SOLUTION: There are 15 stuff toys of two(orange and white) colors in a box, (α+39)% of total stuff toys are orange and remaining are white. A toy is drawn at random. If the toy is orange,

Question 1185547: There are 15 stuff toys of two(orange and white) colors in a box, (α+39)% of total stuff toys are
orange and remaining are white. A toy is drawn at random. If the toy is orange, it is kept out of
the box and a second toy is drawn from the box. If the toy is white, then It is put back in the box
and a orange toy is added to the box. Then a second toy is drawn from the box.
(a) What is the probability that both stuff toys drawn are orange?
(b) If the second drawn toy is orange, what is the probability that the first drawn stuff toy was
white?

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to solve this probability problem:
**1. Calculate the number of orange and white toys:**
* Total toys = 15
* Percentage of orange toys = (α + 39)%
* Number of orange toys = 15 * (α + 39)/100
* Number of white toys = 15 - (Number of orange toys)
**Example (Let α = 1):**
* Orange toys = 15 * (1 + 39)/100 = 15 * 0.4 = 6
* White toys = 15 - 6 = 9
**2. Probability of two orange toys:**
There are two ways to draw two orange toys:
* **Scenario 1:** The first toy drawn is orange, and it is kept out. The second toy is also orange.
* P(Orange 1st) = (Number of orange toys) / 15
* After removing an orange toy, there are (Number of orange toys - 1) orange toys left and 14 total toys left.
* P(Orange 2nd | Orange 1st) = (Number of orange toys - 1) / 14
* P(Two Orange) = P(Orange 1st) * P(Orange 2nd | Orange 1st)
* **Scenario 2:** The first toy drawn is white, and it is put back in. Then an orange toy is added, and the second toy drawn is orange.
* P(White 1st) = (Number of white toys) / 15
* After replacing the white toy and adding an orange toy, the total number of toys remains 15, but the number of orange toys is (Number of orange toys + 1)
* P(Orange 2nd | White 1st) = (Number of orange toys + 1) / 15
* P(Two Orange) = P(White 1st) * P(Orange 2nd | White 1st)
* **Total P(Two Orange):** Add the probabilities from both scenarios.
**Example (α = 1):**
* **Scenario 1:**
* P(Orange 1st) = 6/15 = 2/5
* P(Orange 2nd | Orange 1st) = 5/14
* P(Two Orange) = (2/5) * (5/14) = 1/7
* **Scenario 2:**
* P(White 1st) = 9/15 = 3/5
* P(Orange 2nd | White 1st) = 7/15
* P(Two Orange) = (3/5) * (7/15) = 7/25
* **Total P(Two Orange):** (1/7) + (7/25) = (25 + 49)/175 = 74/175
**3. Probability the first toy was white given the second was orange:**
We want to find P(White 1st | Orange 2nd). We can use Bayes' Theorem:
P(White 1st | Orange 2nd) = [P(Orange 2nd | White 1st) * P(White 1st)] / P(Orange 2nd)
We already calculated all these probabilities in part (b). P(Orange 2nd) is the probability of getting an orange toy on the second draw, which is the sum of the probabilities of drawing two oranges, which is 74/175.
**Example (α = 1):**
P(White 1st | Orange 2nd) = [(7/15) * (3/5)] / (74/175) = (7/25) / (74/175) = (7/25) * (175/74) = 49/74
Substitute your actual value of α into the formulas to get the final answers.

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