fitz Probability question (1 Viewer)
- Thread starter ToO LaZy ^*
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ToO LaZy ^*
n/a
easy question..
3 boys and 5 girls are at a party
for the 'pass the parcel' game the children remain in a circle, but 2 of the boys are asked not to sit together. in how many ways may this occur?
3 boys and 5 girls are at a party
for the 'pass the parcel' game the children remain in a circle, but 2 of the boys are asked not to sit together. in how many ways may this occur?
I wonder if that approach comes from non-HSC experienceArchman said:
hmm.. but it can be from 2U and/or ext2...
anyway the more ext1-ish way of doing these 2 questions would be:
"five cards are drawn at random from a pack of 52 playing cards. what is the probability that they are all from the same suit?"
Total possible outcomes: 52C5
Favourable outcomes: 13C5+13C5+13C5+13C5 = 4*13C5
"a bag contains 5 red balls and 4 white balls. 3 balls are drawn without replacement. find the prob. of drawing at least 2 red balls.."
Total possible outcomes: 9C3
Favourable outcomes: 5C2*4C1 (for when two are red) + 5C3 (when three are red)
you can also consider them as ordered selection, doesn't matter since it's probability.
is it just asking for how many ways they can be seated?ToO LaZy ^* said:
if it's two particular boys from the 3 of them,
no of ways
= boy1 * boy2 * everyone else
= 1 * 5 * 6!
boy1 can be seated in 1 way because it's in a circle so the first person essentially can sit in one way only.
boy2 is 5 because he can sit anywhere not next to boy1.
Managore
Member
7! - 6! I thought..
7! - 2*6! if you prefer this methodManagore said:
7! if the boys can sit anywhere
the first 6! is when boy2 sits to the right of boy1
the second 6! is when boy2 sits to the left of boy1
left and right matters even though it's in a circle
I just sort of found this out
Position or order is relative to a particular person. Right and left for him are different.
Last edited:
Managore
Member
thanks