Out of these four outcomes, $3$ are favorable. So the probcapability need to be $frac34$.

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But have to you take into account the order of their birth? Due to the fact that in that instance it would certainly be $frac78$!

The complement of at least one boy is all three girls

So, $P($ at least one boy$)=1-P(GGG)$

$=displaystyle1-left(frac12 ight)^3$

This is the de facto method of fixing troubles of Probability of at leastern one in instance of Binomial Distribution prefer tossing a coin and so on.

Tbelow are in reality eight feasible outcomes:

$$GGG,,,GGB,,,GBG,,,BGG,,,BBB,,,BBG,,,BGB,,,GBB$$

Of these, only one does not incorporate a boy (B) in the event, and for this reason the probability of all girls is $;dfrac18;$ .

Anvarious other way to look at this is to draw this out

Here I follow the stereotypical association of gender and colors: the blue boxes recurrent boys and the pink boxes recurrent girls. Each time you have a boy or a girl, in the following generation you have the right to have actually a boy or a girl likewise, so the variety of possibilities is doubled each generation.

In regards to your difficulty, as soon as you have a boy, that represents a checkmark against "at least one of them is a boy", so I"ve crossed the box involved. However all the subsequent generations after this boy are additionally families in which there is at least one boy, so I"ve crossed those out as well. You deserve to view that the possibility of having actually at least one boy is $1/2$ in the first generation, $3/4$ in the second, and also $7/8$ in the third. This generalizes to $(2^n-1)/2^n$ in the nth generation.

Conversely the chance of having actually no boys is $1/2$ in the initially generation, $1/4$ in the second, and also $1/8$ in the third. This generalizes to $1/2^n$ in the nth generation.

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(Basically I"ve drawn a probcapacity tree diagram below, which generalizes to much more complex problems).