strawberry115
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This should be really easy but I am struggling to work out how to do it. Ex 1A Q2 I can do the 5 all yellow. What should I be looking for /or asking myself to do (b) and (c)? I really don't understand how this works.
Thank you.
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figureeight
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I would love to help, but you need to reference where your question is from, otherwise we don't know what the question is... unless you feel like typing it out
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hermaphrodite
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eg. 5! = 5 x 4 x 3 x2 x 1

0! = 1
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figureeight
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(Original post by hermaphrodite)
eg. 5! = 5 x 4 x 3 x2 x 1

0! = 1
Ah maybe I should have thought about the thread title before posting... although it's still not really clear what the OP wants to understand. If they knew it was called a factorial, my guess was that they knew what it meant too... meh.
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.ACS.
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In Exercise 1A Question 2, you say you can do (a) which is for all yellow, but why can't you do (b) for three red and two green?

For (a) you basically did this:

- Work out the P(Y), P(Y|Y), P(Y|2Y), P(Y|3Y), P(Y|4Y)
- You multiplied each probability by each other
- Then, due to combinations, you multiplied that by \frac{n!}{x!y!z!}, where n is the total number, and x, y, and z represent as many of each colour used.

So if you actually know what you did in (a), you should be fine for the rest...
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strawberry115
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The question is : A bag contains 6 red, 5 green and 9 yellow beads. Five beads are selected from the bag without replacement. Find the probability that they are a) all yellow, I've done this one.

b) 3 red and 2 green
c) 1 red and 4 yellow
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.ACS.
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(Original post by strawberry115)
The question is : A bag contains 6 red, 5 green and 9 yellow beads. Five beads are selected from the bag without replacement. Find the probability that they are a) all yellow, I've done this one.

b) 3 red and 2 green
c) 1 red and 4 yellow
I've already explained how to do both (b) and (c) above... (I've done Edexcel S2 already, so have done the exact same questions as you're trying now). Do you understand what I've explained or do you want me to go over it again?
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strawberry115
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For the 5 yellow I did
9/20 X 8/19 X 7/18 X 6/17 X 5/16 which gave me the right answer of
21/2584.
Is there an easier way?
___________
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.ACS.
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(Original post by strawberry115)
For the 5 yellow I did
9/20 X 8/19 X 7/18 X 6/17 X 5/16 which gave me the right answer of
21/2584.
Is there an easier way?
___________
No, there's not an easier way... but the actual whole method is:

\frac{9}{20} \times \frac{8}{19} \times \frac{7}{18} \times \frac{6}{17} \times \frac{5}{16} \times \frac{5!}{5!}

It just so happens that \frac{5!}{5!} = 1... so this explains why you got the right answer. Do you understand? If instead of 5!/5! you did as I said above using \frac{n!}{x! \cdot y!} where n is the total amount of beads and x and y are the numbers of beads of each different colour (in this case 3 and 2), you'll get the right answer.

Do you understand why this is?
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strawberry115
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No .ACS. I don't fully understand this.

How do I know which to use when there are 3 red and 2 green out of a total of 6 red and 5 green?

Thanks for being patient with me.
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.ACS.
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Okay, don't worry, I'll go through the (b) very thoroughly with you...


- You must calculate: P(R), P(R|R), P(R|2R), P(G|3R), P(G|G and 3R), in the exact same way you calculated previously for (a) P(Y), P(Y|Y), P(Y|2Y), P(Y|3Y), P(Y|4Y).

- In (a) you ignored \dfrac{5!}{5!} as this is equal to 1. This factorial part shows the possible combinations available, but of course there is only one way of having everything the same. Now you must deal with 3 red and 2 green. The numerator stays the same as you are always dealing with five beads, however the denominator changes as you no longer have five yellow beads but two green and three red. So this becomes \dfrac{5!}{3! \cdot 2!}.

So if you've followed through, you must do the following for (b):

P(R) \cdot P(R|R) \cdot P(R|2R) \cdot P(G|3R) \cdot P(G|G and 3R) \cdot \dfrac{5!}{3! \cdot 2!}
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strawberry115
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Thank you for that .ACS. It is really helpful and I think I know how to do these now.
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