# Probability Questions: Union, Intersection and Combinations

• AN630078
In summary: I'm sorry. I made a mistake in my previous post. Prob##(P|T)## is not what the original problem asked for. It asked for Prob##( P \cap T)##. Both approaches give the same answer.Prob##(P|T) =...?...I am not sure how to solve this. Can you give me a hint?To find Prob(P|T), we need to find the probability that a randomly selected number from the set T is also in the set P. In other words, we need to find the probability of the intersection of the two sets. Can you apply the methods we discussed earlier to calculate this probability?Sure, so the intersection of P and
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

Last edited:
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?

Last edited:
Master1022
AN630078 said:
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.

FactChecker
AN630078 said:
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.

Mark44 said:
T and T' are not events -- they are sets.
Good point. I corrected my sloppy wording.

FactChecker said:
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 ?

Mark44 said:
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

Master1022 said:
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

AN630078 said:
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.

FactChecker said:
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

AN630078 said:
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##.

Last edited:
AN630078
FactChecker said:
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)?

AN630078 said:
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.)

AN630078 and FactChecker
mjc123 said:
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
mjc123 said:
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?

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.

mjc123 said:
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).

FactChecker

## 1. What is the difference between union and intersection in probability?

The union of two events in probability refers to the event that either one or both of the events occur. On the other hand, the intersection of two events refers to the event that both events occur simultaneously.

## 2. How do you calculate the probability of a union of two events?

To calculate the probability of a union of two events, you can use the formula P(A∪B) = P(A) + P(B) - P(A∩B), where P(A) and P(B) are the individual probabilities of the events and P(A∩B) is the probability of their intersection.

## 3. Can the probability of a union be greater than 1?

Yes, the probability of a union can be greater than 1 if the events are not mutually exclusive. This means that the events can occur at the same time, resulting in a higher probability than if they were independent.

## 4. What is the difference between permutations and combinations in probability?

Permutations refer to the number of ways to arrange a set of objects in a specific order, while combinations refer to the number of ways to select a subset of objects from a larger set, regardless of order.

## 5. How do you calculate the number of combinations in probability?

To calculate the number of combinations, you can use the formula nCr = n! / (r!(n-r)!), where n represents the total number of objects and r represents the number of objects in the subset.

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