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A probability question - as level maths
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anonimen_2
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#1
Hello there,
I started doing this question, but I got stuck
Two bags, A and B, each contain balls which are either red or yellow or green. Bag A contains 4 red, 3 yellow and n green balls. Bag B contains 5 red, 3 yellow and 1 green ball. A ball is selected at random from bag A and placed into bag B. A ball is then selected at random from bag B and placed into bag A.The probability that bag A now contains an equal number of red, yellow and green balls is p. Given that p> 0, find the possible values of n and p
I said that in order to make three colours equal, we must have to draw a red ball from bag A with a return of green ball.
so, n+1=3
therefore, n =2, but there has to be one more value for n and I don't understand how to find it.
The values of p are even harder for me to find.
Any help would be appreciated.
Thanks in advance!
I started doing this question, but I got stuck
Two bags, A and B, each contain balls which are either red or yellow or green. Bag A contains 4 red, 3 yellow and n green balls. Bag B contains 5 red, 3 yellow and 1 green ball. A ball is selected at random from bag A and placed into bag B. A ball is then selected at random from bag B and placed into bag A.The probability that bag A now contains an equal number of red, yellow and green balls is p. Given that p> 0, find the possible values of n and p
I said that in order to make three colours equal, we must have to draw a red ball from bag A with a return of green ball.
so, n+1=3
therefore, n =2, but there has to be one more value for n and I don't understand how to find it.
The values of p are even harder for me to find.
Any help would be appreciated.
Thanks in advance!
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mqb2766
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#2
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#2
(Original post by anonimen_2)
Hello there,
I started doing this question, but I got stuck
Two bags, A and B, each contain balls which are either red or yellow or green. Bag A contains 4 red, 3 yellow and n green balls. Bag B contains 5 red, 3 yellow and 1 green ball. A ball is selected at random from bag A and placed into bag B. A ball is then selected at random from bag B and placed into bag A.The probability that bag A now contains an equal number of red, yellow and green balls is p. Given that p> 0, find the possible values of n and p
I said that in order to make three colours equal, we must have to draw a red ball from bag A with a return of green ball.
so, n+1=3
therefore, n =2, but there has to be one more value for n and I don't understand how to find it.
The values of p are even harder for me to find.
Any help would be appreciated.
Thanks in advance!
Hello there,
I started doing this question, but I got stuck
Two bags, A and B, each contain balls which are either red or yellow or green. Bag A contains 4 red, 3 yellow and n green balls. Bag B contains 5 red, 3 yellow and 1 green ball. A ball is selected at random from bag A and placed into bag B. A ball is then selected at random from bag B and placed into bag A.The probability that bag A now contains an equal number of red, yellow and green balls is p. Given that p> 0, find the possible values of n and p
I said that in order to make three colours equal, we must have to draw a red ball from bag A with a return of green ball.
so, n+1=3
therefore, n =2, but there has to be one more value for n and I don't understand how to find it.
The values of p are even harder for me to find.
Any help would be appreciated.
Thanks in advance!
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willa12345
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#3
You first need to think about the two possible ways you could take one ball from bag A and where you would puy an extra ball to make all the balls equal. So first n-2 which you have already found and the second is n=5 as if you take on of the green beads away then add a yellow they would all be 4 balls. Second of all to find the probablitiy I would personally use a tree diagram. The first tree would the probabilities of taking the diffrent couloured balls from bag a and the second would be the probabilities of taking back the diffrent coloured balls from bag b(remember the amount of balls has increased by 1). You would then use this for n=2 and n=5. changing the values appropriately. Then find which probablities you would need. In this case for n=2 4/9*1/10=4/90 and for n=5 5/9*3/10=15/120 4/90 15/120 = 61/360 p=61/360 and n=2,5
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mqb2766
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#4
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#4
(Original post by willa12345)
You first need to think ...
You first need to think ...
Howeve, dont follow/agree with some of your numbers for the second case.
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