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

- mathmath333

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- dan815

=]

- anonymous

P(at least 2 boys|1boy)

- mathmath333

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

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## More answers

- mathmath333

how to solve this P(at least 2 boys|1boy) ?

- dan815

you can click edit and put it in the blue part

- mathmath333

thnks for suggestion

- anonymous

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)

- mathmath333

is this like bayes therem

- anonymous

no just conditional probability
Bayes is a little different

- mathmath333

P(at least 2 boys)/P(at least one boy)=(1/3+1/3)/(3/3)

- anonymous

not quite...
P(At least 2 boys) = P(2 boys) + P(3 boys)
P(at least 1 boy) = 1-P(no boys)

- mathmath333

=(1/3+1/3)/(1-0/3) =?

- anonymous

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

- mathmath333

final answer=(2/3)/(1/8)=1/12

- dan815

|dw:1444505252205:dw|

- dan815

the area of the smallest circle divded by the bigger circle

- anonymous

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

- dan815

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)

- mathmath333

P(at least 2 boys)/P(at least one boy)=(1/2)/(1/8)=1/4 , is it

- dan815

|dw:1444505457548:dw|

- dan815

|dw:1444505579498:dw|

- mathmath333

but the answer given inn book is \(\large \dfrac34\)

- mathmath333

i m very confused

- dan815

i prolly messed up somewhere, trying a slightly weird method

- dan815

pgpilots way is more intuitive

- amistre64

b gg
b gb
b bg
b bb
3 out of 4 in the sample space have at least 2 bs, given 1b

- dan815

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

- dan815

oh amistres way makes sense xD

- dan815

however im not too sure if thats all there is to it, like in his scneario the first one is always b

- mathmath333

i m dipresed

- dan815

lmao

- dan815

why

- mathmath333

which answer is correct

- dan815

we can see :)

- dan815

with a program

- dan815

hey its 4/7 ya

- dan815

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

- mathmath333

oh ok 4/7

- mathmath333

i hope ur program is correct

- freckles

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...

- freckles

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

- freckles

like is it 3 kids +1 boy that is given
or is it saying 3 kids where 1 of those 3 is a boy
?

- ybarrap

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:
$$
P(B_2)=\cfrac{2}{8}\\
P(B_3)=\cfrac{1}{8}\\
P(B_2)+P(B_3)=\cfrac{3}{8}
$$

- freckles

BBBB
BBBG
BBGG
BGGG
GGGG
this would be:
\[\frac{P(2 \text{ or } 3 \text{ or } 4)}{P(1 \text{ or } 2 \text{ or } 3 \text{ or } 4)}=\frac{3}{4}\]

- freckles

BBB;BBG;BGB;GBB;GGB;GBG;BGG;GGG
aren't some of these possibilities equal?
for example
BBG=BGB=GBB
or
GGB=GBG=BGG

- amistre64

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

- freckles

bbg=bgb
I listed all the possibilities for 3 kids
notice the answer didn't even care about the possibility that is GGG

- amistre64

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?

- freckles

when you guys list the 8 possibilities
does that mean we are concerned with which one popped out first, second, and third...

- freckles

what is the difference between bbg and bgb

- freckles

both of those say 2 boys and one girl

- amistre64

the sample space is preserved, in my mind

- ybarrap

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.

- amistre64

i tend to do card probabilities with P instead of C as well ... dont know why

- freckles

how do we know age matters?

- dan815

thats not how it works i thnk

- dan815

u are only considering b first then

- dan815

when u consider b atleast in all places and then see the other possibilties , the bbb case becomes repeated in them

- dan815

so its not exactly 3/4 its actually a bit less

- freckles

hey P(B_2)=3/7 right @ybarrap if you go with your new list you have

- dan815

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

- dan815

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

- dan815

you can manually check its usually less than 3/4 and around 4/7

- freckles

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

- freckles

this is what @amistre64 was saying earlier I believe
to far up to scroll

- freckles

I think I agree with @amistre64 now

- dan815

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

- dan815

the book answer is wrong

- freckles

but isn't your answer suggesting B doesn't have to come first?

- dan815

B doesnt ahve to come first though

- dan815

they just want B anywhere

- freckles

Then how did you all those extra possibility

- dan815

watcha mean

- freckles

Like if B has to come first then you only have:
the possibilities I just suggested

- dan815

yeah that is true

- dan815

but B doesnt haev to come first so we gotta consider everything

- freckles

oh I thought you said B had to come first

- dan815

"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

the answer suggest the book doesn't care about age
your answer suggest the book cares about age

- freckles

I think

- dan815

i dunno lemme think lol

- dan815

i was just think about this picture

- dan815

|dw:1444514572870:dw|

- dan815

yeah i did consider the different permutations
BGB
BBG
GBB

- freckles

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

- freckles

and you can omit the GGG from the list above

- freckles

oops nope I'm confused again

- freckles

because BGB is the same as BBG

- freckles

err...

- amistre64

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 ....

- amistre64

i think since the kids are already borned ... that age is not relevant

- freckles

so @amistre64 you are saying the first kid has to by a boy
but the order doesn't matter for the other two right?

- amistre64

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.

- dan815

kinddd offfff

- amistre64

is this a correct interpretation? lol ... in my case i never know

- dan815

lol kind of i mean whats stopping us from just looking at everything

- dan815

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

- dan815

thats kind of the way i approached it

- dan815

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

- dan815

i removed the ones where it was all 3 girls i mean

- dan815

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

- Empty

http://math.stackexchange.com/questions/697433/a-family-has-three-children-what-is-the-probability-that-at-least-one-of-them-i

- Empty

Do we agree that there is a higher probability of there being 2 boys than there is of being 3 boys?

- dan815

yes

- Empty

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.

- dan815

right

- ybarrap

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

- anonymous

probabilities amistre gave are not all equal... if they were, then it would be 3/4.

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