Question

A couple wants to start a family. A doctor tells the couple based on many
factors, the probability distribution of possible births is given. G is for girl and B is for
boy. Assume the probabilities don't change for any reason.

x G B
P(X=x) 5/9 3/8

What is the probability that within the first 8 births, the family has at least 2 boys?

Answer 1

remarkable that the probabilities do not add to one. is there some other choice besides boy or girl, or is there a typo?

Answer 2

Um... first of all...

Answer 3

the first 8?? jesus they need a tv

Answer 4

P(transvestite children) = 5/72?

Answer 5

this is the probability is:
1-p(0)-p(1)
where \[p(X=x)=\left(\begin{matrix}8 \\ x\end{matrix}\right).p^{x}.(1-p)^{8-x}\]

Answer 6

actually transvestite would count as male i think. maybe hermaphrodite

Answer 7

I don't understand biology. =(

Answer 8

question makes no sense, so i would not complete it

Answer 9

maybe it is supposed to be \(\frac{5}{8}\) and \(\frac{3}{8}\) respectively

Answer 10

I mean, maybe it's just a really gender-progressive question.

Answer 11

she type it wrong 5/8

Answer 12

It's for a stats class, and it's probability. it DOES make sense, I just don't know how to set it up.
It's not typed wrong.

Answer 13

well, maybe my teachers a retard. i guess that makes sense.

Answer 14

no it does not. the probability has to add to one, because the complete sample space is G, B for each child

Answer 15

Pretty box... the stats you gave mean that 5 out of 72 times, the child will be neither male nor female.

Answer 16

we can pretend to complete the problem in any case.
at least two boys means
not one boy and
not two boys

probability of no boys in 8 births is
\[(\frac{5}{8})^8\]
probability of one boy is... well here is where we get stuck

Answer 17

You mean not one boy and not 0 boys. But yes, I like this method.

Answer 18

yeah that is what i mean. but the problem here is that \(1-p\) i don't know what to use

Answer 19

Let's just pretend 5/8 G and 3/8 B.... Oi.

Answer 20

probability of getting a boy is \(\frac{3}{8}\) so probability of not getting a boy should be
\[1-\frac{3}{8}=\frac{5}{8}\] but it seems to have morphed in to
\[\frac{5}{9}\] so it is not clear what number to use

Answer 21

well then we can do it
no boys = all girls
probability is therefore
\[(\frac{5}{8})^8\]
one boy, seven girls
probability is
\[8\times (\frac{5}{8})^7\times \frac{3}{8}\] reasoning you have one boy, 7 girls, and 8 different ways to do this (first child is a girl, second child is a girl etc

Answer 22

at least 2 boys means not all girls and not 7 girls so compute
\[1-(\frac{5}{8})^8-8\times(\frac{5}{8})^7\times \frac{3}{8}\]

Answer 23

thank you, that makes sense and i should be able to follow this pattern to figure out the other questions. <3