Probability Questions: Union, Intersection and Combinations

[AN630078]

Homework Statement

Hello, I very much struggle with probability problems no matter how much I revise. Nonetheless, i am determined to hone my comprehension of the topic and to develop my overall dexterity. Consequently while revising below I came across the following questions, would anyone be able to advise whether the methods I have used to solve them would be correct or if I can improve upon my workings in any manner? I think I stumbled more so on question 2 to be perfectly candid.

Question 1; The integers 1 to 12 are placed in one or more of the following categories;
S: square numbers
P: prime numbers (where 1 is not a prime number)
E:even numbers
T: multiples of 3

If one of the numbers is chosen at random find the probability of it falling within the following sets:

a) T'
b) S ∪ E
c)P⋂T

Question 2; A shortlist of 3 parliamentary candidates is to be selected from a group of 5 men and 6 women. What is the probability that the shortlist consists only of men?
Assume that each selection of 3 candidates is equally likely

Relevant Equations

P(A ∪ B)=P(A)+P(B)-P(A⋂B)
P(A⋂B) = P(A)P(B)

Question 1:

a) T' is the complementary event of T
Therefore, T'=1-T
In set T = {3,6,9,12}
P(T)=4/12 =1/3
P(T')=1-1/3=2/3

b) The addition rule states; P(A ∪ B)=P(A)+P(B)-P(A⋂B)
Therefore, P(S ∪ E) = P(S)+P(E)-P(S⋂E)
S={1,4,9}
P(S)=3/12=1/4
E={2,4,6,8,10,12}
P(E)=6/12=1/2
(S⋂E)={4}
P(S⋂E)=1/12

P(S ∪ E) =3/12+6/12-1/12=8/12=2/3

c)P(A⋂B) = P(A)P(B)
Therefore, P(P⋂T)=P(P)P(T)
P={2,3,5,7,11}
P(P)=5/12
T={3,6,9,12}
P(T)=4/12
P(P⋂T)=5/12*4/12=20/144=5/36

Question 2;
This is a combination problem so the order of selection does not matter.
The total number of ways the constituency can be formed;
5+6=11 in total
3 of 11 = 11 C 3 = 11!/3!*8!=165 ways
The number of ways a constituency of 3 men can be formed;
5 men in total
3 of 5 = 5 C 3 =5!/3!*2!=10
The probability that the constituency is only formed of men is 10/165=2/33

Would this be correct, I feel that there is perhaps more to this problem which I have neglected? Thank you to anyone who replies

 

[FactChecker]

Everything looks ok to me except (c). What is $P\cap T$? Multiplying probabilities as you did is only valid if the events are independent. Are the events "the selected number is in $P$" and "the selected number is in $T$" independent?

 

[Mark44]

Question 1:
a) T' is the complementary event of T
Therefore, T'=1-T
In set T = {3,6,9,12}
P(T)=4/12 =1/3
P(T')=1-1/3=2/3

T and T' are not events -- they are sets.
T' = {1, 2, 4, 5, 7, 8, 10, 11}, so it makes no sense to say that T' = 1 - T.

You could say that T' = U - T, where U is the universal set here, the integers from 1 through 12.

I didn't look at the rest of your work, so have no comment on it.

 

[Master1022]

would anyone be able to advise whether the methods I have used to solve them would be correct or if I can improve upon my workings in any manner

I agree with the posts above.

However, I think using the probability theorems here slightly overcomplicates the problems as you can run into problems about independence (e.g. in (c) as pointed out by @FactChecker ). Instead, I think the easiest way to do the problems in 1 is just to list out the event spaces and count (as you did for part (a)). I.e. instead of using a theorem for (c), you could just write out the intersection of P and T and count how many members there are.

 

[FactChecker]

T and T' are not events -- they are sets.

Good point. I corrected my sloppy wording.

 

[AN630078]

Everything looks ok to me except (c). What is P∩T? Multiplying probabilities as you did is only valid if the events are independent. Are the events "the selected number is in P" and "the selected number is in T" independent?

Thank you for your reply, no they are not independent since 3 is a member of both sets. Would the sets be dependent, so this is a conditional probability problem?
P(P⋂T)=P(P)*P(T|P)
P(P)=5/12
How would I find P(T|P)?

Or would it be simpler just to find the intersecting members of the two sets?
P={2,3,5,7,11}
P(P)=5/12
T={3,6,9,12}
P(T)=4/12
(P⋂T)={3}
P(P⋂T)=1/12 ?

