An easy statistics question, maybe too easy

[Nitemare]

I had a teacher give this question on a test, I gave an answer that is technically correct yet still get it marked incorrect.

A couple plan on having 4 children. What are the odds of at least 3 of them being girls?

I will give both correct answers after a couple of guesses.

 

[Legendary]

Originally posted by: Nitemare
I had a teacher give this question on a test, I gave an answer that is technically correct yet still get it marked incorrect.

A couple plan on having 4 children. What are the odds of at least 3 of them being girls?

I will give both correct answers after a couple of guesses.

Probability 3 girls 1 boy or probability 4 girls
P(3 girls 1 boy) + P(4 girls) = 2 / 2^4 = 1/8
I think that's right.

 

[joepa99]

I would guess 1 to 3 odds...

GGGG
GGGB
GBGG
BGGG

 

[dullard]

Here are your possible results that meet the criteria:
4 girls
3 girls, boy is eldest
3 girls, boy is 2nd eldest
3 girls, boy is 2nd youngest
3 girls, boy is youngest.
Thus 5 possible outcomes that meet the criteria. Now how many total outcomes are there? 2^4 = 16. Thus the odds you are looking for are 5/16.

 

[Haircut]

Originally posted by: Legendary

Originally posted by: Nitemare
I had a teacher give this question on a test, I gave an answer that is technically correct yet still get it marked incorrect.

A couple plan on having 4 children. What are the odds of at least 3 of them being girls?

I will give both correct answers after a couple of guesses.

Probability 3 girls 1 boy or probability 4 girls
P(3 girls 1 boy) + P(4 girls) = 2 / 2^4 = 1/8
I think that's right.

Not quite right.
Probability of at least 3 girls is P(3 girls) + P(4 girls) as you said.

Assuming 0.5 chance of getting a boy or girl

p(3 girls) is 0.5^3 * 0.5 * 4C3 = .25
p(4 girls) is 0.5^ 4 = .0625

Probability of getting at least 3 girls is .25+.0625 = 0.3125

 

[dullard]

Originally posted by: Haircut

Assuming 0.5 chance of getting a boy or girl

p(3 girls) is 0.5^3 * 0.5 * 4C3 = .25
p(4 girls) is 0.5^ 4 = .0625

Probability of getting at least 3 girls is .25+.0625 = 0.3125

Good answer, just 1 minute too late . By the way, does anyone here know the actual % chance of getting a boy or girl? I once read it but have forgotten the statistic. It isn't quite 50%/50% (even though I assumed it in my answer above).

By the way, if the first 3 were girls, there is a very good chance that the father has imperfect male sperms thus the chance of having a 4th girl is much greater than 50%. The same goes with 3 boys born first...

 

[dullard]

Are we ever going to see your answer and the teacher's answer?

 

[HomeBrewerDude]

4*.5^4

the above is correct unless you are in a more advance probability class, in which case, you would want to use a baysian approach to take into account the probability of concieving.

 

[dullard]

Originally posted by: yamahaXS
4*.5^4

No that is the probability of getting 3 girls out of 4 kids. Nitemare wants to know the probability of getting AT LEAST 3 girls. You forgot the 4 girls in 4 births option.

 

[Nitemare]

Originally posted by: dullard
Are we ever going to see your answer and the teacher's answer?

since you asked yes

My approach was similar to joepa99's except for the fact that he did not include the possibility of having 4 boys
GGGG
GGGB
GBGG
BGGG
BBBB

Therefore the possibility I gave was 2/5 or .4

However, I had not accounted for the birth order, did not see where in the question it said to account for the birth order, so I just did the final outcome.

The answer that he wanted was 5/16

5 correct possibilities in 16 possible outcomes.

bbbb
bbbg
bbgb
bbgg
bgbb
bgbg
bggb
bggg
gbbb
gbbg
gbgb
gbgg
ggbb
ggbg
gggb
gggg

was my 2/5 answer wrong? I can understand how both are right, but the teacher has yet to explain to me how my answer is wrong.

 

[HomeBrewerDude]

Originally posted by: dullard

Originally posted by: yamahaXS
4*.5^4

No that is the probability of getting 3 girls out of 4 kids. Nitemare wants to know the probability of getting AT LEAST 3 girls. You forgot the 4 girls in 4 births option.

you're right.

its 5*.5^4

 

[HomeBrewerDude]

Originally posted by: Nitemare

Originally posted by: dullard
Are we ever going to see your answer and the teacher's answer?

since you asked yes

My approach was similar to joepa99's except for the fact that he did not include the possibility of having 4 boys
GGGG
GGGB
GBGG
BGGG
BBBB

Therefore the possibility I gave was 2/5 or .4

However, I had not accounted for the birth order, did not see where in the question it said to account for the birth order, so I just did the final outcome.

The answer that he wanted was 5/16

5 correct possibilities in 16 possible outcomes.

bbbb
bbbg
bbgb
bbgg
bgbb
bgbg
bggb
bggg
gbbb
gbbg
gbgb
gbgg
ggbb
ggbg
gggb
gggg

was my 2/5 answer wrong? I can understand how both are right, but the teacher has yet to explain to me how my answer is wrong.

Yes your answer was wrong. Both cannot be right. IMO, the question did not need to account for different "orders" because that is what you are supposed to learn. Namely, the probability of getting 3 heads out of 4 coin flips is differnt than the probability of observing the sequence of "head, head, head, tail" out of 4 coin flips.

 

[dullard]

Therefore the possibility I gave was 2/5 or .4

However, I had not accounted for the birth order, did not see where in the question it said to account for the birth order, so I just did the final outcome.

The answer that he wanted was 5/16

5 correct possibilities in 16 possible outcomes.

bbbb
bbbg
bbgb
bbgg
bgbb
bgbg
bggb
bggg
gbbb
gbbg
gbgb
gbgg
ggbb
ggbg
gggb
gggg

was my 2/5 answer wrong? I can understand how both are right, but the teacher has yet to explain to me how my answer is wrong.

You just showed the teacher's reasoning and there is nothing you can deny about it. There are 16 possible outcomes, 5 match your criteria of at least 3 girls, thus there is a 5/16 chance of getting at least 3 girls. Anything else is wrong. Thus your first question is answered: 2/5 is wrong. But the reason why it is wrong is your second question. You said there were 5 possible outcomes: 4 girls, 3 girls, 2 girls, 1 girl, or 0 girls. Of those 5 outcomes two meet the criteria. So far everything is correct. But (and here is the key) each outcome isn't equally likely. Thus there isn't a 2/5 chance of having at least 3 girls.

Think about this example. If I play a dice game and throw two dice (the game of craps does this). I am far more likely to get a total of 7 on the dice than I am to get a total of 2. Thus the game of craps is based entirely around this fact. (Note: there is a 1/6 chance of getting a total of 7 and a 1/36 chance of getting a total of 2.) Do you agree that you will get more 7s than 2s? If not, try rolling a pair of dice and see how many more 7s you get than 2s. Now lets use the failed 'logic' that you just did on the test. There are 11 possible outcomes for the total of two dice: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Using your logic there is one outcome with a total of 2 and one outcome with a total of 7. Thus using your logic, all combinations are equally likely with a 1/11 chance of occuring. Common sence says that won't happen, the odds at Vegas says it won't happen, any test you do yourself says it won't happen.

Anytime one outcome is more likely to occur than another you cannot use your logic. If and only if all outcomes are equally likely will your logic work (and in those rare cases, it works great).