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 ,    P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6         P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4        P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0           P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Part(a): The required joint probability of W and Z is ,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

Part(b): The pmf (marginal distribution) of W is,

[tex]P(w=0)=0.36,P(w=1)=0.48,P(w=2)=0.16[/tex]

Part(c): The pmf (marginal distribution) of Z is,

[tex]P(z=0)=0.6,P(z=1)=0.4[/tex]

Part(a):

The joint distribution is,

[tex]P(w=0\z=0)=0.6,P(w=1|z=0)=0.4,P(w=2|z=0)=0[/tex]

Also,

[tex]P(w=0\z=1)=0,P(w=1|z=1)=0.6,P(w=2|z=1)=0.4[/tex]

Therefore,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

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