Probability.

**Oasis1993** · May 29th, 2010, 02:04 AM

Hey guys, Im having trouble with this question..I'd appreciate the help.

A study of the numbers of male and female children in families in a certain population is being carried out.

a) A sample model is that each child in any family is equally likely to be male or female, and that the sex of each child is independent of the sex of any previous children in the family. Using this model calculate the probability that, in a randomly chosen family of 4 children, there are 2 males and 2 females.

b) An alternative model is the first child in any family is equally likely to be male or female, but that, for any subsequent children, the probability that tthey will be of the same sex as the previous child is 3/5.
Usin tis model, calculate the probability that, in a randomly chosen family of 4 children,

i) all four will be the same sex
ii) no two consecutive children will be of the same sex,
iii) there will be 2 males and 2 females

Thanks.

**TKHunny** · May 29th, 2010, 04:14 AM

$(p + q)^{4}$

p = q = 1/2, but don't substitute these until AFTER you expand the expression

**Archie Meade** · May 29th, 2010, 07:20 AM

Originally Posted by Oasis1993

Hey guys, Im having trouble with this question..I'd appreciate the help.

A study of the numbers of male and female children in families in a certain population is being carried out.

a) A sample model is that each child in any family is equally likely to be male or female, and that the sex of each child is independent of the sex of any previous children in the family. Using this model calculate the probability that, in a randomly chosen family of 4 children, there are 2 males and 2 females.

b) An alternative model is the first child in any family is equally likely to be male or female, but that, for any subsequent children, the probability that tthey will be of the same sex as the previous child is 3/5.
Usin tis model, calculate the probability that, in a randomly chosen family of 4 children,

i) all four will be the same sex
ii) no two consecutive children will be of the same sex,
iii) there will be 2 males and 2 females

Thanks.

a)

Since the probabilities are always 0.5, you can use the binomial expansion,
choosing the term with 2 of both.

Alternatively, 2 girls and 2 boys can be born in the following sequences

BBGG
BGBG
BGGB
GGBB
GBGB
GBBG

Hence the probability is $6(0.5)^4$

b)

This one is not binomial due to the varying probabilities

(i) The sequences are

BBBB
GGGG

For the 4 boys, the probability is

$\frac{1}{2}\ \frac{3}{5}\ \frac{3}{5}\ \frac{3}{5}=\frac{1}{2}\left(\frac{3}{5}\right)^3$

same for the 4 girls, hence the solution is

$\left(\frac{3}{5}\right)^3$

(ii) The sequences are

BGBG
GBGB

$\frac{1}{2}\ \frac{2}{5}\ \frac{2}{5}\ \frac{2}{5}$ for both sequences, giving

$\left(\frac{2}{5}\right)^3$

(iii) The sequences are

GGBB
GBGB
GBBG
BGBG
BBGG
BGGB

The probability is

$\frac{1}{2}\ \frac{3}{5}\ \frac{2}{5}\ \frac{3}{5}$

$+\frac{1}{2}\ \frac{2}{5}\ \frac{2}{5}\ \frac{2}{5}$

$+\frac{1}{2}\ \frac{2}{5}\ \frac{3}{5}\ \frac{2}{5}$

$+\frac{1}{2}\ \frac{2}{5}\ \frac{2}{5}\ \frac{2}{5}$

$+\frac{1}{2}\ \frac{3}{5}\ \frac{2}{5}\ \frac{3}{5}$

$+\frac{1}{2}\ \frac{2}{5}\ \frac{3}{5}\ \frac{2}{5}$

$=\frac{1}{2}\left(2\frac{2}{5}\left(\frac{3}{5}\ri ght)^2+2\frac{3}{5}\left(\frac{2}{5}\right)^2+2\le ft(\frac{2}{5}\right)^3\right)$

$=\frac{2}{5}\left(\frac{3}{5}\right)^2+\frac{3}{5} \left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right )^3$

**Oasis1993** · May 29th, 2010, 10:20 AM

Hey, thanks for your reply.

I dont understand why we took the cube of 3/5 for question b) i ?
because there are 4 children...?

**Archie Meade** · May 29th, 2010, 10:37 AM

Originally Posted by Oasis1993

Hey, thanks for your reply.

I dont understand why we took the cube of 3/5 for question b) i ?
because there are 4 children...?

This is because in both cases all 4 children are of the same sex, but the probability of the first child being a girl (or a boy) is 0.5.
Hence, if the first child is a girl, then the probability of the 2nd child being a girl is 0.6, the probability of the 3rd child also being a girl is 0.6 and the probability of the 4th child being a girl is 0.6.

The same situation holds for the boys.
At the start of either of those sequences, the probability of a boy or girl being born is still 0.5 (equally likely to be male or female).

When we sum the probabilities for the boys and the girls we have 2(0.5) cancelling.

**Oasis1993** · May 30th, 2010, 05:42 AM

ohhh thank you, i undertand now.
But i have another question...sorry im kind of confused..why do we use 2/5 for b) ii ?

**Archie Meade** · May 30th, 2010, 05:56 AM

Originally Posted by Oasis1993

ohhh thank you, i undertand now.
But i have another question...sorry im kind of confused..why do we use 2/5 for b) ii ?

In this case, the probability of the first child being a boy or girl is $\frac{1}{2}$

If the probability of the next child being the same sex as the previous child is $\frac{3}{5}$

then the probability of the next child having opposite sex to the previous child is $1-\frac{3}{5}=\frac{5}{5}-\frac{3}{5}=\frac{2}{5}$ .

No 2 consecutive children have the same sex means we are alternating between boy and girl,

hence the subsequent probabilities in theses sequences are $\frac{2}{5}$

**Oasis1993** · May 30th, 2010, 06:33 AM

Thank you so much..i really appreciate the help!

**TKHunny** · June 1st, 2010, 01:40 PM

Just for the record, if you had done what I suggested in the first place, it is VERY likely that you would have been able to answer your own questions.

Learn the binomial theorem and recognize it when it pops up.

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