 

[AN630078]

T and T' are not events -- they are sets.
T' = {1, 2, 4, 5, 7, 8, 10, 11}, so it makes no sense to say that T' = 1 - T.

You could say that T' = U - T, where U is the universal set here, the integers from 1 through 12.

I didn't look at the rest of your work, so have no comment on it.

Thank you for your reply, yes you are correct they are not events but sets I had overlooked my error.
Yes, indeed I could write T' = U - T thank you for the suggestion

 

[AN630078]

I agree with the posts above.

However, I think using the probability theorems here slightly overcomplicates the problems as you can run into problems about independence (e.g. in (c) as pointed out by @FactChecker ). Instead, I think the easiest way to do the problems in 1 is just to list out the event spaces and count (as you did for part (a)). I.e. instead of using a theorem for (c), you could just write out the intersection of P and T and count how many members there are.

Thank you for your reply, ok in which case taking a simpler approach:

a) T' = U - T
In set T = {3,6,9,12}
In set T'={1,2,4,5,7,8,10,11}
In set U={1,2,3,4,5,6,7,8,9,10,11,12}
P(T)=8/12 =2/3
P(T')=4/12=1/3

b) S={1,4,9}
P(S)=3/12=1/4
E={2,4,6,8,10,12}
P(E)=6/12=1/2
(S ∪ E)={1,2,4,6,8,9,10,12}
P(S ∪ E) =8/12=2/3

c)P={2,3,5,7,11}
P(P)=5/12
T={3,6,9,12}
P(T)=4/12
(P⋂T)={3}
P(P⋂T)=1/12

 

[FactChecker]

Thank you for your reply, no they are not independent since 3 is a member of both sets. Would the sets be dependent, so this is a conditional probability problem?
P(P⋂T)=P(P)*P(T|P)
P(P)=5/12
How would I find P(T|P)?

Since you are given that the number was in P, the original problem has changed. Now there are 5 possible values, {2,3,5,7,11} all with equal probability. The probability that the selected number is 3 is 1/5.

Or would it be simpler just to find the intersecting members of the two sets?
P={2,3,5,7,11}
P(P)=5/12
T={3,6,9,12}
P(T)=4/12
(P⋂T)={3}
P(P⋂T)=1/12 ?

I don't know about you, but I think that finding the intersection is more direct and dirt simple. Of course, it is good to understand your first method because it will be needed often.

 

[AN630078]

Since you are given that the number was in P, the original problem has changed. Now there are 5 possible values, {2,3,5,7,11} all with equal probability. The probability that the selected number is 3 is 1/5.I don't know about you, but I think that finding the intersection is more direct and dirt simple. Of course, it is good to understand your first method because it will be needed often.

So would P(P|T)=5/12*1/5=1/12?

Yes, I agree that I do need to improve my understanding of probability problems, especially applying theorems to more complex problems

 

[FactChecker]

So would P(P|T)=5/12*1/5=1/12?

Yes. CORRECTION: No. And Prob$(P|T)$ is not what the original problem asked for. It asked for Prob$( P \cap T)$. Both approaches give the same answer.

CORRECTION. Prob$(P|T) = 1/5$.

 

[AN630078]

Yes. And isn't it nice that both approaches give the same answer!

Indeed it is, one query, does that mean that P(P⋂T) here equals P(T|P)?

 

[mjc123]

Indeed it is, one query, does that mean that P(P⋂T) here equals P(T|P)?

No. P(P⋂T) = P(P)*P(T|P). (Your statement in post #10 is wrong.)

 

[FactChecker]

No. P(P⋂T) = P(P)*P(T|P). (Your statement in post #10 is wrong.)

You are correct. Sorry. I was sloppy in my response, which I have now corrected.

 

[AN630078]

No. P(P⋂T) = P(P)*P(T|P). (Your statement in post #10 is wrong.)

Thank you for your reply. I am sorry but I think I have rather confused myself here. What would be the best method to find P(P⋂T) here?

 

[mjc123]

In the case of relatively small sets like this, the easiest and safest way is simply to enumerate the members of each set. P∩T is the set of integers from 1 to 12 that are both prime and multiples of 3. It is easily seen that 3 is the only member of this set, therefore the probability is 1/12.
Generally, you would need to evaluate both terms on the RHS of my equation. There are 5 primes below 12, so P(P) = 5/12. Only one of them is divisible by 3, so P(T|P) is 1/5.

 

[Mark44]

P(P) = 5/12

Is anyone else bothered by the use of P to mean two different things -- probability and the name of a set?

It would be less confusing to write Pr(P) or Prob(P) rather than P(P).