# Math Help - AQA probability - gcse

1. ## AQA probability - gcse

I should start by saying I am a mature student (embarrassed!!) and it isn't strictly speaking homework but I am using the AQA website to work through past papers as I am self taught. I am struggling with a probability question and I find that using probability trees helps me. Well, the question is:
You have 100 people divided into 14 male managers and 26 male teachers, 8 female managers and 52 female teachers. What is the probability that you will select a male teacher and any female.

On my "tree" I have put the probability of males / females is 2/5 and 3/5 respectively.
I have then on my branch off the "males" I have divided males into 7/20 probability for manager and 13/20 probability for teacher. On my branch off the "females" I have divided females into 2/15 managers and 13/15 teachers.

I worked out then that probability of male teacher = 2/5 x 13/20 and any female = 3/5 x 2/15 plus 3/5 x 13/15

2. Originally Posted by bauer_rules
i should start by saying i am a mature student (embarrassed!!) and it isn't strictly speaking homework but i am using the aqa website to work through past papers as i am self taught. I am struggling with a probability question and i find that using probability trees helps me. Well, the question is:
You have 100 people divided into 14 male managers and 26 male teachers, 8 female managers and 52 female teachers. What is the probability that you will select a male teacher and any female.

On my "tree" i have put the probability of males / females is 2/5 and 3/5 respectively.
I have then on my branch off the "males" i have divided males into 7/20 probability for manager and 13/20 probability for teacher. On my branch off the "females" i have divided females into 2/15 managers and 13/15 teachers.

I worked out then that probability of male teacher = 2/5 x 13/20 and any female = 3/5 x 2/15 plus 3/5 x 13/15

(26 + 8 + 52)/100.

3. ## sorry - still need help

The "model answer" gives 52/165 so they want to see the tree method being used. Can anyone tell me where I went wrong or even give me a "starter"!!

4. Originally Posted by Bauer_rules
The "model answer" gives 52/165 so they want to see the tree method being used. Can anyone tell me where I went wrong or even give me a "starter"!!
Please post the question exactly as it's written. Are you making two selections without replacement from the group of 100?

5. ## Probability

Hello Bauer_rules
Originally Posted by Bauer_rules
I should start by saying I am a mature student (embarrassed!!) and it isn't strictly speaking homework but I am using the AQA website to work through past papers as I am self taught. I am struggling with a probability question and I find that using probability trees helps me. Well, the question is:
You have 100 people divided into 14 male managers and 26 male teachers, 8 female managers and 52 female teachers. What is the probability that you will select a male teacher and any female.

On my "tree" I have put the probability of males / females is 2/5 and 3/5 respectively.
I have then on my branch off the "males" I have divided males into 7/20 probability for manager and 13/20 probability for teacher. On my branch off the "females" I have divided females into 2/15 managers and 13/15 teachers.

I worked out then that probability of male teacher = 2/5 x 13/20 and any female = 3/5 x 2/15 plus 3/5 x 13/15

I don't think you have given us the full question. I think that you should have said "Two people are selected. What is the probability ...?"

The answer, then, is that the probability of selecting a male teacher first is $\frac{26}{100}$. There are then 99 people left, of whom 60 are female. The probability that the second person selected is female is therefore $\frac{60}{99}$. So the probability that you select a male teacher, and then a female is

$\frac{26}{100} \times \frac{60}{99}$

In exactly the same way, you could select the female first and then a male teacher, with probability

$\frac{60}{100} \times \frac{26}{99}$

So the probability that one of these occurs is

$2\times \frac{26}{100} \times \frac{60}{99} = \frac{52}{165}$

6. Originally Posted by mr fantastic
Please post the question exactly as it's written. Are you making two selections without replacement from the group of 100?
The question gave the table of figures for males/females and managers/teachers. The only text then said "Two of these employees are chosen at random to attend a meeting. Calculate the probability that the chosen employees are a male teacher and any female." It says nothing about replacing people.

I have thought about it since and think that it means male teachers have a 26/100 probability multiplied by the probability of selecting any female which is 60/99 then multiplied by 2 because they go back to select again after the first choice. However, if I had done this in an exam I think I definitely would have got this wrong.

Hello Bauer_rulesI don't think you have given us the full question. I think that you should have said "Two people are selected. What is the probability ...?"

The answer, then, is that the probability of selecting a male teacher first is $\frac{26}{100}$. There are then 99 people left, of whom 60 are female. The probability that the second person selected is female is therefore $\frac{60}{99}$. So the probability that you select a male teacher, and then a female is

$\frac{26}{100} \times \frac{60}{99}$

In exactly the same way, you could select the female first and then a male teacher, with probability

$\frac{60}{100} \times \frac{26}{99}$

So the probability that one of these occurs is

$2\times \frac{26}{100} \times \frac{60}{99} = \frac{52}{165}$

Yes - I thought I had worded the whole question but - more haste less speed. Oh well.

In terms of your reply it was excellent thank you very much. I had forgotten the most obvious bit - to multiply by 2 because they could have selected a female first.

8. ## Probability

Hello again Bauer_rules
Originally Posted by Bauer_rules
Yes - I thought I had worded the whole question but - more haste less speed. Oh well.

In terms of your reply it was excellent thank you very much. I had forgotten the most obvious bit - to multiply by 2 because they could have selected a female first.
In that case, you might find it instructive to re-draw your tree, so that it splits twice:

• the first time, where the tree splits into three branches, to represent the first person chosen;
• the second time, where the three branches each split into three, giving nine branches in all, to represent the second person chosen.

On the first split, label the three branches:

• male teacher
• female
• other

and write the probabilities alongside each one: $\frac{26}{100}$, $\frac{60}{100}$ and $\frac{14}{100}$, respectively.

You could then label all nine smaller branches similarly, but you don't need all that level of detail. So, along the first level 'male teacher' branch, you're only interested in the outcome 'female' . The probability of this is now $\frac{60}{99}$ (since there are now only 99 people left, of whom 60 are female). And along the initial 'female' branch, label one of the last three small branches 'male teacher' and its probability $\frac{26}{99}$.

Now can you see that there are two 'routes' through the tree network each of which leads to the outcome you want? So work out their probabilities and add them together, as I did in my first post. This gives the correct answer.

Although you don't need to, it may also be helpful to you to complete all nine branches, and work out their respective probabilities. If you then add them up, you'll find that the total is 1. This should help you understand how all the possible outcomes fit together.