A person has 3 children with at least one boy. Find the probablitiy of having atleast 2 boys among the children ?

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A person has 3 children with at least one boy. Find the probablitiy of having atleast 2 boys among the children ?

Mathematics
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P(at least 2 boys|1boy)
\(\large \color{black}{\begin{align} & \normalsize \text{A person has 3 children with at least one boy.} \hspace{.33em}\\~\\ & \normalsize \text{Find the probablitiy of having atleast 2 boys } \hspace{.33em}\\~\\ & \normalsize \text{among the children ?} \hspace{.33em}\\~\\ \end{align}}\)

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how to solve this P(at least 2 boys|1boy) ?
you can click edit and put it in the blue part
thnks for suggestion
P(at least 2 boys | at least 1boy) = P(at least 2 boys and at least 1 boy)/P(at least one boy) P(at least 2 boys)/P(at least one boy)
is this like bayes therem
no just conditional probability Bayes is a little different
P(at least 2 boys)/P(at least one boy)=(1/3+1/3)/(3/3)
not quite... P(At least 2 boys) = P(2 boys) + P(3 boys) P(at least 1 boy) = 1-P(no boys)
=(1/3+1/3)/(1-0/3) =?
P(no boys) = P(all girls) how many children does the couple have? 3. => P(all girls) = P(GGG) = (1/2)(1/2)(1/2) = 1/8
final answer=(2/3)/(1/8)=1/12
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the area of the smallest circle divded by the bigger circle
no P(At least 2 boys) = P(2 boys) + P(3 boys) P(3 boys) = P(3 girls) = 1/8 P(2 boys) = 3C2 * (1/8) = 3/8 P(At least 2 boys) = 1/2
Prob of atleast 2 boys = prob of atleast 1 boy - prob of only 1 boy prob of @2boys/prob of @ 1 boy = (prob of atleast 1 boy - prob of only 1 boy)/prob of @ 1 boy = 1- prob_of_1_boy/prob of @ 1 boy = 1- prob of 1 boy/(1-prob of no boys)
P(at least 2 boys)/P(at least one boy)=(1/2)/(1/8)=1/4 , is it
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but the answer given inn book is \(\large \dfrac34\)
i m very confused
i prolly messed up somewhere, trying a slightly weird method
pgpilots way is more intuitive
b gg b gb b bg b bb 3 out of 4 in the sample space have at least 2 bs, given 1b
p @least 2boys =p(2boys)+p(3boys)= 3/8 + 1/8=4/8 p@1boy = p@2 boys + p(1boy) = 4/8+3/8 herefore 4/8/7/8 = 4/7
oh amistres way makes sense xD
however im not too sure if thats all there is to it, like in his scneario the first one is always b
i m dipresed
lmao
why
which answer is correct
we can see :)
with a program
hey its 4/7 ya
d=randi(2) a=[] t=1000 for i =1:t for j=1:3 d=randi(2); if d==1 a(i,j)=1; else a(i,j)=0; end end end a; hold=[] for i = 1:t if sum(a(i,:))==0; hold(i)=1; end end b=sum(hold) %number of no 1 hold=[] for i=1:t if sum(a(i,:))==1; hold(i)=1; end end c=sum(hold)% number of only 1 prob=(t-(b+c))/(t-b) OUTPUT prob = 0.5716 and 4/7 = 0.5714
oh ok 4/7
i hope ur program is correct
I was thinking this as first: BBB BGG BBG GGG are the only possibilities because the following list give repeats: GBG since this is equal to BGG GGB since this is still BGG GBB since this is equal to BBG BGB since this is equal to BBG So we have 4 possibilities for a group 3 children: BBB BGG BBG GGG Now we have to find the probabilities of least 2 boys given at least 1 boy this means we are looking for \[P(2 \text{ or } 3|1 \text{ or } 2 \text{ or } 3) \\ =\frac{P( [2 \text{ or } 3] \cap [1 \text{ or } 2 \text{ or } 3] )}{P(1 \text{ or } 2 \text{ or } 3)}=\frac{P(2 \text{ or } 3)}{P(1 \text{ or } 2 \text{ or } 3)} =\frac{\frac{2}{4}}{\frac{3}{4}}=\frac{2}{3}\] .. but I'm still thinking...
