Bob and Jane have three children. Given that one child is their daughter Mary, what is the probability that Bob and Jane have at least two daughters?
(I am also interested in wordings of this problem where additional given information such as "name is Mary" or "born on a Tuesday" are shown to alter the probability.)
Evidently there are eight equally probable families (here lower numbers are older children e.g.). There are 24 children who we could meet. But we met Mary, who is a girl. There are 12 girls.
(G1, G2, G3)
(G4, G5, B6)
(G7, B8, G9)
(B10, G11, G12)
(G13, B14, B15)
(B16, G17, B18)
(B19, B20, G21)
(B22, B23, B24)
Of the 12 girls we could have met (each equally probable), 9 of them are in a family with 2+ daughters. 3 are alone.
So the probability is 9/12 = 3/4.
Evidently there are eight equally probable families. We met one of the seven families that contain a girl.
(G, G, G)
(G, G, B)
(G, B, G)
(B, G, G)
(G, B, B)
(B, G, B)
(B, B, G)
(B, B, B)
Of the 7 families we could have met (each equally probable), 4 of them have 2+ daughters. 3 are single-daughter families.
So the probability is 4/7.
"Given a daughter" in a 3-child family is logically equivalent to "not three boys." "Bob and Jane have 3 children; the children are not all boys." There are seven equally possible families:
Four of these have 2+ girls; hence, 4/7.
Since this can be easily verified experimentally by using three coins and tallying P(2+ Heads, given 1+ Heads), answers that claim "3/4" may be up against a wall.