Probability of drawing chosen three balls from bag
- Thread starter Fonda1971
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What formula did they give you for choosing "n" particular elements out of "m" total elements?
When you reply, please include a clear listing of your thoughts and efforts so far, so we can "see" where you're getting stuck. Thank you!
I do apologise, it’s a completely hypothetical question but an answer would help a real life situation.
In my sons math class, 3 boys out of eight boys were chosen by their teacher to be given an opportunity to be taken to a museum. A parent complained that this was unfair - that this opportunity should have been open to everyone and not just the teachers ‘favorite’ kids.
this was put to the teacher who then said he would put the names of the eight boys in class into a hat and pull three names out randomly.
Incredibly, the outcome was the same ~ the names of the three boys he had originally chosen were pulled out of the hat.
Of course, this was done in private.
My son cried ‘fix’ so, I’d like to show him the probability of this occurring in real life.
In my sons math class, 3 boys out of eight boys were chosen by their teacher to be given an opportunity to be taken to a museum. A parent complained that this was unfair - that this opportunity should have been open to everyone and not just the teachers ‘favorite’ kids.
this was put to the teacher who then said he would put the names of the eight boys in class into a hat and pull three names out randomly.
Incredibly, the outcome was the same ~ the names of the three boys he had originally chosen were pulled out of the hat.
Of course, this was done in private.
My son cried ‘fix’ so, I’d like to show him the probability of this occurring in real life.
Well, the number of ways to choose three balls from a set of 8 is the binomial coefficient \(\displaystyle \binom{8}{3}\), which can also be written \(\displaystyle _8C_3\). This number is
\(\displaystyle \displaystyle
\binom{8}{3} = \frac{8!}{(8-3)!\,3!} = \frac{8\cdot7\cdot6}{3\cdot2} = 56.
\)
And the number of ways to choose three particular balls is \(\displaystyle \binom{3}{3} = 1\): you just pick out those balls and that's that. So the probability of randomly choosing 1,2,3 on the second draw would be
\(\displaystyle \displaystyle
\frac{\binom{3}{3}}{\binom{8}{3}}
=\frac{1}{56}
\approx0.0179,
\)
which is about 1.8%. So not terribly likely but certainly not impossible.
\(\displaystyle \displaystyle
\binom{8}{3} = \frac{8!}{(8-3)!\,3!} = \frac{8\cdot7\cdot6}{3\cdot2} = 56.
\)
And the number of ways to choose three particular balls is \(\displaystyle \binom{3}{3} = 1\): you just pick out those balls and that's that. So the probability of randomly choosing 1,2,3 on the second draw would be
\(\displaystyle \displaystyle
\frac{\binom{3}{3}}{\binom{8}{3}}
=\frac{1}{56}
\approx0.0179,
\)
which is about 1.8%. So not terribly likely but certainly not impossible.