educationn.
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Even though I read through the mark scheme, I still do not understand the answer and the process to get to the answer.
Here is the question:
There are 9 counters in a bag.
7 of the counters are green.
2 of the counters are blue.
Ria takes at random two counters from the bag.
Work out the probability that Ria takes one counter of each colour.
You must show your working.
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mqb2766
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(Original post by itsjustrahel)
Even though I read through the mark scheme, I still do not understand the answer and the process to get to the answer.
Here is the question:
There are 9 counters in a bag.
7 of the counters are green.
2 of the counters are blue.
Ria takes at random two counters from the bag.
Work out the probability that Ria takes one counter of each colour.
You must show your working.
Why not post the mark scheme and be a bit more specific about what you're confused about?
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skent6
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Here’s how I worked it out by putting it on a probability tree. The chance of getting Blue then green is 2/9 x 7/8 then the chance of getting Green then Blue is 7/9 x 2/8 so you just add these two together for your answer
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mohamadjamil03
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You know that Ria takes one counter of each colour, so there are two possible scenarios: either you take a green counter followed by a blue counter, or you take a blue counter followed by a green counter. Let's consider each scenario separately:

Probability of taking green counter first = 7/9
Probability of taking blue counter second = 2/8 (because the total number of counters has now decreased from 9 to 8)
Probability of green, then blue: 7/9 * 2/8 = 14/72 = 7/36

Probability of taking blue counter first = 2/9
Probability of taking green counter second = 7/8
Probability of blue, then green: 2/9 * 7/8 = 14/72 = 7/36

You can then add the probabilities together to find the overall probability that she takes different coloured counters from the bag:
i.e. 7/36 + 7/36 = 14/36 = 7/18
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help_me_learn
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(Original post by itsjustrahel)
Even though I read through the mark scheme, I still do not understand the answer and the process to get to the answer.
Here is the question:
There are 9 counters in a bag.
7 of the counters are green.
2 of the counters are blue.
Ria takes at random two counters from the bag.
Work out the probability that Ria takes one counter of each colour.
You must show your working.
I am not 100% sure if I am right because I am still learning myself but my answer is 7/8 simplified.

I got this by doing the tree diagram. in the first two branches, for green, it is 7/9 and for blue it is 2/9.
the tricky part is the second branch because we have to consider if the person puts the counter back or not. in this case they do NOT.
so the two branches off of green should have 6/8 (green) and 2/8(blue). the denominator is 8 because u have already taken a counter out so the total number changes.
for the second branches for the blue are 7/8(green) and 1/8 (blue)
now we have to use the AND or OR rule. if it is AND then we multiply and if it is OR we add.
the possible ways of getting taking out one counter of each colour could only be (G&B) OR (B&G)
thus using the tree u plug the values in
(7/9 X 2/8) + (2/9 X 7/8)
this simplified will give u 28/72 but this can be further simplified into 7/18

sorry for the long explanation...

ps: I am unsure if I am correct so can u let me know if I got it right?
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mohamadjamil03
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Considering we got the same answers, I'm pretty sure you did it right (assuming they don't put the counter back into the bag)
(Original post by help_me_learn)
I am not 100% sure if I am right because I am still learning myself but my answer is 7/8 simplified.

I got this by doing the tree diagram. in the first two branches, for green, it is 7/9 and for blue it is 2/9.
the tricky part is the second branch because we have to consider if the person puts the counter back or not. in this case they do NOT.
so the two branches off of green should have 6/8 (green) and 2/8(blue). the denominator is 8 because u have already taken a counter out so the total number changes.
for the second branches for the blue are 7/8(green) and 1/8 (blue)
now we have to use the AND or OR rule. if it is AND then we multiply and if it is OR we add.
the possible ways of getting taking out one counter of each colour could only be (G&B) OR (B&G)
thus using the tree u plug the values in
(7/9 X 2/8) + (2/9 X 7/8)
this simplified will give u 28/72 but this can be further simplified into 7/18

sorry for the long explanation...

ps: I am unsure if I am correct so can u let me know if I got it right?
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Anonymouspsych
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(Original post by itsjustrahel)
Even though I read through the mark scheme, I still do not understand the answer and the process to get to the answer.
Here is the question:
There are 9 counters in a bag.
7 of the counters are green.
2 of the counters are blue.
Ria takes at random two counters from the bag.
Work out the probability that Ria takes one counter of each colour.
You must show your working.
No tree diagram is required

There are only two possibilities:

Ria takes a green counter first and then a blue counter ---> in this case the probability is 7/9 *2/8

OR

Ria takes a blue counter first and then a green counter -----> in this case the probability is 2/9 * 7/8

So the overall probability that Ria takes one counter of each colour is simply the above results added together which in fact is also the same as one of the above probabilities being doubled.
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Muttley79
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(Original post by skent6)
Here’s how I worked it out by putting it on a probability tree.
Edit your post - it is against the rules to post solutions
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Muttley79
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(Original post by Anonymouspsych)
No tree diagram is required
Edit your post - it is against the rules to post solutions
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