Answers (3)
I am corrected.
"You may have heard that if a couple have two children, both of the same gender, say girls, their chances of conceiving a third female baby are higher than conceiving a male. This idea has lived on for centuries but it’s simply a myth. Another myth related to gender are that men are more “manly” if they produce a line of boys and that women are more “feminine” if they produce all girls.
With each pregnancy, a couple has a 50% chance of conceiving either a boy or a girl. It’s no different than flipping a coin two times in a row and getting tails each time; the third time, the likelihood of getting tails is still 50%. The previous two coin flips won’t influence the outcome of the third. With each pregnancy, a couple has the same 50% chance of conceiving a girl.
It’s human nature to believe that somehow a cause underlies such events, but not true in the game of chance or in gender selection.
Although numerous people can attest to examples of families with only girls or only boys — sometimes six or 10!— semen samples from the fathers in these cases reveal a 50:50 ratio of male to female sperm. Remember, all semen contains approximately 50% of female-carrying (X) sperm cells and 50% male-carrying (Y) sperm cells."
b := boy
g := girl
Let's build our event-space. What do we know?
There are two different sexes (b, g) , and for all practical purposes the probabilities P for the 1st child are: P(b) = 0.5, P(g) = 0.5 .
The probabilities for the 2nd child are also: P(b) = 0.5, P(g) = 0.5 .
The question refers to the outcome of both independent events. For both children we get this event space S(E):
S(E) = {b,g} × {b,g} = { (b,b), (b,g), (g,b), (g,g) }, each with a probability of 0.25.
The condition "At least one girl" matches 3 states of the event space. The probability of having at least one girl is then 0.75 or 75%.
Every sequence of events which come in only two flavours ( sexes, flipped coins, on-off switches, ... you name it) can be represented as a binary sequence. If you know the number n of such consecutive binary events, 2^n is the number of all possible sequences.
In your example there are two children. For 2 children to be born consecutively the possible sequences are: bb, bg. gb, gg. ( 00, 01, 10, 11 in binary). There is only one sequence with no g in it; that is bb (00). This is one out of four numbers. So, the probability to pick this number is 1/4 = 25%. In all other cases (2^2 -1 = 3) there is AT LEAST one g. And the probability for this is then: 1 - 1/4 = 3/4 = 75%.
In the family with n children the probability P that there is at least one girl g (or one boy for that matter) is:
P = (2^n - 1) / 2^n = 1 - 2^-n.
Actually, three of them qualify, so the probability is 75%.