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

[sanctifier]

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:

$p= \frac{1}{3} \frac{3}{10} + \frac{1}{3} \frac{7}{15} + \frac{1}{3} \frac{5}{25}$

Answer 2:

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

Are these answers correct? Thank you in advance!

 

[maajdl]

I don't understand why these denominators in the second answer.

 

[sanctifier]

$\frac{7}{10} \frac{6}{9}$ denotes the case the 1st drew product is unflawed and the 2nd drew product is also unflawed.

$\frac{6}{9}$ means after the 1st withdrawal there are 9 products left in the Box 1 and 6 of the left products are unflawed

 

[maajdl]

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?

 

[Ray Vickson]

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:

$p= \frac{1}{3} \frac{3}{10} + \frac{1}{3} \frac{7}{15} + \frac{1}{3} \frac{5}{25}$

Answer 2:

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

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
$$P(F|U) = \frac{P(F \cap U)}{P(U)} = \frac{\sum_{i=1}^3 P(F \cap U | B_i) P(B_i)}{\sum_{i=1}^3 P(U|B_i) P(B_i)}$$
or should we use
$$\sum_{i=1}^3 P(B_i) P((F|U)|B_i) ?$$
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,

 

[sanctifier]

Ray is professional.

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

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

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

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

The question 2 requires the conditional probability and the above should meet the need.

 

[haruspex]

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

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.