In a certain country, it has been found over many years that $55$% of the babies born there are males. For a family in that country with five children, what is the probability that
(i) the two youngest children will be female and the three oldest children male
I am not sure about this part of the question. I originally thought that it might use the formula $n_1!/n_2!n_3!$ $*$ $(p)^x$ $*$ $(1-p)^{n-x}$, but I still can not get the answer. Can someone please show me how to do this.
(ii) exactly three children will be male
My attempt:
$5 \choose3$ $(0.55)^3(0.45)^2$ $=$ $0.33691$
(iii) at least three children will be male
My attempt:
$P(X \geq 3)$ $=$ $5 \choose3$ $(0.55)^3(0.45)^2$ + $5 \choose4$ $(0.55)^4(0.45)^1$ + $5 \choose5$ $(0.55)^5(0.45)^0$ $=$ $0.5931$
I tried to work out these problems. Are there any other ways of solving these parts of this problem. I am not sure about (i). Can someone please show and explain to me how they would get it? I tried to do this one but I am still not sure.
The question cannot be answered without further information (or without making some assumptions). For example, suppose each family that has three boys followed by two girls stops having children, while every family that reaches 5 children in a different pattern continues having children. Then in every family that has 5 children, the three oldest are male and the two youngest are female, so the answer is 100%. But you can also construct a scenario where the answer is 0%, or any number in between.