# Stuck on this probability QWatch

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#1
A bag contains 100 balls, 20 red, 30 green and 50 blue.

If you take out three balls what is the probability of getting 1 red, 1 green and 1 blue?
1
1 month ago
#2
(Original post by HighFructose)
A bag contains 100 balls, 20 red, 30 green and 50 blue.

If you take out three balls what is the probability of getting 1 red, 1 green and 1 blue?
What are you stuck with?
0
#3
(Original post by mqb2766)
What are you stuck with?
I can work out red then green then blue but not 1 red x 1 green x 1 blue.
0
4 weeks ago
#4
So what do you get for the bit you understand?
(Original post by HighFructose)
I can work out red then green then blue but not 1 red x 1 green x 1 blue.
0
4 weeks ago
#5
(Original post by HighFructose)
A bag contains 100 balls, 20 red, 30 green and 50 blue.

If you take out three balls what is the probability of getting 1 red, 1 green and 1 blue?
The chance of getting a red ball the first time is 2/10, then assuming you put the ball back, the chance of getting a green ball the second time is 3/10, then assuming you put the ball back, the chance of getting a blue ball the third time is 5/10.

Can you remember what you need to do with the 2/10, 3/10 and 5/10?
0
4 weeks ago
#6
(Original post by harrysbar)
The chance of getting a red ball the first time is 2/10, then assuming you put the ball back, the chance of getting a green ball the second time is 3/10, then assuming you put the ball back, the chance of getting a blue ball the third time is 5/10.

Can you remember what you need to do with the 2/10, 3/10 and 5/10?
Do the balls have to be in a specific order
Last edited by stewai; 4 weeks ago
0
4 weeks ago
#7
(Original post by stewai)
Do the balls have to be in a specific order
It doesn't matter which order you picked them out in, the odds would still be 5/10 blue, 2/10 red and 3/10 green (as long as you put the ball back each time)
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4 weeks ago
#8
(Original post by harrysbar)
It doesn't matter which order you picked them out in, the odds would still be 5/10 blue, 2/10 red and 3/10 green (as long as you put the ball back each time)
Then all you need to do is times the fractions together and the times it by how many ways there are to get the outcome you want
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4 weeks ago
#9
It might seem a bit obvious, but drawing a tree diagram is always my go-to for probability questions like these. Draw it big and clear and you'll be able to see every outcome very easily. Try it if you're still stuck!
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4 weeks ago
#10
(Original post by harrysbar)
The chance of getting a red ball the first time is 2/10, then assuming you put the ball back, the chance of getting a green ball the second time is 3/10, then assuming you put the ball back, the chance of getting a blue ball the third time is 5/10.

Can you remember what you need to do with the 2/10, 3/10 and 5/10?
If you take out three that usually implies no replacement.
0
4 weeks ago
#11
(Original post by HighFructose)
A bag contains 100 balls, 20 red, 30 green and 50 blue.

If you take out three balls what is the probability of getting 1 red, 1 green and 1 blue?
RGB - how many different ways can you end up with these three colours?
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