There are 3 boxes on stage that appear identical, but one is Lucky. The boxes are full of tickets; some are labeled "win" and the others are labeled "lose." In the Lucky box, ninety percent of the tickets are winners. In each of the other two boxes, only twelve percent of the tickets are winners. 1. You will pick a box at random and draw one ticket from it at random.2. What is the probability you will draw a winning ticket? 3. If you do draw a winning ticket, what is the chance it came from the Lucky box?

2.-P = 0.38

3.-P [ Lb | Wt ]  =  0.788

Step-by-step explanation:

1.-Probability of choosing any box is, 1/3. So the probability of choosing the lucky box is 1/3

Let´s say the lucky box is the number 2 box  ( that consideration does not in any way change the problem generality)

Then we have

p₁ probability of choosing box  1 is 1/3   p₁´ Probability of win ticket is  0.12

p₂ probability of choosing box 2 is 1/3  p₂´Probability of win ticket is  0.90

p₃ probability of choosing box 3 is 1/3  p₃´ Probability of win ticket is  0.12

Then

P (of choosing a winning ticket is) = p₁*p₁´ + p₂*p₂´ +  p₃*p₃´

P  =  1/3*0.12 + 1/3*0.9 + 1/3*0.12

P  =  0.04 + 0.3 + 0.04

P = 0.38

3.- if I draw a winning ticket what is the probability it came from Lucky box

According to Bayes theorem

P [ Lb | Wt ]  =   P(Lb) * P[ Wt|Lb]/ P(Wt)

P(Lb) = 1/3  =  0.33333

P[Wt|Lb]  = 0.9

P(Wt) = 0.38

Then By substitution

P [ Lb | Wt ]  =  0.333 *  0.9 / 0.38

P [ Lb | Wt ]  =  0.788

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

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step-by-step explanation:

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