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I have 2 specific problems, one 'requiring' me to use a probability tree, and the other a Venn diagram. I know that apparently the Venn diagrams can be converted into probability trees and vice versa, so I have attempted to use the probability diagrams for both questions, which wasn't successful. However, I do not know what I am doing wrong and whether I am assuming something which is not actually true, e.g. that certain events are independent. In some questions I am obliged to assume the latter, whereas in others this gives me erroneous results.

Problem 1:

38% of the students in a Year 12 IB Mathematics class are female. Of the female students in this class, 13% are left-handed, whereas 24% of the male students are left handed.

a. Find the prob. that a randomly chosen student from this class is left handed. b. Find the prob. that a randomly chosen student is female, given that the student is left-handed.

For this problem, the mark scheme offers a probability diagram: Prob diagram

However, I do not understand why in part a it is assumed that the events 'being female' and 'being left-handed' and 'being male' and 'left handed' are independent (as P(A and B)=P(A)P(B)). (It does seem intuitive though.) But then the part b answer implies that the events 'being female' and 'being left-handed are independent' as P(A given B) doesn't equal P(A). What is going on?

Problem 2:

In the town of Expiet, 71% of the population are right-handed. 44% are either right handed or have blonde hair but not both, and 21% do not have blonde hair.

A member of the population is selected at random. Find the likelihood that the person:

a. is right handed but not blonde b. is both right handed and has blonde hair c. is right handed or has the blonde hair

Although the markscheme has used a venn diagram with 4 sets of 4-variable equations, I have attempted to use a prob tree, which was this:

Prob tree 2

In this case, I know that this tree is clearly wrong, but I dont know why. The 2nd equation evidently gives me the same result as the probability mentioned in the question, so this is weird. Also, I think that I can't assume that each of the right-handed/ non-right handed branch will have the same prob. of being blonde/non-blonde, but I am unsure. So what is the flaw in my reasoning and what is the correct prob. tree diagram?

If someone could clarify the above issues (and perhaps suggest how to convert a Venn diagram into a prob. tree diagram), I would be grateful.

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  • $\begingroup$ In problem 1, there is no assumption of independence. In part a, what is really used is $P(A \text{ and } B) = P(A) P(B|A)$. For example the $13$% mentioned in the problem is the probability of a random student to be left handed given that the student is female. $\endgroup$ – Matthias Klupsch Mar 26 '16 at 10:28
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How do I know when to use a Venn diagram or a probability tree? Also, when can I assume that the events are independent?

Thank you for asking this question. It helped me realise the following ! :

There is no strict rule for using either one or other technique. You need to try and model the question in both ways and more (probability tables, equations etc). The model that can satisfy the question best with, is good for you, for that question. In fact mostly the question is worded in a way where it encourages you to use a certain model.

Do not assume anything unless it is specifically mentioned. Just try to model the information in question with the techniques you know.

For question 1 I felt "Tree diagram" is the best way to model the problem. And for second one I felt more comfortable with Venn diagrams and then algebraic equations to solve for unknowns. I tried but could not create a satisfactory Venn model of first problem.

In general I find Tree diagrams easy to deal with in conditional probability and Venn diagrams in sets or very simple probability questions....but maybe that is just me ! Question 1 (Tree would be as follows) [1]-----0.38----F-----0.13---L

[2]-----0.38----F-----0.87---R

[3]-----0.62----M-----0.24---L

[4]-----0.62----M-----0.76---R

P(L) = [1]+[4] = 0.13x0.38+0.24x0.62 P(F|L)=[1]/([1]+[3]) = (0.13x0.38)/((0.13x0.38)+(0.24x0.62))

Question 2 x = Only R not B y = R and B z= Only B not R x+y = 0.71 x+z = 0.44 1-y-z = 0.21 solving i get x = 0.18, y=0.53,z=0.26, and that gives you the answer to the second problem.

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  • $\begingroup$ Welcome to math.SE! I see that you understand the main issues with the question. Your answer could be improved if you edit the language for clarity, and you also directly address all the specific questions of the OP. For example, you can help the OP with their confusion on the "independence" of events. $\endgroup$ – Thanassis Mar 27 '17 at 12:59
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Problem 1

As @Matthias-Klupsch has noted, there is no assumption that the events are independent. The possibility of multiplying $P(A)$ and $P(B|A)$ to obtain $P(A \cap B)$ is not exclusive to independent events. You can do that even with dependent events. But if $P(A) \cdot P(B) = P(A \cap B) \implies P(B|A)=P(B)$ then A and B are independent.

Problem 2

R := right handed

B := blonde

We know the following:

  • $P(R) = 0.71$
  • $P(R \sqcup B)=0.44$
  • $P(not B)=0.21 \implies P(B)=0.79$

The key here is the following equation. We know both B and R and the intersection of them is part of both, so when we add B and R we need to subtract the intersection twice to get the disjoint union. Here we don't know the intersection value, so we use $x$ $$ P(B)+P(R) - 2 \cdot P(B \cap R) = P(R \sqcup B) \\ 0.79+0.71 - 2 x = 0.44 \\ x = 0.53 \\ \implies P(R \cap B)=0.53 $$

Now that we know $P(R \cap B)$, we can calculate P(R \cap not B) as follows:

$$ P(R \cap not B)=P(R)-P(R \cap B) \\ P(R \cap not B)=0.71-0.53=0.18 $$

And the union of R and B is the disjoint union plus the intersection.

$$ P(R \cup B)=P(R \sqcup B)+P(R \cap B) \\ P(R \cup B)=0.44+0.18 $$

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