Question

Find the probability that when a couple has sixsix ​children, at least one of them is a girlgirl. ​(assume that boys and girls are equally​ likely.)

Answer 1

Answer:
`(63)/(64)`

Explanation:

"At least one" means:

(1g & 5b) or (2g & 4b) or (3g & 3b) or (4g & 2b) or (5g & 1b) or (6g & 0b)

which is the same as: not(0g & 6b)

not (0g & 6b)

= 1 -
`[((1)/(2))^(0)*((1)/(2))^(6) ]`

= 1 -
`1 * (1)/(64)`

= 1 -
`(1)/(64)`

=
`(63)/(64)`