Mathematics, 07.07.2019 20:30, goob92

The probability that a randomly selected box of a certain type of cereal has a particular prize is .2. suppose you purchase box after box until you have obtained two of these prizes. a. what is the probability that you purchase x boxes that do not have the desired prize? b. what is the probability that you purchase four boxes? c. what is the probability that you purchase at most four boxes? d. how many boxes without the desired prize do you expect to purchase? how many boxes do you expect to purchase?

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answered: Manuel2019

$P_{1}$ = Probability that a randomly selected box of a certain type of cereal has a particular prize is .2 .

$P_{2}$ = Probability that you purchase x boxes that do not have the desired prize =1-0.2=0.8

Let $E_{1} and E_{2}$ denote the events of winning two of these prizes.

$E_{1} =1. 0.8\times (0.2) +2. (0.8)^2\times(0.2)+3. (0.8)^3\times(0.2) + 4.(0.8)^4\times(0.2)+.....\\$ ..................(1)

Multiplying both sides of equation (1) by 0.8,we get

$0.8 E_{1}= (0.8)^2\times 0.2+ 2. (0.8)^3\times 0.2+3. (0.8)^4\times 0.2+ 4.(0.8)^5\times 0.2+5. (0.8)^6\times 0.2.....$ ..........(2)

(1) - (2)

we get

$0.2 E_{1} = 0.8 \times 0.2+ (0.8)^2\times 0.2+ (0.8)^3\times 0 .2+.......$

$0.2 E_{1}=0.2\times [ 0.8 + (0.8)^2+ (0.8)^3+ (0.8)^4+........]$

Cancelling 0.2 from both sides , as well as applying the rule of sum of an infinite geometric progression

$E_{1} = \frac{0.8}{1-0.8}$

$E_{1} =\frac{0.8}{0.2}$

$E_{1} = 4$

Now , you have got one of the Prizes.

So , $E_{2}$ =4

Total expected value = 4+4= 8 which is expected number of draws that you will get 2 prizes.

(a). Probability that you purchase x boxes that do not have the desired prize

= Probability of not getting prize x times

= $0.8\times0.8\times0.8.......x times\\(0.8)^x$

(b) Probability of purchasing four boxes =

$(P_{1})^4+ (P_{1})^3(P_{2}) + (P_{1})^2(P_{2})^2+(P_{1})(P_{2})^3+(P_{2})^4$

= .5232

1 - [Probability of purchasing four boxes]

= 1 - 0.5232

=0.4768

(d) it is not given in the question that you have to buy certain number of boxes of this kind. But as written in the question that you have to get two prize boxes , So total expected number of purchase without the desired prize is 8 and then i am getting 2 prize boxes.

Number of boxes without the desired prize =8

So, Total number of boxes that i expect to purchase=8+2=10

answered: jhashknkughb6759

Here we want to analyze some properties of the probability of getting a prize in a cereal box.

a) p(x) = (0.8)^x

b) 0.0768

c) 0.1808

d) 30 boxes in total, 28 of them without the prize.

We know that the probability for a box of having a prize is 0.2.

a) The probability of not having the prize is p = 1 - 0.2 = 0.8

And this is the probability for each one of the x boxes, remember that the joint probability will be the product of the individual probabilities, then the probability that in x boxes you don't have a prize is:

p(x) = (0.8)^x

b) The probability of only purchasing two boxes is:

p = (0.2)^2*(0.8)^2

So two have the prize and two dont, one of the boxes with a prize must be the last one, and the other box can by any of the first 3. Then we have 3 permutations, and the actual probability is:

P = 3*(0.2)^2*(0.8)^2 = 0.0768

c) This is the probability of purchasing two plus the probability of purchasing 3 plus the probability of purchasing four.

For two boxes the probability is:

p₂ = (0.2)^2 = 0.04

For 3 boxes is:

p₂ = 2*(0.2)^2*(0.8) = 0.064

Where the factor "2" represents the permutations of the first box with the prize.

For 4 boxes we know:

p₄ = 0.0768

The total probability is:

P(4)= = p₂ + p₃ + p₄ = 0.04 + 0.064 + 0.0768 = 0.1808

d) before we got the probability for getting the two prizes in a maximum of four boxes, and we got something around 18%.

We just need to keep adding boxes to that sum, until it reaches the 99.9%.

for x boxes the total probability of getting the prize witin the X boxes is:

P(x) = (0.2)^2*(1 + 2*(0.8) + 3*(0.8)^2 + 4*(0.8)^3 + ... + (x-1)*(0.8)^(x-2))

We must find X such that:

P(x) = 1.

Or, we can just find x such that the last term does not contribute anymore, this means that we can solve:

(x-1)*(0.8)^(x - 2) ≈ 0

Notice that this never does become actually equal to zero, but there is a tendency that we can see graphically below:

In the graph we can see that for x = 30, the tendency to zero starts.

This means that for x = 30 the expression (x-1)*(0.8)^(x- 2) becomes almost zero, we can evaluate that:

(30-1)*(0.8)^(30 - 2) = 0.06

This may be large compared with the numbers we saw before, but remember that this is multiplied by (0.2)^2.

This means that for around 30 boxes you are almost sure to get the two prizes you want (there will always be a really small probability that this does not happen, smaller than 5%.).

So you will get 30 boxes in total, and in 28 without prize.

If you want to learn more, you can read:

link