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

answer: sorry i can't you.

step-by-step explanation:

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.