Bernoulli trials and permutations

Question

While solving questions using the Binomial probability theorem, the result is the probability of r successes among n trials, given that there are only two outcomes of probability p and q where p represent success(occurrence of desired event) and q represents failure(nonoccurrence of desired event), yes?

My doubt is this; does this theorem take into account all the different possibilities? For example, if I were applying this theorem to the situation of a family having three children:

If p is the probability of success of the birth of one boy, p=1/2 and q=1/2. Can we obtain the probability of 1 boy and 2 girls by

$³C1$x$(1/2)²$x$1/2=3/8?$ Does this account for all the permutations, viz. BBG, BGB and GBB? I am confused because we have used nCr here and not nPr.

Please share your insights and help me to understand this. Much thanks in advance :) Regards.

Answer 1

The binomial distribution is a distribution of a sum of $n$ iid random variables with the Bernoulli distribution, i.e. if $X_1,\ldots,X_n$ are iid random variables with the Bernoulli distribution, then $$ S_n=X_1+\ldots+X_n $$ has the binomial distribution. The random variable $S_n$ counts the number of success. The order is not important, only the number of success is important.

If we want to calculate the probability that we obtain $k$ successes out of $n$ trials, we first have to find the number of ways we can obtain $k$ successes. This is given by the binomial coefficient $$ {n\choose k}=\frac{n!}{k!(n-k)!}. $$ The probability to obtain $k$ successes and $n-k$ failures in any particular order is given by $$ p^k(1-p)^{n-k}. $$ If we combine the number of ways we can obtain $k$ success with the probability, we get $$ P(S_n=k)={n\choose k}p^k(1-p)^{n-k} $$ for $k\ge0$.

We took into account all different possibilities since we multiplied by the number of ways we can obtain $k$ successes. If $n=3$ and $k=2$, then there are $3$ ways to obtain $2$ successes.

I hope this helps.

Answer 2

Here when we calculate P(2 Girls and 1 boy) using Combination nCr, as we donot care about the order in which the children are born. If we had used nPr, we are actually considering all possible orders(BGG, GGB, GBG..), which is not needed, so while using nCr we are actually considering combination of 1 Boy and 2 Girls.

P(A and B) = P(A) * P(B|A) P(2 Girls and 1 Boy) = P(2 Girls) * P(1 boy| 2 Girls were born)

P(2 Girls) = Number of ways in which 2 girls are born out of 3 children(in any order) * P(Girl being born)^r * P(Boy being born)^(n-r)

P(2 Girls) = 3C2 * 0.5^2 * 0.5^1 = 3 * 0.25 * 0.5 = 0.375

P(1 boy | 2 Girls were born) = 1, as the total children are 3 and 2 are girls

P(2 Girls and 1 Boy) = P(2 Girls) * P(1 boy | 2 Girls were born)= 0.375 *1 = 0.375