# Conditional probability problem - good solution?

• diracdelta
In summary, the probability of at least one product being new is 1/3, the probability of at least one product being new if we knew that the first draw yielded two used products is 5/9, and the probability of at least one product being new if we knew that the first draw yielded one new and one used product is 24/90.
diracdelta

## Homework Statement

We have 6 new and 4 used products in a box.
We draw randomly 2 products and used them for a while. then return them back into the box.
After that, again we draw 2 products.

a) What is probability that at least one is new?
b) What is probability that at least one is new if we found out that in first draw two used were not drawn out.

## The Attempt at a Solution

a)
A={ we drawn out 2 used products in second draw}
H1={ we drawn out 2 new products in first draw}
H2={ we drawn out 1 new and 1 used product in first draw}
H3={ we drawn out 1 used and 1 new product in first draw}
H4={ we drawn out 2 used product in first draw}

P(H1)=6/10 * 5/9 = 1/3
P(H2)=6/10 * 4/9 = 24/90
P(H3)=4/10 * 6/9 = 24/90
P(H4)=4/10 * 3/9 = 12/90

P(A|H1)=6/10 * 5/9 = 1/3
P(A|H2)=5/10 * 4/9 = 2/9
P(A|H3)=4/10 * 5/9 = 2/9
P(A|H4)=4/10 * 3/9= 12/90

P(A) = 1/3 * 1/3 + 24/90 * 2/9 + 24/90 * 2/9 + 12/90 * 12/90
P(A) = 0.24

P(at least one new) = 1- P(A) = 0.76

b)
A={ we drawn out 2 used products in second draw}
H1={ we drawn out 2 new products in first draw}
H2={ we drawn out 1 new and 1 used product in first draw}
H3={ we drawn out 1 used and 1 new product in first draw}

P(H1)=6/10 * 5/9 = 1/3
P(H2)=6/10 * 4/9 = 24/90
P(H3)=4/10 * 6/9 = 24/90

P(A|H1)=6/10 * 5/9 = 1/3
P(A|H2)=5/10 * 4/9 = 2/9
P(A|H3)=4/10 * 5/9 = 2/9

P(A) = 1/9 + 2/9 * 24/90 + 2/9 * 24/90
P(A) = 0.2296

P(at least one new) = 1- P(A) = 0.7703Is this correct solution?
My apologies if bad english, and if unpractical in absence of latex.

Thanks!

The first part looks good since you covered all possibilities.
On the second part, your scale factor for the condition has to account for the fact you know that H4 didn't happen. So you need to multiply by a factor so that P(H1|~H4) + P(H2|~H4) + P(H3|~H4) = 1. Then you can multiply what you found for P(A|H1), etc. by that scaling factor. This will ensure you are preserving the known information.

diracdelta
RUber said:
The first part looks good since you covered all possibilities.
On the second part, your scale factor for the condition has to account for the fact you know that H4 didn't happen. So you need to multiply by a factor so that P(H1|~H4) + P(H2|~H4) + P(H3|~H4) = 1. Then you can multiply what you found for P(A|H1), etc. by that scaling factor. This will ensure you are preserving the known information.
I am not sure I understand exactly.
I am supposed to find factor which preserves P(H1|~H4) + P(H2|~H4) + P(H3|~H4) = 1 and later on multiply every P(A|H1) with it.
What does tilda ~ means?

The tilde means "not"
Essentially, you know that H4 didn't happen. So the sum of the other probabilities has to add up to 1. As it stands, your probabilities for H1 to H3 sum to 78/90=13/15, so if you multiplied each one by 15/13, they would sum to one. Then use that scaled probability for the calculation of P(A).

diracdelta
RUber said:
The tilde means "not"
Essentially, you know that H4 didn't happen. So the sum of the other probabilities has to add up to 1. As it stands, your probabilities for H1 to H3 sum to 78/90=13/15, so if you multiplied each one by 15/13, they would sum to one. Then use that scaled probability for the calculation of P(A).
Oh I get it now. Thanks!

diracdelta said:
P(A) = 1/3 * 1/3 + 24/90 * 2/9 + 24/90 * 2/9 + 12/90 * 12/90
P(A) = 0.24
I make it closer to 0.25.

## 1. What is conditional probability?

Conditional probability is the probability of an event occurring given that another event has already occurred. It takes into account additional information or conditions that may affect the likelihood of an event happening.

## 2. How do you solve a conditional probability problem?

To solve a conditional probability problem, you need to first identify the given conditions and events. Then, use the formula P(A|B) = P(A∩B) / P(B) to calculate the probability of event A occurring given that event B has already occurred. You may also need to use other probability rules such as the addition and multiplication rules to simplify the problem.

## 3. What is a good solution in a conditional probability problem?

A good solution in a conditional probability problem is one that accurately calculates the probability of an event occurring given the given conditions. It should also clearly explain the reasoning and steps used to arrive at the solution.

## 4. What are some common mistakes to avoid when solving a conditional probability problem?

Some common mistakes to avoid when solving a conditional probability problem include forgetting to incorporate all given conditions, using the wrong probability rules, and incorrectly calculating probabilities. It is important to carefully read and understand the problem, and double check your calculations to avoid these mistakes.

## 5. Can conditional probability be applied in real-world situations?

Yes, conditional probability is commonly used in real-world situations, such as in insurance, medicine, and finance. For example, it can be used to calculate the probability of a patient having a certain disease given a positive test result, or the probability of a customer purchasing a product given their demographic information.

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