Answer to Question #168738 in Statistics and Probability for tejendra

Question

n a game show contestants choose one of three doors to determine what prize they win. History shows that the three doors, 1, 2, and 3, are chosen with probabilities 0.30, 0.45, and 0.25, respectively. It is also known that given door 1 is chosen, the probabilities of winning prizes of $ 0, $100, and $1000 are 0.10, 0.20, and 0.70. For door 2 the respective probabilities are 0.50, 0.35, and 0.15, and for door 3 they are 0.80, 0.15, and 0.05. If X is a random variable describing dollars won, and D describing the door selected (value of D are D1 = 1, D2 = 2, and D3 = 3). 

Find: (a) FX(x|D = D1) and fX(x|D = D1), (b) fX(x|D = D2)

Answer 1

Given probability of doors are-

$P(D_1)=0.30,P(D_2)=0.45,,P(D_3)=0.25$

Let A denote the $0 won, B denote the $100 won and C denote the $1000 won.

Probability of winning the $0 by door $1 P(A/D_1)=0.10$

Probability of winning the $0 by door $2 P(A/D_2)=0.50$

Probability of winning the $0 by door $3 P(A/D_3)=0.80$

Probability of winning the $100 by door $1 P(B/D_1)=0.20$

Probability of winning the $100 by door $2 P(B/D_2)=0.35$

Probability of winning the $100 by door $3 P(B/D_3)=0.15$

Probability of winning the $1000 by door $1 P(C/D_1)=0.70$

Probability of winning the $1000 by door $2 P(C/D_2)=0.15$

Probability of winning the $1000 by door $3 P(C/D_3)=0.05$

(a) Probability $P(x/D=D_1)=\dfrac{P(D_1)\times P(A/D_1)}{ P(D_1)\times P(A/D_1)+P(D_2)\times P(A/D_2)+P(D_3)\times P(A/D_3)}$

$=\dfrac{(0.30)(0.10)}{(0.30)(0.10)+(0.45)(0.50)+(0.25)(0.80)}$

$=\dfrac{0.03}{0.03+0.225+0.20}=\dfrac{0.03}{0.455}=0.06659$

$P(x/D=D_2)=\dfrac{P(D_2)\times P(A/D_2)}{ P(D_1)\times P(A/D_1)+P(D_2)\times P(A/D_2)+P(D_3)\times P(A/D_3)}$

$=\dfrac{(0.45)(0.50)}{(0.30)(0.10)+(0.45)(0.50)+(0.25)(0.80)}$

$=\dfrac{0.225}{0.03+0.225+0.20}=\dfrac{0.225}{0.455}=0.4945$

$P(x/D=D3)=\dfrac{P(D_3)\times P(A/D_3)}{ P(D_1)\times P(A/D_1)+P(D_2)\times P(A/D_2)+P(D_3)\times P(A/D_3)}$

$=\dfrac{(0.25)(0.80)}{(0.30)(0.10)+(0.45)(0.50)+(0.25)(0.80)}$

$=\dfrac{0.2}{0.03+0.225+0.20}=\dfrac{0.20}{0.455}=0.4395$

$f(x/D=D_1)=P(X\le x)=P(0<X\le 3)$

$=\sum_{x_i<x}P(X=x)=P(X=1)+P(X=2)+P(X=3)\\ =0.06659+0.4945+0.4395\\=1.00059$

(b) Probbility

$P(x/D=D_2)=\dfrac{P(D_2)\times P(A/D_2)}{ P(D_1)\times P(A/D_1)+P(D_2)\times P(A/D_2)+P(D_3)\times P(A/D_3)}$

$=\dfrac{(0.45)(0.50)}{(0.30)(0.10)+(0.45)(0.50)+(0.25)(0.80)}$

$=\dfrac{0.225}{0.03+0.225+0.20}=\dfrac{0.225}{0.455}=0.4945$