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.)

Respuesta :

Answer: [tex]\frac{63}{64}[/tex]

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 - [tex][(\frac{1}{2})^{0}*(\frac{1}{2})^{6} ][/tex]

= 1 - [tex]1 * \frac{1}{64}[/tex]

= 1 - [tex]\frac{1}{64}[/tex]

= [tex]\frac{63}{64}[/tex]

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