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: $\frac{63}{64}$

Step-by-step 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 - $[(\frac{1}{2})^{0}*(\frac{1}{2})^{6} ]$

= 1 - $1 * \frac{1}{64}$

= 1 - $\frac{1}{64}$

= $\frac{63}{64}$