# In a family of 5 children, determine the following probabibilities?

#### crownvicman

Hi, I'm a bit confused with one part of this probability problem. I'm pretty sure I got part a and b, but I'm not sure about c.

A family has 5 children, determine the following probabilities

a.) All 5 are the same gender

b.) 3 are boys, and 2 are girls

c.) The 3 oldest are boys, and the 2 youngest are girls.

For a. I'm pretty sure it's 1/32 since it would be .5^5 b/c there's a 1/2 probability of being either gender for each of the 5 children.
For b. I think it's also 1/32 since the probability of either gender doesn't change per birth.
As far as c. is concerned I think it's 5/16 because C(5,3) = 10 ways to have 3 boys and 2 girls multiplied by 1/32 but I'm not sure

#### romsek

MHF Helper
Hi, I'm a bit confused with one part of this probability problem. I'm pretty sure I got part a and b, but I'm not sure about c.

A family has 5 children, determine the following probabilities

a.) All 5 are the same gender

b.) 3 are boys, and 2 are girls

c.) The 3 oldest are boys, and the 2 youngest are girls.

For a. I'm pretty sure it's 1/32 since it would be .5^5 b/c there's a 1/2 probability of being either gender for each of the 5 children.
For b. I think it's also 1/32 since the probability of either gender doesn't change per birth.
As far as c. is concerned I think it's 5/16 because C(5,3) = 10 ways to have 3 boys and 2 girls multiplied by 1/32 but I'm not sure
a) is correct

b) you neglect to count the various ways you can have 3 boys and 2 girls, think of a 5 digit binary string. There are more than one with 3 1's and 2 0's

c) did you mix up b and c? In this case there is only one string, if you order it by age, that matches the criterion, 11100, out of the 32 possible strings. So in this case the probability is just

$\left(\frac{1}{2}\right)^5=\frac{1}{32}$

If you think of this as a probability tree you are being forced down a single path through the tree in contrast with b.

#### crownvicman

Thank you, sorry about that I did flip flop what I meant to say for b and c. Is 5/16 the correct answer for 3 are boys and 2 are girls?

#### romsek

MHF Helper
yes 5/16 is right

#### Plato

MHF Helper
A family has 5 children, determine the following probabilities
a.) All 5 are the same gender
To model this think of a truth table with five inputs.
Replace the T's with g's & the F's with b's. Now that is a model of the possible the birth-order and gender of five children. There are thirty-two rows, the first is all g's and the last is all b's.

Thus, there are two out of thirty-two of $\frac{1}{16}$ probability of same gender.

1 person

#### crownvicman

So does this mean the answer to part a.) is actually 1/16 since the probability is 1/32 for only 1 gender yet there are two gender possibilities so that 1/32 is multiplied by 2?

then b.) is still 5/16

and c.) is still 1/32 for 3 boys, 2 girls in any order?

#### romsek

MHF Helper
So does this mean the answer to part a.) is actually 1/16 since the probability is 1/32 for only 1 gender yet there are two gender possibilities so that 1/32 is multiplied by 2?

then b.) is still 5/16

and c.) is still 1/32 for 3 boys, 2 girls in any order?
man I really screwed the pooch on this one.

a) is as Hollywood noted 2 x 1/32 = 1/16

b) is 5/16

for (c) there are 3!=6 ways to arrange the 3 boys as the 3 oldest, and then 2 ways to arrange the girls in the remaining 2 slots. So there are 12 ways altogether to arrange the 3 boys and 2 girls such that the 3 boys are the oldest.

So the overall probability of this configuration is 12/32 = 3/8

#### Plato

MHF Helper
(c) there are 3!=6 ways to arrange the 3 boys as the 3 oldest, and then 2 ways to arrange the girls in the remaining 2 slots. So there are 12 ways altogether to arrange the 3 boys and 2 girls such that the 3 boys are the oldest.

So the overall probability of this configuration is 12/32 = 3/8
There is only one way to have three b's followed by two g's.
So $\dfrac{1}{32}~.$

#### romsek

MHF Helper
that's not true. we're not given any rule for assigning ages so we can order them however we like.

let the boy's be b1 b2 b3, the girls g1 and g2

b1 b2 b3 g1 g2

and

b3 b2 b1 g1 g2

for example both fit the bill

#### Plato

MHF Helper
that's not true. we're not given any rule for assigning ages so we can order them however we like.
What I posted is correct

Suppose we have blue balls and green balls that are identical except for color.
Say ten of each color, or maybe a hundred.

We select five of them and randomly arrange them into a string.
What is the probability that the string turns out to be $BBBGG~?$

That is the model for this question. It is like flipping a coin five times.
$\mathcal{P}(HHHTT)=~?$

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