Choose a ball at random from a randomly selected box - alternative way

In summary, the author discusses how adding probabilities can lead to the number of ways being multiple of the number of possibilities.
  • #1
Rikudo
120
26
Homework Statement
There are 2 boxes. box A contains 20 white balls and 20 black balls. Box B contains 10 white balls and 5 black balls. A box is chosen at random, then a ball is taken at random from the chosen box. What is the probability that the ball is white?
Relevant Equations
##P = \frac {n(A)}{n(S)}##
https://www.physicsforums.com/threa...-random-from-a-randomly-selected-box.1034377/

First of all, I would like to point out that this is the same exact question from what is being discussed in the thread above.
In that thread, the problem is solved by adding the probability of taking white from box 1 and taking white from box 2.

Now, I am trying to solve this problem with a different and straightforward way.

We all know that probability is the ratio of the number of favorable outcomes ##n(A)## to the total number of outcomes ##n(S)##of an event.
Or, simply put: $$P = \frac {n(A)}{n(S)}$$
So, let's find the quantity of those two variables.

1. Find ##n(A)##
This one is obvious. This is just the total number of white balls in both box. So, $$n(A)= 20+10=30$$

2. Find ##n(S)##
Now, this one is quite complicated. First, we can choose the 1st box and then choose ANY balls from both boxes. Second, we can choose the 2nd box and then choose ANY balls from both boxes.
So, $$n(S) = the\,number\,of\,box \,\,X\,\,Total\, number \,of \,balls = 2 (20+20+10+5) = 110$$

Hence, the answer is $$P = \frac {30} {110}$$
But, this is not the correct answer. What mistakes did I make?
 
Physics news on Phys.org
  • #2
Rikudo said:
Now, I am trying to solve this problem with a different and straightforward way.

We all know that probability is the ratio of the number of favorable outcomes ##n(A)## to the total number of outcomes ##n(S)##of an event.
This is your mistake. This is only true if all the outcomes are equally likely, which is not the case here.
Consider this extreme example: Select one of two equally-likely boxes, one with only one thousand white balls and the other with only one black ball. The black ball has a 0.5 probability of being selected. Each white ball has 1/2,000 probability of being selected.
 
  • Like
Likes DrClaude and PeroK
  • #3
Rikudo said:
We all know that probability is the ratio of the number of favorable outcomes n(A) to the total number of outcomes n(S)of an event.
But, this is the definition of probability itself. So, isn't that should be true for any cases?
 
  • #4
Rikudo said:
But, this is the definition of probability itself. So, isn't that should be true for any cases?
That is true if you are talking about the rate that the outcomes will occur given a large number of independent trials. But the OP is counting the number of possible outcomes that would be counted as event A versus the total set of results, not their rate. If you read the entire post, he is counting the number of white balls in the experiment versus the total number of balls in the experiment. He is not talking about how many white ball outcomes there are in N independent trials.
 
  • #5
##\displaystyle P(\text{white})=P(\text{box 1 AND white})\,\text{OR}\,P(\text{box 2 AND white})=\frac{1}{2}\cdot\frac{20}{40}+\frac{1}{2}\cdot\frac{10}{15}=\frac{7}{12}##

That is the correct answer from the other thread. From the denominator of this final result, we can conclude that the possible total number of ways is multiple of 12. Is this assumption correct?

If yes, then:
In post #1, we get that the total ways is 110, which is not the multiple of 12. How can this happen?
 
  • #6
Rikudo said:
##\displaystyle P(\text{white})=P(\text{box 1 AND white})\,\text{OR}\,P(\text{box 2 AND white})=\frac{1}{2}\cdot\frac{20}{40}+\frac{1}{2}\cdot\frac{10}{15}=\frac{7}{12}##

That is the correct answer from the other thread. From the denominator of this final result, we can conclude that the possible total number of ways is multiple of 12. Is this assumption correct?
No. You are confusing the number of possible outcomes with the probability of those outcomes. They are not the same. You got in the habit of simply counting up the number of ways that something can happen without considering the probabilities. That only works for a certain class of "counting" problems. This is not that type of problem. Most of its parts are simple counting, but this problem introduces some simple probability calculations also. You should take a hard look at what they are doing and why. Look at my example on post #2 and notice how different the probabilities are from simple counting of the balls. One ball has a 1/2 probability and the others have a 1/2000 probability.
 
Last edited:
  • Like
Likes DrClaude and PeroK
  • #7
FactChecker said:
No. You are confusing the number of possible outcomes with the probability of those outcomes. They are not the same.
I found something that discussed the relationship between addition of probabilities and number of ways.
Here is what it says:Suppose the event can happen in two ways which cannot concur and let ##\frac {a_1}{b_1}##, ##\frac {a_2}{b_2}## be the chances of the happening of the event in these two ways respectively. Then out of ##b₁ b₂## cases, there are ##a_1 b_2##, in which the event may happen in the first way , and ##a_2 b_1## ways in which the event may happen in the second; and these ways cannot concur . Therefore, the chance that the event will happen in one or other of the two ways is: $$\frac {a_1 b_2 + a_2b_1} {b_1b_2}$$ Which is equal to $$\frac {a_1}{b_1} +\frac {a_2}{b_2}$$I do not understand why the total number of ways is ##b₁ b₂## instead of ##b₁+ b₂##, though...
 
  • #8
Rikudo said:
But, this is the definition of probability itself. So, isn't that should be true for any cases?
Suppose you attempt a probability problem. There are two possible outcomes: you get the right answer; or, you get the wrong answer. Only one outcome is favourable. Hence, the probability that you get the right answer is ##\frac 1 2##. Always. Regardless of how hard you study, and regardless of how easy or difficult the problem is.
 
