Ten people, 6 boys and 4 girls are playing on the playground. Miss Tweedle selects a group of 5 at random. What is the probability that the group has 3 boys and 2 girls?
C(5,3) .6^3 .4^2 maybe?
do i multiply.......?
.3456 if i did it right :)
its a binomial distribution; b=.6 and g=.4 bbbbb bbbbg bbbgb bbgbb bgbbb gbbbb bbbgg bbgbg bgbbg gbbbg bbggb bgbgb gbbgb etc
ohhh so is there any formula for this? like nCr?
yes
C(n,r) tends to be a simpler way to express it
soo its 6C3 and 4C2?
5 pick 3 boys = C(5,3) b^3 g^2
5C3, 5C2, 6C3, 4C2 ?
not quite .... its a binomial distribution
6/10 * 5/9* 4/8 * 4/7 * 3/6 is more like it
ohh thankyou, so after i solve all of that, what do i do?
Not quite, that's assuming you pick three boys in row, followed by two girls
you multiply it by the number of ways there are to pick that
5C3 ways to do so
4/10 * 3/9 * 6/8 * 5/7 * 4/6, two girls followed by three boys, slightly different probability
so i do 5C3 then 5C2?
bbbgg bbgbg bgbbg gbbbg bbggb bgbgb gbbgb bggbb gbgbb ggbbb 10 ways
"4/10 * 3/9 * 6/8 * 5/7 * 4/6, two girls followed by three boys, slightly different probability" -so do i do that and add them?
so to recap; the P(b & b & b & g & g) = \(\cfrac{6.5.4.4.3}{10.9.8.7.6}\) and there are 10 ways to pick that so \(\cfrac{\cancel{10}.\cancel{6}.5.\cancel{4}.\cancel{4}2.\cancel{3}}{\cancel{10}.\cancel{9}3.\cancel{8}\cancel{2}.7.\cancel{6}}\)=\(\cfrac{10}{21}\)
that is such a pain to type :)
oh im sorry :(
10*9*8*7*6 comes from the 5 group i get that 6,5,4 for boys and 4,3 for girls...
ohh i get it :D thankyou!!!
That makes sense. I wouldn't call it a binomial distribution though, because the probability of picking a boy or girl is not constant (i.e., no independence between trials).
yay!! :)
yeah, i saw that it wasnt binomial after i started decresing the boy girl count lol
thankyou so so so so much :) biggest help i've ever gotten today :)
binomial-ish at best ;)
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