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vladimir1956 [14]

2 years ago

13

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

re equally likely.)

1 answer:

Vaselesa [24]2 years ago

6 0

Answer: $\frac{63}{64}$

<u>Step-by-step explanation:</u>

"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)

<u>not (0g & 6b)</u>

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

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

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

= $\frac{63}{64}$

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<u>Answer:</u>

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Answer:

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Step-by-step explanation:

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Table B

For $x=-2, y=-4$ ------> $k=-4/-2=2$

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For $x=1, y=3$ ------> $k=3/1=3$

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For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=2$ ------> $k=2/1=2$

For $x=2, y=4$ ------> $k=4/2=2$

This table has the same value of k

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Table D

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=2$ ------> $k=2/1=2$

For $x=2, y=6$ ------> $k=6/2=3$

This table has different values of k

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