  • Like
  • Love
Likes Steve4Physics, FactChecker and pbuk
  • #9
Hi @Rikudo. A simple probabilIty-tree diagram may help:
prob.jpg

Probability of picking Box A and then a white ball is 0.5*20/40.
Probability of picking Box B and then a white ball is 0.5*10/15.
Overall probability of picking a white ball is 0.5*20/40 + 0.5*10/15.
 
  • Like
Likes PeroK and FactChecker
  • #10
Rikudo said:
Suppose the event can happen in two ways which cannot concur and let ##\frac {a_1}{b_1}##, ##\frac {a_2}{b_2}## be the chances of the happening of the event in these two ways respectively. Then out of ##b₁ b₂## cases, there are ##a_1 b_2##, in which the event may happen in the first way , and ##a_2 b_1## ways in which the event may happen in the second; and these ways cannot concur . Therefore, the chance that the event will happen in one or other of the two ways is: $$\frac {a_1 b_2 + a_2b_1} {b_1b_2}$$ Which is equal to $$\frac {a_1}{b_1} +\frac {a_2}{b_2}$$
Again, you seem to be ASSUMING that all the possible final results are equally likely and you only have to count them up. That is the starting point of the subject of probability but it will not get you very far.

Look at the tree diagram OF PROBABILITIES in post #9 of @Steve4Physics . He labeled the choice of each box as 1/2 probability. But what if there is some preference for choosing the first box? Suppose it has a ribbon that people like and it is chosen 2/3 of the time, leaving 1/3 for the other box. Better yet, what if the ribbon made that box probability 0.654321? Then the problem solution can not be obtained by simply counting possible outcomes. You need to start learning more techniques than simply counting the possible outcomes. You need to learn how to deal with calculations of conditional probabilities without relying on counting.
Rikudo said:
I do not understand why the total number of ways is ##b₁ b₂## instead of ##b₁+ b₂##, though...
Because it is not the total number of distinct possible outcomes. It is just the result of mathematical manipulation.
 
Last edited:
  • Like
Likes Rikudo
  • #11
Rikudo said:
But, this is the definition of probability itself. So, isn't that should be true for any cases?
You have to get the sample space S right for this to work: every outcome has to be equally likely.

For this problem, if you choose some outcome in S at random, there should be a 50% chance that it corresponds to choosing a ball from box 1. That means half of the outcomes in S should be for box 1 and the other half for box 2. That can't happen if you're just pooling all the balls together and choosing one at random since box 1 has more balls than box 2.

Suppose we adjust the number of balls in each box so that each box has 120 balls while keeping the ratio of white balls to black balls in each box unchanged. Now if we pool all the balls together and choose a ball at random, there's a 50% chance it came from one of the boxes. Box 1 will have 60 white balls and 60 black balls so that the probability of choosing a white ball from box 1 is still 1/2. Similarly, box 2 will have 80 white balls and 40 black balls so that the probability of choosing a white ball from box 2 is 2/3. We could express the sample space for this latter configuration as {(1, w1), ..., (1, w60), (1, b1), ... , (1, b60), (2, w1), (2, w2), ..., (2, w80), (2, b1), ... , (2, b40)}. The first number of the ordered pair corresponds to which box a ball belongs to, and the second item labels the specific ball.

The total number of white balls is 140, and the total number of balls is 240. The probability of choosing a white ball is therefore 140/240 = 7/12. This calculation works because the sample space was explicitly constructed so that each outcome is equally likely.

The way you were looking at it, the sample space was something like {(1, w1), ..., (1, w20), (1, b1), ..., (1, b20), (2, w1), ..., (2, w10), (2, b1), ..., (2, b5)}. Here, the outcomes aren't equally likely. If they were n(box 1) = 40 while n(S) = 55, so p(box 1) = 8/11. But the problem assumed choosing either box was equally likely, that is, p(box 1)=1/2.

I also wanted to point out your initial calculation for n(S) is wrong. There are no outcomes corresponding to selecting a ball that came from box 1 when you initially chose box 2, and vice versa.
 
  • Like
Likes Rikudo

Related to Choose a ball at random from a randomly selected box - alternative way

1. What does it mean to choose a ball at random from a randomly selected box?

Choosing a ball at random means that each ball in the box has an equal chance of being selected. Selecting a box at random means that each box has an equal chance of being chosen.

2. Is there a difference between choosing a ball at random and selecting a box at random?

Yes, there is a difference. Choosing a ball at random means that you are selecting one ball from a specific box. Selecting a box at random means that you are choosing a box from a group of boxes, and then selecting a ball from that chosen box.

3. How does this method of choosing a ball at random from a randomly selected box work?

This method works by assigning each ball in the box a number or label, and then using a random number generator to select a number. The selected number corresponds to the ball that is chosen. Similarly, each box is assigned a number or label, and a random number generator is used to select a box. The chosen box is then opened and a ball is selected from it.

4. What is the purpose of choosing a ball at random from a randomly selected box?

This method is often used in scientific experiments to ensure that the results are unbiased and representative of the entire population. By choosing at random, we eliminate any potential bias or influence in the selection process.

5. Are there any limitations to this method of choosing a ball at random from a randomly selected box?

Yes, there are limitations. This method assumes that all balls and boxes are identical and that the selection process is truly random. In reality, there may be slight variations in the balls and boxes, and the selection process may not be completely random. Additionally, this method may not be suitable for certain situations where specific criteria need to be met in the selection process.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
25
Views
3K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
8
Views
517
  • Precalculus Mathematics Homework Help
Replies
24
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
2K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
Back
Top