Probability of Vacation Employment During Christmas Holidays

[Saracen Rue]

Homework Statement

During the Christmas holidays, 42 students from a group of 85 students found vacation employment while 73 students went away for holidays. Assuming that every student had at least a job or went on a holiday, what is the probability that a randomly selected student worked through the holidays (that is, did not go away on holidays), given that he/she had a job.

Homework Equations

I'm not sure if all of these are needed for this specific question, but I'm going to post every equation I was given just in case:

• Pr(A) = Pr(A|B)Pr(B) + Pr(A|B')Pr(B')
• Pr(A) = Pr(A∩B) + Pr(A∩B')
• Pr(A|B) = Pr(A∩B)/Pr(B)
• Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
• Pr(A∪B) = n(A∪B)/n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif)

The Attempt at a Solution

First, I wrote out the information I knew;
n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif) = 85
A = students who worked = 42
B = students who went away = 73
Pr(A) = 42/85
Pr(B) = 73/85

Next, I solved for Pr(A∪B):
Pr(A∪B) = n(A∪B)/n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif)
Pr(A∪B) = 115/85

I then attempted to plug 115/85 into the equation Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B) and solve for Pr(A∩B), but I encountered a problem...
Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
115/85 = 42/85 + 73/85 - Pr(A∩B)
115/85 = 115/85 - Pr(A∩B)
0 = Pr(A∩B)

As you can see, using the equation results in Pr(A∩B) = 0. However, I know this cannon be true as A+B > n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif).
So, I tried to just use common sense to determine Pr(A∩B); I added A and B together and took away n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif). This left me with Pr(A∩B) = 30/85. I then used this to determine that Pr(A∩B') = 12/85 (by taking 30 away from A) and Pr(A'∩B) = 43/85. From here, I decided to attempt to solve the actual equation, which I believed would use the equation Pr((A∩B')|A).
Pr(A|B) = Pr(A∩B)/Pr(B)
Pr((A∩B')|A) = Pr((A∩B')∩A)/Pr(A)

Now, I needed to find Pr((A∩B')∩A), but to do that I needed to know Pr((A∩B')∪A).
Pr((A∩B')∪A) = n((A∩B')∪A)/n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif)
Pr((A∩B')∪A) = 12+42/85
Pr((A∩B')∪A) = 54/85

Next, I tried to use the equation Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B) again, but the same thing happened...
Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
Pr((A∩B')∪A) = Pr(A∩B') + Pr (A) - Pr((A∩B')∩A)
54/85 = 12/85 + 42/85 - Pr((A∩B')∩A)
54/85 = 54/85 - Pr((A∩B')∩A)
0 = Pr((A∩B')∩A)

As you can see, this formula once again resulted giving me an answer of 0. I also just can't solve this with common sense as it's getting too complicated for me to actually understand what's going on. I think that there's something wrong with this formula, so if anyone could fix it or give me a correct one it would be much appreciated.

Also, I'm sorry if this is really obvious and I'm just being an idiot. I'm good at most areas of maths but for some reason I always struggle immensely with probability... anyway, thank you all for taking the time to read this.

 

[vishwesh]

Hi!

Whenever you face problems with probability questions of this sort, try to build a venn diagram (like the one I have attached).

Use the venn diagram to calculate n(A∩B) and then Pr(A∩B). Then, you will have to use your third formula because here a condition is given ---> he/she has a job. So, that means you have to calculate conditional probability.

Hope this helps

 

[PeroK]

Another idea is to work out some scenarios to see what's happening:

85 students: 42 worked (43 didn't work); 73 went on holiday (12 didn't go on holiday).

Scenario 1: All 42 who worked went on holiday. So, nobody who worked didn't go on holiday. In this case, your probability would be 0.

Scenario 2: All 43 who didn't work went on holiday. So, only 30 of those who did work went on holiday. So, p = 12/42.

Scenario 3: Something in between. Of those who worked somewhere between 0 and 12 didn't go on holiday.

So, it's not the sort of problem that you thought it was. It's not a conditional probablity problem.

