M&Ms Probability Help: Understanding P(B|A) in Random Selection Scenarios"

In summary: This is a problem to test your understanding of independence. In probability, two events(A,B) are said to be independent if P(B|A)=P(B), that is, A and B have nothing to do with each other.Although this answer provides a brief explanation of the concept of independence, it does not answer the question. In summary, the bowl contains a large number of M&Ms, and a single M&M is chosen at random. The probability of a yellow M&M being chosen is not affected by the probability of a blue M&M being chosen.
  • #1
Mark53
93
0

Homework Statement


[/B]
A bowl contains a large number of M&Ms. A single M&M is chosen at random, its colour is observed, and then it is returned to the bowl. A second M&M is chosen at random and its colour is observed. Let A be the event that the first M&M is yellow and B be the event that the second M&M is blue.
i. Provide a reason why you think P(B|A) = P(B) is true or not.

(d) An adult member of the Australian population is chosen at random. Let A be the event that chosen person was born in Australia and let B be the event that the person speaks more than one language.
i. Provide a reason why you think P(B|A) = P(B) is true or not.

The Attempt at a Solution



not sure if this is correct:

a) It is true as once you take it out you are putting it back in meaning that the probability is still the same

b) it is not true as the probability of selecting someone who speaks another language in the Australian population is different to selecting a person who was born in Australia and speaks another language
 
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  • #2
a) I agree with your answer.
b) Although I don't have statistics to back it up, I think it is probably true that the percentage of the population that speaks more than one language varies depending on where one is born. So I agree with you on b also.
EDIT: I want to change "probably true" to "almost certainly true" for part b.
 
  • #3
Mark53 said:
a) It is true as once you take it out you are putting it back in meaning that the probability is still the same
This is a bit tricky. It depends how the probability distribution is defined.

If it means the distribution based on the actual numbers of each colour in the pot (even if you do not know these numbers), you are right.

But you could read the question in terms of your own estimate of the probability without knowing what the actual numbers are. In this case, you would need to consider that you have an a priori belief about what the numbers might be. Whatever probabilities you assign to those possible distributions, the fact that you drew a yellow one the first time should shift those estimates in favour of distributions with more yellows. Consequently, your estimate of the likelihood of picking a blue next decreases slightly.
If you have heard of Bayesian statistics, you should seriously consider this interpretation.
 
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Likes Mark53
  • #4
You're correct. This is a problem to test your understanding of independence. In probability, two events(A,B) are said to be independent if P(B|A)=P(B), that is, A and B have nothing to do with each other.

Solution:

a)
Because the M&M is returned, choosing a yellow is independent of choosing a blue.

b) Nationtionality has nothing to do with language plurality.
 
  • #5
Matthew314159271828 = MatthewPie :smile:
 
  • #6
Matthew314159271828 said:
b) Nationtionality has nothing to do with language plurality.
I'm inclined to agree with Tom Hart. I don't have any statistics to back this up, but my sense is that if the person selected was not born in Australia, that person is more likely to speak another language than one who was born in Australia.
 
  • #7
TomHart said:
Matthew314159271828 = MatthewPie :smile:
e=
2.718281828459045235360287471352662497757247093699959574966967627724076630353
547594571382178525166427427466391932003059921817413596629043572900334295260
595630738132328627943490763233829880753195251019011573834187930702154089149
934884167509244761460668082264800168477411853742345442437107539077744992069
551702761838606261331384583000752044933826560297606737113200709328709127443
747047230696977209310141692836819025515108657463772111252389784425056953696
770785449969967946864454905987931636889230098793127736178215424999229576351
482208269895193668033182528869398496465105820939239829488793320362509443117
301238197068416140397019837679320683282376464804295311802328782509819455815
301756717361332069811250996181881593041690351598888519345807273866738589422
879228499892086805825749279610484198444363463244968487560233624827041978623
209002160990235304369941849146314093431738143640546253152096183690888707016
768396424378140592714563549061303107208510383750510115747704171898610687396
965521267154688957035035402123407849819334321...
π=
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 ...
 
  • #8
MatthewPie, I think we all appreciate that you used an abbreviated version for your username. :)
 
  • #9
No Problem!
 
  • #10
Matthew314159271828 said:
a) Because the M&M is returned, choosing a yellow is independent of choosing a blue.
I feel sure that is the intended answer, but it's not quite that simple, for the reason I gave in post #3.

It is reasonable to suppose that not all pots of M&Ms have the same blue:yellow ratio. The fact of getting a yellow the first time very slightly pushes the odds in favour of this pot having lower such ratio than average, so slightly depresses the odds of a blue next.
 
Last edited:

What is the definition of P(B|A)?

P(B|A) represents the conditional probability of event B occurring given that event A has already occurred. This means that the probability of B happening is dependent on the occurrence of A.

How is P(B|A) calculated?

P(B|A) is calculated by taking the probability of both events A and B occurring together, and dividing it by the probability of event A occurring. This can be written as P(A and B) / P(A).

Can you give an example of P(B|A) in a random selection scenario?

For example, if you have a bag of M&Ms with 10 red candies and 20 blue candies, the probability of selecting a red candy (event B) given that you have already selected a blue candy (event A) would be 10/30 or 1/3. This is because after selecting a blue candy, there are now 9 red candies out of 29 total candies left in the bag.

What is the difference between P(B|A) and P(A|B)?

P(B|A) and P(A|B) represent the same conditional probability, but in reverse order. P(B|A) is the probability of event B occurring given that event A has happened, while P(A|B) is the probability of event A occurring given that event B has happened.

How does P(B|A) relate to independence of events?

If two events A and B are independent, then P(B|A) would be equal to the overall probability of event B happening. This is because the occurrence of event A does not affect the probability of event B happening. In other words, the knowledge of event A does not change the probability of event B occurring.

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