Probability: What is the probability of drawing items from containers?

In summary: This is because the first object is not replaced. The probability of drawing a perfect object on the second draw is$\frac{1}{3} \frac{7}{10} + \frac{1}{3} \frac{8}{15} + \frac{1}{3} \frac{20}{25} = \frac{29}{60}$Therefore, the probability of drawing a flawed object on the first, and an unflawed on the second is:$P(F \cap U) = P(F) P(U|F) = \frac{31}{60} \frac{33}{59}$The probability of drawing an unflawed object on the second draw is:$P
  • #1
sanctifier
58
0

Homework Statement



There are 3 boxes containing some products of same type.

Box 1 contains 10 products and 3 of them are flawed.

Box 2 contains 15 products and 7 of them are flawed.

Box 3 contains 25 products and 5 of them are flawed.

Randomly choose one box from the 3 then draw one product from the chosen box, continue to draw one product from rest products in the chosen box, i. e., drew product will not be returned to the box.

Question 1 : What is the probability that the 1st drew product is a flawed one?

Question 2: What is the probability that the 1st drew product is a flawed one if we know the 2nd drew product is unflawed?

Homework Equations



Nothing special.

The Attempt at a Solution



Answer 1:

[itex] p= \frac{1}{3} \frac{3}{10} + \frac{1}{3} \frac{7}{15} + \frac{1}{3} \frac{5}{25} [/itex]

Answer 2:

[itex] p= \frac{1}{3}* \frac{ \frac{3}{10} \frac{7}{9} }{ \frac{7}{10} \frac{6}{9} +\frac{3}{10} \frac{7}{9}} +\frac{1}{3}* \frac{ \frac{7}{15} \frac{8}{14} }{ \frac{8}{15} \frac{7}{14} +\frac{7}{15} \frac{8}{14}} +\frac{1}{3}* \frac{ \frac{5}{25} \frac{20}{24} }{ \frac{20}{25} \frac{19}{24} +\frac{5}{25} \frac{20}{24}} [/itex]

Are these answers correct? Thank you in advance!
 
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  • #2
I don't understand why these denominators in the second answer.
 
  • #3
[itex] \frac{7}{10} \frac{6}{9} [/itex] denotes the case the 1st drew product is unflawed and the 2nd drew product is also unflawed.

[itex] \frac{6}{9} [/itex] means after the 1st withdrawal there are 9 products left in the Box 1 and 6 of the left products are unflawed
 
  • #4
That's clear.
But why do you use these denominators?
What is the reasoning behind, and what are the rules/theorems you are using?
It is more important to explain.

My translation of question 2 is: what is the probability of drawing first a flawed and in second an unflawed.
The probability to do so via the first box, for example, how do you calculate it, based on which rule?
 
  • #5
sanctifier said:

Homework Statement



There are 3 boxes containing some products of same type.

Box 1 contains 10 products and 3 of them are flawed.

Box 2 contains 15 products and 7 of them are flawed.

Box 3 contains 25 products and 5 of them are flawed.

Randomly choose one box from the 3 then draw one product from the chosen box, continue to draw one product from rest products in the chosen box, i. e., drew product will not be returned to the box.

Question 1 : What is the probability that the 1st drew product is a flawed one?

Question 2: What is the probability that the 1st drew product is a flawed one if we know the 2nd drew product is unflawed?

Homework Equations



Nothing special.

The Attempt at a Solution



Answer 1:

[itex] p= \frac{1}{3} \frac{3}{10} + \frac{1}{3} \frac{7}{15} + \frac{1}{3} \frac{5}{25} [/itex]

Answer 2:

[itex] p= \frac{1}{3}* \frac{ \frac{3}{10} \frac{7}{9} }{ \frac{7}{10} \frac{6}{9} +\frac{3}{10} \frac{7}{9}} +\frac{1}{3}* \frac{ \frac{7}{15} \frac{8}{14} }{ \frac{8}{15} \frac{7}{14} +\frac{7}{15} \frac{8}{14}} +\frac{1}{3}* \frac{ \frac{5}{25} \frac{20}{24} }{ \frac{20}{25} \frac{19}{24} +\frac{5}{25} \frac{20}{24}} [/itex]

Are these answers correct? Thank you in advance!

