In a survey of nuclear families with exactly four children each, n such familes are selected. For n = 10, find the probability that at least one of the n familes selected has all four children as girls?
Can anyone pls help? Thanks!
2006-11-10 17:51:29 · 4 answers · asked by sky_blue 1 in Science & Mathematics ➔ Mathematics
First the probability that a family of 4 is all girls is:
1/16 (and the chance that they are not is 15/16).
Rather than figuring the chance of at least one family having this arrangement, figure the probability that *none* of the families do. Then subtract from 1.
The answer will be 1 - (15/16)^10
or about 47.6%
It's not a simple multiplication by n... if it were, then with 16 families you should have a 100% chance. And with 32 families you would have a 200% chance. That logic would be correct if you couldn't repeat a combination, but you can. Just because family A has four boys, for example, doesn't mean family B can't also have four boys.
Label the 10 families as A, B, C, D, ... , I, J. The probability that A has four girls is 1/16. The probability that B has four girls is 1/16, etc.
The chance that *none* of them have four girls is 15/16 x 15/16 x 15/16 x ... x 15/16. This is (15/16)^10 or about 52.4%. Therefore the chance that at least one of them has four girls is 100 - 52.4 or 47.6%
2006-11-10 17:55:33 · answer #1 · answered by Puzzling 7 · 1⤊ 0⤋
chance of one kid being girl is 1/2
for each extra one you multiply this chance
so four girls would be 1/2 x 1/2 x1/2 x 1/2 = 1/16
so and given family would have this likelihood of all girls
if you have 10 families it is ten times as likely as one family
so 10 x 1/16 = 0.625
So there are 5 in 8 odds that there will be 4 girls if looking at 10 families.
2006-11-10 17:59:25 · answer #2 · answered by ignoramus 7 · 0⤊ 0⤋
danger is oftentimes expressed as a fragment: danger of happening/all available strategies it could ensue. there are 8 equivalent sized section, so all and sundry is the two probable. so there are 8 available strategies the spinner can end. 4 5 6 are 3 of em. so P(4, 5, or 6) = 3/8
2016-12-28 18:34:30 · answer #3 · answered by ? 3 · 0⤊ 0⤋
We must assume the probability of a boy vs. girl. I will take it 0.5.
Now (independent) probability that all for cildren are girls is 0.5^4 = 0.0625. The probability that any of 10 families has 4 girls is 10*0.0625 = 0.625.
2006-11-10 18:20:21 · answer #4 · answered by fernando_007 6 · 0⤊ 0⤋