Probability Question
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- dhanda.arun
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This is how I approached the problem:
Let's say you first pick a red, then another red, then a green apple.
The probability would be (7/10)(6/9)(3/8)=7/40
You must next consider the number of different arrangements, and I believe this is where most make mistakes.
RRG
RGR
GRR
or if you like combinations,
The number of ways to pick 2 red apples: 3C2 or 3
Or the number of ways to pick 1 green apple: 3C1 or 3...same thing.
There you have 3 possible arrangement, hence (3)(7/10)=21/40.
Let's say you first pick a red, then another red, then a green apple.
The probability would be (7/10)(6/9)(3/8)=7/40
You must next consider the number of different arrangements, and I believe this is where most make mistakes.
RRG
RGR
GRR
or if you like combinations,
The number of ways to pick 2 red apples: 3C2 or 3
Or the number of ways to pick 1 green apple: 3C1 or 3...same thing.
There you have 3 possible arrangement, hence (3)(7/10)=21/40.
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Number of ways to pick 2 red apples from 3ddm wrote:The number of ways to pick 2 red apples: 3C2 or 3
Number of ways to pick 2 red apples should be 7c2?
or am i making some mistake?
To simplify, multiply 7/40 by 3
because you have 3 opportunities to get gree apple - i can be 1st, 2nd and 3rd in the group of three
smithpa2's post:
RRG
RGR
GRR
- cubicle_bound_misfit
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the answer is
(7c2*3c1)/10c3
= ((7*6/2)*3)/(10*9*8/6)
which comes down to 21/40
(7c2*3c1)/10c3
= ((7*6/2)*3)/(10*9*8/6)
which comes down to 21/40
Cubicle Bound Misfit