There is a subtlety here which you need to think about, and to explain it we need a better and more explicit notation (which is what you should get in the habit of using anyway). So, let the events be ##F## = flawed first, ##U## = unflawed second and ##B_1, B_2, B_3## = choose box 1,2,3, respectively. It is uncontroversial to say that ##P(F) = \sum_{i=1}^3 P(F|B_i) P(B_i).## However, the conditional probability issue is different. Should we use
[tex] P(F|U) = \frac{P(F \cap U)}{P(U)} =
\frac{\sum_{i=1}^3 P(F \cap U | B_i) P(B_i)}{\sum_{i=1}^3 P(U|B_i) P(B_i)} [/tex]
or should we use
[tex] \sum_{i=1}^3 P(B_i) P((F|U)|B_i) ? [/tex]
The logic behind the second one is that we we are drawing both balls from the same box, and if it is box i the conditional probability is ##P((F|U)|B_i)##. You decide,
 
  • #6
Ray is professional.

Indeed, I should use some notation to explicitly describe the answer.

You have given the correct answer, it should be [tex] P(F|U) = \frac{P(F \cap U)}{P(U)} =
\frac{\sum_{i=1}^3 P(F \cap U | B_i) P(B_i)}{\sum_{i=1}^3 P(U|B_i) P(B_i)} [/tex]

where [itex] P(U) = \frac{1}{3}( \frac{7}{10} \frac{6}{9} + \frac{3}{10} \frac{7}{9} )+\frac{1}{3}( \frac{8}{15} \frac{7}{14} + \frac{7}{15} \frac{8}{14} )+\frac{1}{3}( \frac{20}{25} \frac{19}{24} + \frac{5}{25} \frac{20}{24} ) [/itex]

and [itex] P(F \cap U)= \frac{1}{3} ( \frac{3}{10} \frac{7}{9} ) + \frac{1}{3} ( \frac{7}{15} \frac{8}{14} )+ \frac{1}{3} ( \frac{5}{25} \frac{20}{24} ) [/itex]

The question 2 requires the conditional probability and the above should meet the need.
 
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  • #7
sanctifier said:
where [itex] P(U) = \frac{1}{3}( \frac{7}{10} \frac{6}{9} + \frac{3}{10} \frac{7}{9} )+\frac{1}{3}( \frac{8}{15} \frac{7}{14} + \frac{7}{15} \frac{8}{14} )+\frac{1}{3}( \frac{20}{25} \frac{19}{24} + \frac{5}{25} \frac{20}{24} ) [/itex]
True, but a little more complicated than necessary.
The probability of drawing a perfect object on the second draw is the same as for drawing it on the first.
 

FAQ: Probability: What is the probability of drawing items from containers?

1. What is the definition of probability?

Probability is the measure of the likelihood that an event will occur. It is typically expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

2. How is probability calculated?

The probability of an event can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This is known as the classical definition of probability.

3. What is the difference between theoretical and experimental probability?

Theoretical probability is based on calculations and assumptions, while experimental probability is determined by conducting experiments and recording the results. Theoretical probability is used to predict outcomes, while experimental probability is used to analyze and understand real-world data.

4. How does the number of items in a container affect the probability of drawing a specific item?

The more items there are in a container, the lower the probability of drawing a specific item. This is because there are more possible outcomes and therefore the chance of drawing a specific item decreases.

5. What is the difference between with replacement and without replacement when drawing items from containers?

With replacement means that an item is put back into the container after it is drawn, while without replacement means that the item is not put back in. This affects the probability of drawing the same item multiple times, as it is possible with replacement but not without replacement.

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