# issues with probability

a man has $4$ children, given that atleast one of whom is a girl.Find the probability that he has $3$ girls and $1$ boy.

MY TRY :

probability of girl=$1/4$ and probability of boy=$3/4$ (my doubt is here, are these two probabiity correct? because it is said that atleast one of whom is a girl , not exactly 1)

=> 4c3 * (1/4)^4 * 3/4

• en.wikipedia.org/wiki/Boy_or_Girl_paradox Feb 26, 2014 at 15:52
• For each child, the probability of a boy = probability of a girl = 1/2. Feb 26, 2014 at 15:55
• do you have any problem understanding my solution?? @ritabrata Gautam?? Feb 26, 2014 at 16:12

I would imagine that the following assumptions hold: each child is either male or female; each child has the same chance of being male as of being female; and the sex of each child is independent of the sex of the other.

If at least one of the four children is a girl, then there are 15 equally-likely possibilities: BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG. Of these, there are four equally-likely cases where he has 3 girls and 1 boy (BGGG, GBGG, GGBG, GGGB). Thus, the probability that he has 3 girls and 1 boy given that he has at least 1 girl is 4/15.

• why we need ordering?? BBBG, BBGB and BGBB are all same .... Feb 26, 2014 at 16:01
• The 15 possibilities I listed are all equally likely (assuming the probabilities of having a boy and of having a girl are the same). The probability of having, say, four girls is not the same as the probability of having, say, three girls and one boy.
– JRN
Feb 26, 2014 at 16:04
• but all these BBBG, BBGB and BGBB refers to having 3 boys and 1 girl....so why we need all these case?? Feb 26, 2014 at 16:06
• @JoelReyesNoche what is wrong in joyentanuj das's procedure? Feb 26, 2014 at 16:07
• You can imagine the sample space of having four children as the set {four boys, three boys and one girl, two boys and two girls, one boy and three girls, four girls}. But these events are not equally likely. Would you say that the probability of having four girls is the same as the probability of having three girls and one boy?
– JRN
Feb 26, 2014 at 16:08

You can also see it like this. Define the events

• $A$ = at least one child is a girl
• $B$ = three children are girls and one is a boy

There are 16 cases in the whole, of which $15$ satisfy $A$ and $4$ satisfy $B$. Furthermore $B \cap A = B$. Thus $p(A)=\frac{15}{16}$ and $p(B)=\frac{4}{16}$, and $$p(B|A)=\frac{p(B \cap A )}{p(A)}=\frac{p(B)}{p(A)}=\frac{4}{15}.$$

• Sorry, my mistake. I forgot to mention that in this case $B \cap A = B$. It might be $B|A$ instead of $A|B$, it is just a matter of how you write "the event $B$ given the event $A$". It's a convention. But my formula is right, and by the way it confirms your result.
– alex
Feb 26, 2014 at 16:50

I would condition on all the possible # of girls:

$P(3G & 1B)= P(2G & 1B among three children|#G=1)P(#G=1)+ +P(1G & 1B among two children| #G=2)P(#G=2) + +P( 1B the fourth children |#G=3 )P(#G=3) + +P(-1G |#G=4)P(#G=4)$

the last term is zero because if he has four girls he can't have a boy

$P(#G=1)=1$

$P(#G=2)=Bin(n=3,k=1,p=1/2)$

$P(#G=3)=Bin(n=3,k=2,p=1/2)$

$P(2G & 1B among three children|#G=1)= Bin(n=3,k=2,p=1/2)$

$P(1G & 1B among two children| #G=2)= Bin(n=2,k=1,p=1/2)$

$P( 1B the fourth children |#G=3 )=1/2$

The man has four children of which have have atleast one girl so he may have 1 girl and three boys or two girls and two boys or three girls and one boy or four girls. so the probability that he has 3 girls and one boy is 1/4.

• do you have any problem understanding my solution?? @ritabrata Gautam?? Feb 26, 2014 at 16:11
• as joel mentioned according to your solution the probability of having four girls is the same as the probability of having three girls and one boy Feb 26, 2014 at 16:14