Can you see how to solve it? You need to make a key assumption.

 

[HallsofIvy]

The first thing I note is that 42+ 73= 115 which is larger than 85 so some students must be counted twice. In fact, 115- 85= 30 students both had a job and went away on a holiday. Of the 73 students who had a job, 73- 30= 43 also went away on a holiday.

 

[Ray Vickson]

Homework Statement

During the Christmas holidays, 42 students from a group of 85 students found vacation employment while 73 students went away for holidays. Assuming that every student had at least a job or went on a holiday, what is the probability that a randomly selected student worked through the holidays (that is, did not go away on holidays), given that he/she had a job.

Homework Equations

I'm not sure if all of these are needed for this specific question, but I'm going to post every equation I was given just in case:

• Pr(A) = Pr(A|B)Pr(B) + Pr(A|B')Pr(B')
• Pr(A) = Pr(A∩B) + Pr(A∩B')
• Pr(A|B) = Pr(A∩B)/Pr(B)
• Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
• Pr(A∪B) = n(A∪B)/n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif)

The Attempt at a Solution

First, I wrote out the information I knew;
n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif) = 85
A = students who worked = 42
B = students who went away = 73
Pr(A) = 42/85
Pr(B) = 73/85

Next, I solved for Pr(A∪B):
Pr(A∪B) = n(A∪B)/n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif)
Pr(A∪B) = 115/85

I then attempted to plug 115/85 into the equation Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B) and solve for Pr(A∩B), but I encountered a problem...
Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
115/85 = 42/85 + 73/85 - Pr(A∩B)
115/85 = 115/85 - Pr(A∩B)
0 = Pr(A∩B)

As you can see, using the equation results in Pr(A∩B) = 0. However, I know this cannon be true as A+B > n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif).
So, I tried to just use common sense to determine Pr(A∩B); I added A and B together and took away n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif). This left me with Pr(A∩B) = 30/85. I then used this to determine that Pr(A∩B') = 12/85 (by taking 30 away from A) and Pr(A'∩B) = 43/85. From here, I decided to attempt to solve the actual equation, which I believed would use the equation Pr((A∩B')|A).
Pr(A|B) = Pr(A∩B)/Pr(B)
Pr((A∩B')|A) = Pr((A∩B')∩A)/Pr(A)

Now, I needed to find Pr((A∩B')∩A), but to do that I needed to know Pr((A∩B')∪A).
Pr((A∩B')∪A) = n((A∩B')∪A)/n(http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i1/s4univ.gif)
Pr((A∩B')∪A) = 12+42/85
Pr((A∩B')∪A) = 54/85

Next, I tried to use the equation Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B) again, but the same thing happened...
Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)
Pr((A∩B')∪A) = Pr(A∩B') + Pr (A) - Pr((A∩B')∩A)
54/85 = 12/85 + 42/85 - Pr((A∩B')∩A)
54/85 = 54/85 - Pr((A∩B')∩A)
0 = Pr((A∩B')∩A)

As you can see, this formula once again resulted giving me an answer of 0. I also just can't solve this with common sense as it's getting too complicated for me to actually understand what's going on. I think that there's something wrong with this formula, so if anyone could fix it or give me a correct one it would be much appreciated.

Also, I'm sorry if this is really obvious and I'm just being an idiot. I'm good at most areas of maths but for some reason I always struggle immensely with probability... anyway, thank you all for taking the time to read this.

I generally find that it make everything so much easier if you use a more informative notation. In this case, say $V =${students who went on vacation} and $E =${students who were employed}. Also, let $H =${students who stayed home (did not go on a vacation)}. You want to know $P(H|E)$. You are given $P(V) = 73/85$ and $P(V \cap E) = 42/85$. Of course, $P(H) = 1 - P(V)$.

From inclusion-exclusion we have $1 = P(V \cup E) = P(V) + P(E) - P(V \cap E)$, so you can find $P(E)$. Now apply the definition of $P(V|E)$ and recall that $P(V|E) + P(H|E) = 1$.