however if I read that one thing as we have 3+1 kids where that 1 kid is boy and the other children or whatevers.. then following that one route from above we do get 3/4
like is it 3 kids +1 boy that is given or is it saying 3 kids where 1 of those 3 is a boy ?
If you don't consider order there are four distinct possibilities; BBB;GGG;GGB;BBG But we must consider order and so there are really eight possibilities: BBB;BBG;BGB;GBB;GGB;GBG;BGG;GGG Let \(B_i\) represent the event that there are \(i\) boys born out of 3. We are then looking for \(P(B_2)+P(B_3)\), the probability of 2 or more boys among the children: $$ P(B_2)=\cfrac{2}{8}\\ P(B_3)=\cfrac{1}{8}\\ P(B_2)+P(B_3)=\cfrac{3}{8} $$
BBBB BBBG BBGG BGGG GGGG this would be: \[\frac{P(2 \text{ or } 3 \text{ or } 4)}{P(1 \text{ or } 2 \text{ or } 3 \text{ or } 4)}=\frac{3}{4}\]
BBB;BBG;BGB;GBB;GGB;GBG;BGG;GGG aren't some of these possibilities equal? for example BBG=BGB=GBB or GGB=GBG=BGG
A person has 3 children with at least one boy. b gg or b bg or b gb or b bb ggg is not even in the sample space
bbg=bgb I listed all the possibilities for 3 kids notice the answer didn't even care about the possibility that is GGG
lets assume we are wanting to kill socrates :) we grab 3 of stones from a jar of black stones and white stones ... we show one stone, it is black. what is the probability that we have at least 2 black stones in our grips?
when you guys list the 8 possibilities does that mean we are concerned with which one popped out first, second, and third...
what is the difference between bbg and bgb
both of those say 2 boys and one girl
the sample space is preserved, in my mind
That's right! Given one boy was born, the sample space should be: BBB;BBG;BGB;GBB;GGB;GBG;BGG $$ P(B_2)=\cfrac{2}{7}\\ P(B_3)=\cfrac{1}{7}\\ P(B_2)+P(B_3)=\cfrac{3}{7} $$ Order is important in this sample space because there are more than one way that a group of 3 children can have two girls or two boys, but only one way to have all boys.The relative age of the child is relevant.
i tend to do card probabilities with P instead of C as well ... dont know why
how do we know age matters?
thats not how it works i thnk
u are only considering b first then
when u consider b atleast in all places and then see the other possibilties , the bbb case becomes repeated in them
so its not exactly 3/4 its actually a bit less
hey P(B_2)=3/7 right @ybarrap if you go with your new list you have
here's some trial space a = 0 0 0 1 0 1 1 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 0 1 0 a = 0 1 1 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 1 1
a = 0 0 0 0 1 0 1 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 a = 0 0 1 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 a = 1 1 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 0 0 1 1
you can manually check its usually less than 3/4 and around 4/7
ok if B has to come first then we have this list: B|BB B|GG B|BG B|GB There are 4 possibilities P(2)+P(3)=2/4+1/4=3/4
this is what @amistre64 was saying earlier I believe to far up to scroll
I think I agree with @amistre64 now
p @least 2boys =p(2boys)+p(3boys)= 3/8 + 1/8=4/8 p@least 1boy = p@2 boys + p(1boy) = 4/8+3/8 p(@least2boys|@1boy) = 4/8/7/8 = 4/7
the book answer is wrong
but isn't your answer suggesting B doesn't have to come first?
