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mathmath333
 one year ago
A person has 3 children with at least one boy.
Find the probablitiy of having atleast 2 boys
among the children ?
mathmath333
 one year ago
A person has 3 children with at least one boy. Find the probablitiy of having atleast 2 boys among the children ?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0P(at least 2 boys1boy)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{A person has 3 children with at least one boy.} \hspace{.33em}\\~\\ & \normalsize \text{Find the probablitiy of having atleast 2 boys } \hspace{.33em}\\~\\ & \normalsize \text{among the children ?} \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0how to solve this P(at least 2 boys1boy) ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2you can click edit and put it in the blue part

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0thnks for suggestion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0P(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)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is this like bayes therem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no just conditional probability Bayes is a little different

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0P(at least 2 boys)/P(at least one boy)=(1/3+1/3)/(3/3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not quite... P(At least 2 boys) = P(2 boys) + P(3 boys) P(at least 1 boy) = 1P(no boys)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0=(1/3+1/3)/(10/3) =?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0P(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

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0final answer=(2/3)/(1/8)=1/12

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the area of the smallest circle divded by the bigger circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2Prob 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/(1prob of no boys)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0P(at least 2 boys)/P(at least one boy)=(1/2)/(1/8)=1/4 , is it

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but the answer given inn book is \(\large \dfrac34\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i prolly messed up somewhere, trying a slightly weird method

dan815
 one year ago
Best ResponseYou've already chosen the best response.2pgpilots way is more intuitive

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3b gg b gb b bg b bb 3 out of 4 in the sample space have at least 2 bs, given 1b

dan815
 one year ago
Best ResponseYou've already chosen the best response.2p @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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh amistres way makes sense xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.2however im not too sure if thats all there is to it, like in his scneario the first one is always b

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0which answer is correct

dan815
 one year ago
Best ResponseYou've already chosen the best response.2d=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))/(tb) OUTPUT prob = 0.5716 and 4/7 = 0.5714

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i hope ur program is correct

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \[P(2 \text{ or } 31 \text{ or } 2 \text{ or } 3) \\ =\frac{P( [2 \text{ or } 3] \cap [1 \text{ or } 2 \text{ or } 3] )}{P(1 \text{ or } 2 \text{ or } 3)}=\frac{P(2 \text{ or } 3)}{P(1 \text{ or } 2 \text{ or } 3)} =\frac{\frac{2}{4}}{\frac{3}{4}}=\frac{2}{3}\] .. but I'm still thinking...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1however 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like is it 3 kids +1 boy that is given or is it saying 3 kids where 1 of those 3 is a boy ?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1If 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: $$ P(B_2)=\cfrac{2}{8}\\ P(B_3)=\cfrac{1}{8}\\ P(B_2)+P(B_3)=\cfrac{3}{8} $$

freckles
 one year ago
Best ResponseYou've already chosen the best response.1BBBB BBBG BBGG BGGG GGGG this would be: \[\frac{P(2 \text{ or } 3 \text{ or } 4)}{P(1 \text{ or } 2 \text{ or } 3 \text{ or } 4)}=\frac{3}{4}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1BBB;BBG;BGB;GBB;GGB;GBG;BGG;GGG aren't some of these possibilities equal? for example BBG=BGB=GBB or GGB=GBG=BGG

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3A 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1bbg=bgb I listed all the possibilities for 3 kids notice the answer didn't even care about the possibility that is GGG

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3lets 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1when you guys list the 8 possibilities does that mean we are concerned with which one popped out first, second, and third...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what is the difference between bbg and bgb

freckles
 one year ago
Best ResponseYou've already chosen the best response.1both of those say 2 boys and one girl

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3the sample space is preserved, in my mind

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1That'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.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3i tend to do card probabilities with P instead of C as well ... dont know why

freckles
 one year ago
Best ResponseYou've already chosen the best response.1how do we know age matters?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2thats not how it works i thnk

dan815
 one year ago
Best ResponseYou've already chosen the best response.2u are only considering b first then

dan815
 one year ago
Best ResponseYou've already chosen the best response.2when u consider b atleast in all places and then see the other possibilties , the bbb case becomes repeated in them

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so its not exactly 3/4 its actually a bit less

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hey P(B_2)=3/7 right @ybarrap if you go with your new list you have

dan815
 one year ago
Best ResponseYou've already chosen the best response.2here'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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2a = 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2you can manually check its usually less than 3/4 and around 4/7

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok if B has to come first then we have this list: BBB BGG BBG BGB There are 4 possibilities P(2)+P(3)=2/4+1/4=3/4

freckles
 one year ago
Best ResponseYou've already chosen the best response.1this is what @amistre64 was saying earlier I believe to far up to scroll

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think I agree with @amistre64 now

dan815
 one year ago
Best ResponseYou've already chosen the best response.2p @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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the book answer is wrong

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but isn't your answer suggesting B doesn't have to come first?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2B doesnt ahve to come first though

dan815
 one year ago
Best ResponseYou've already chosen the best response.2they just want B anywhere

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Then how did you all those extra possibility

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Like if B has to come first then you only have: the possibilities I just suggested

dan815
 one year ago
Best ResponseYou've already chosen the best response.2but B doesnt haev to come first so we gotta consider everything

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oh I thought you said B had to come first

dan815
 one year ago
Best ResponseYou've already chosen the best response.2"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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1the answer suggest the book doesn't care about age your answer suggest the book cares about age

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i dunno lemme think lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i was just think about this picture

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yeah i did consider the different permutations BGB BBG GBB

freckles
 one year ago
Best ResponseYou've already chosen the best response.1BBB 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and you can omit the GGG from the list above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops nope I'm confused again

freckles
 one year ago
Best ResponseYou've already chosen the best response.1because BGB is the same as BBG

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3there 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 ....

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3i think since the kids are already borned ... that age is not relevant

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so @amistre64 you are saying the first kid has to by a boy but the order doesn't matter for the other two right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3given that we know one if the kids is a boy .... so its assumed that we have this prior knowledge of the sample space yes.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3is this a correct interpretation? lol ... in my case i never know

dan815
 one year ago
Best ResponseYou've already chosen the best response.2lol kind of i mean whats stopping us from just looking at everything

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2thats kind of the way i approached it

dan815
 one year ago
Best ResponseYou've already chosen the best response.2and 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i removed the ones where it was all 3 girls i mean

dan815
 one year ago
Best ResponseYou've already chosen the best response.2d=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))/(tb) %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

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Do we agree that there is a higher probability of there being 2 boys than there is of being 3 boys?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1If 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.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1Thanks @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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0probabilities amistre gave are not all equal... if they were, then it would be 3/4.
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