# Thread: Bernoulli Family

1. ## Bernoulli Family

A family has 4 children. Assuming that boys and girls are equally likely, calculate the probability, using fractions, and Bernoulli's formula (please use this specific formula when necessary)

Probability that:
(i) there are exactly 2 boys.
(ii) there are exactly 3 girls.
(iii) there is at least one boy.

As far as I know, 'k' means the number of successes, 'n' is the number of Bernoulli trials, while P is the probability, hope i am right.

Any input will do >_< Thanks...

2. Yes, you're right.

If we denote the birth of a girl as a succes, that is we let X denote the number of girls,...then $\displaystyle P(X=k)= {n \choose k}p^k(1-p)^{n-k}$ where $\displaystyle p= \frac{1}{2}$. (In this case the formula is ofcourse the same for the number of boys)

Thus calculate:

(i) $\displaystyle P(X=2)$
(ii) $\displaystyle P(X=3)$
(iii) $\displaystyle P(X\geq 1) = 1- P(X< 1) = 1- P(X=0)$ (why?)

3. Originally Posted by Aeonitis
A family has 4 children. Assuming that boys and girls are equally likely, calculate the probability, using fractions, and Bernoulli's formula (please use this specific formula when necessary)

Probability that:
(i) there are exactly 2 boys.
(ii) there are exactly 3 girls.
(iii) there is at least one boy.

As far as I know, 'k' means the number of successes, 'n' is the number of Bernoulli trials, while P is the probability, hope i am right.

Any input will do >_< Thanks...
Let x be the random variable of the number of boys in the family,

X-B(4,0.5)

(1) Calculate P(X=2)

(2) Calculate P(X=1)

(3) Calculate 1-P(X=0)