B doesnt ahve to come first though
they just want B anywhere
Then how did you all those extra possibility
watcha mean
Like if B has to come first then you only have: the possibilities I just suggested
yeah that is true
but B doesnt haev to come first so we gotta consider everything
oh I thought you said B had to come first
"A person has 3 children with at least one boy.Find the probablitiy of having atleast 2 boys among the children ?" the question just wants 1 boy anywhere
the answer suggest the book doesn't care about age your answer suggest the book cares about age
I think
i dunno lemme think lol
i was just think about this picture
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yeah i did consider the different permutations BGB BBG GBB
BBB BGG BGB BBG GGG GBB GBG GGB is the list I think you are using and for example BGG is the same as GGB; like both say two girls and one boy But we included them both because BGG is saying the boy came first then that one girl than that other girl --- BBB BGB BBG BGG this one doesn't care about age at all
and you can omit the GGG from the list above
oops nope I'm confused again
because BGB is the same as BBG
err...
there are 3 kids standing behind a curtain .. lol the first one steps out ... its a boy!! what is the probability that there were at least 2 boys standing behind the curtain? i just listed the ways that the other 2 kids could come forth ....
i think since the kids are already borned ... that age is not relevant
so @amistre64 you are saying the first kid has to by a boy but the order doesn't matter for the other two right?
given that we know one if the kids is a boy .... so its assumed that we have this prior knowledge of the sample space yes.
kinddd offfff
is this a correct interpretation? lol ... in my case i never know
lol kind of i mean whats stopping us from just looking at everything
like forget they even say its atleast 1 boy, cant we just solve it like its 1/2 chance for each and see every single possilbility and then check for ourselves
thats kind of the way i approached it
and to check i made a code that simply spat out 3 kids, boys or girls, and i removed the all Gs from them, and i looked at the ratio @least 2 bots/@atleast 1 boy
i removed the ones where it was all 3 girls i mean
d=randi(2) a=[] t=10000, %The number of sets of 3 children %Creats the Random Trial space for i =1:t for j=1:3 d=randi(2); if d==1 a(i,j)=1; else a(i,j)=0; end end end a; %Find the number of cases where all 3 children were girls hold=[] for i = 1:t if sum(a(i,:))==0; hold(i)=1; end end b=sum(hold) %#of all G's set %Finds the number of case where there is only 1 boy hold=[] for i=1:t if sum(a(i,:))==1; hold(i)=1; end end c=sum(hold) % #of only 1 boy sets %ANSWER prob=(t-(b+c))/(t-b) %atleast2boys/atleast1boy %atleast 2 boys = #of Total sets- #of all G's set-#of only 1 boy sets %atleast 1 boy = #of Total sets - #of all G's set
http://math.stackexchange.com/questions/697433/a-family-has-three-children-what-is-the-probability-that-at-least-one-of-them-i
Do we agree that there is a higher probability of there being 2 boys than there is of being 3 boys?
yes
If you agree then you know that it can't possibly be 3/4 since this is simply considering only: GGG, BGG, BBG, BBB but we know BBG has a higher probability of happening than BBB, so 3/4 is wrong. This isn't a combinatorics question.
right
Thanks @freckles! This probability should be: $$ P(B_2)=\cfrac{3}{7}\\ P(B_3)=\cfrac{1}{7}\\ P(B_2)+P(B_3)=\cfrac{4}{7} $$ The reason is that 2 boys is more likely than 3 boys is the same reason that 2 heads is more likely than 3 heads. For 2 heads you can have hht, hth or thh, while for three heads you only have only hhh as a possibility. So the age/order of the boys matters since the order of heads matters. The sample spaces for 2 or more tails/boys given at least one tail/boy is: BBB;BBG;BGB;GBB;GGB;GBG;BGG TTT;TTH;THT;HTT;HHT;HTH;THH
probabilities amistre gave are not all equal... if they were, then it would be 3/4.

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