Probability maths question

[WhotheHellisEdgar]

Trying to help DS understand probability in his maths homework. Can anyone help explain how to solve the problem below? Just having a mental block 😢

Bag with 9 snooker balls, 3 red and 6 non-red.
What is the probability of pulling out 2 non- red on your first try?

Thanks!

one non - red = 6 out of nine chances, or. Or Two in three chances

Two balls - 6 divided by 2, so one in three chances

Actualky I may ge totally wrong, ignore me.

Depends if you put them back or not.

If you don’t then the chance for the first one is 6/9 and the chance for the second is 5/8. So 6/9 * 5/8.

If you do put them back it’s 6/9 * 6/9.

This is correct.

To clarify is pulling two balls out together.

Thanks for responses so far, will think it through.

Draw a tree. Multiply along the branches.
For putting them back the probabilities don't change, for not returning as in the question, they do change as described by @DelurkingAJ .
If you need more than one you add the leaves (eg if question was probability of one of each colour)

If you put the ball back it's 1/9

If you don't put the first ball back in:
Probability of the first ball not being red = 3/9
Probability of the second ball not being red = 2/8
Probability of both not being red = 3/9 * 2/8 = 1/12

The probability the first ball is non red is 6 out of 9, or 6/9 = 2/3 = 0.666666...

There are now 8 balls remaining, 3 red and 5 non red.

The probability of drawing a non red ball from this is then 5/8 = 0.625.

So the probability of both of these occurring, ie of the first ball being non red and the second ball being non red is
6/9 times 5/8 = (6 x 5)/(9 x 8) = 30/72 = 0.416666....

It should be expressed as fractions rather than decimals, but I put the decimals in in order to make it clear what actual calculation I was doing.

Probability tree diagram

Sorry, I though 6 were red, 3 non-red, but it's the other way round.

If you put the ball back it's 4/9 (6/9 x 6/9)

If you don't put the first ball back in:
Probability of the first ball not being red = 6/9
Probability of the second ball not being red = 5/8
Probability of both not being red = 6/9 * 5/8 = 5/12

Note to self - READ THE OP PROPERLY not when The Archers is on

Thank you all so much!
I've explained how he should do it, so he's worked it out as 5/12 - same as lots of you 😍
Fingers crossed he remembers come exam time!

Remind him to read every question properly.

You could try him on rolling a double with 2 dice.

Versus rolling a double 6
And given you have rolled a 6, what's the chance of rolling another 6..

Cake's dice question is best done by a table rather than a tree diagram. So a 6x6 grid with column and row headers for each number. The 36 squares give the options.
That will show easily chance of double is 6/36.
Useful for totals eg probability total of 8 or whatever.
2 dice rolls are independent, so if you have a 6 already, chance now of another 6 is still 1/6.

Rolling a double
die 1: probability of it being any number = 6/6
die 2: probability of it being the same as die 1 = 1/6

Rolling a double 6
probability of die 1 being 6: 1/6
probability of die 2 being 6: 1/6
probability of both being 6: 1/6 x 1/6 = 1/36

For sure@WhoppingBigBackside !!

If both dice need to add up to a specific number, it depends what the number is:
2 has to be 1+1 (1/36)
3 can be 1+2 , 2+1 (1/36 +1/36 -= 1/18)
4 can be 1+3, 3+1, 2+2 (1/12)
5 can be 4+1, 1+4, 3+2, 2+3 (1/9)
6 can be 4+2, 2+4, 3+3, 1+5, 5+1 (5/36)
7 can be 6+1, 1+6, 2+5, 5+2, 3+4, 4+3 (1/6)
8 can be 6+2, 2+6, 3+5, 5+3, 4+4 (5/36)
9 can be 6+3, 3+6, 4+5, 5+4, (1/9)
10 can be 6+4, 4+6, 5+5 (1/12)
11 can be 6+5, 5+6 (1/18)
12 has to be 6+6 (1/36)

@WhoppingBigBackside is right, but you would do it as in my picture:

I wouldn't but your version is probably easier.

My version is how it would normally be taught. It stops you accidentally missing out something.

Also @WhotheHellisEdgar , unless it says to give the answer in its lowest form, they don't need to simplify the fraction. So an answer of 6/36 would be accepted, even though it simplifies to 1/6.
In my opinion simplifying could lead to 2 problems:
. Getting the simplification wrong
. Not being able to check so easily that total probabilities add to 1.

It looks nice - you can clearly see the distribution of the numbers with the combinations.

Lucky 7 and all that!!

It's dark and you have a drawer with 10 red socks and 10 blue socks. You can't see the colours but you randomly pick 2. What are the chances you get a matching pair?

Pick 3, you'll definitely get a pair then!

Or

@TeenDivided , I'm not a teacher and I tend to work things out in my head. My GCSE days are long behind me. The tree and graph would have taken me longer to figure out than the answer, and we're in Chat not Secondary Education.

I've probably done my bit for OP's son by pointing out how important it is to read the question carefully.

Good luck to @WhotheHellisEdgar 's DS in his exams.