Recently I came across a problem similar to this,
What if you you are planning for your future family of 3 girls and 3 boys, how difficult it is or is it 50/50 chance?
Need few minutes to think of it? Go for it…
Amidst different methods of solving this problem I decided to use Pascal’s triangle. This is trivial but it was fun, therefore I decided to note it.
The above events distribute along a binomial distribution. Theoretically it satisfies four conditions:
- There are n trials
- Outcome of each trial is binary (two values, Girl or Boy)
- Through out the trials, probability of having a girl or boy is constant.
- The trials are independent to each others.
To calculate the coefficients we use Pascal’s triangle. As you all know the rows of the Pascal’s triangle relates to binomial expansion. Check the below Pascal’s triangle.
For example: elements of the 3rd row relates to the coefficient of the binomial expansion of (x + y)^3 = (x^3 + 3x^2y+ 3xy^2 + y^3) or in general nth row of the triangle relates to (x + y)^n.
Lets take having a boy is denoted by x and having a girl is denoted by y. In that case (x + y)^6 would give all the possibilities. However we are interested of having 3 girls and 3 boys. In order to solve that we can take the 6th row of the triangle which is;
1 6 15 20 15 6 1
It can be denoted as follow too:
1.x^6 + 6.x^5.y + 15.x^4.y^2 + 20.x^3.y^3 + 15.x^2.y^4 + 6.x.y^5 + 1.y^6
The coefficient of having 3 boys and 3 girls is 20 (20.x^3.y^3)
How about the total number of events with different combinations? Since there are two outcomes and 6 trials, it should be simply 2^6 = 64 in other words the sum of all the coefficients of (x + y)^6 , again it is equal to the sum of numbers in the 6th row.
So the probability of having 3 boys and 3 girls in your future family is
20/64 = 31.25%