Question 548281: If a couple has 7 kids what is the probability they will have 3 boys or 3 girls?
Answer by SwiftAlbatross(13)   (Show Source):
You can put this solution on YOUR website! Every birth is an independent event so we can simply multiply the probabilities. To have 3 boys from 7 kids, then the other 4 children must be girls. Likewise, to have 3 girls, the other 4 children must be boys. Since the probability of having a boy is the same as the probability of having a girl, 1/2, these two scenarios are exactly the same (P(3 boys/7 kids) = P(3 girls/7 kids) so let's just calculate the probability for one and double it.
Probability(3 boys/7 kids) = Proability(Exactly 3 boys and 4 girls) =
Here is the tricky part. Whenever given a probability problem, you should visualize in your mind a sample space with uniform probabilites from which to start basing your events upon. For example, it would be really hard for me to wrap my mind around the probability of rolling snake eyes on two 6-sided dice. However, if I construct a sample space with uniform probabilites, this becomes easy. In this case, the sample space is:
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
where every outcome's probability is the same = 1/36
Now finding the probability of snake eyes is easy. Just locate it on the samples space, (1,1), and put the number of outcomes (in this case, 1) over the number of outcomes in the sample space (36). So rolling snake eyes has a chance of 1/36.
Now getting back to our original problem, let's simplify the scenario so it is easier to see the math. Let's say we had only 2 kids, what would the sample space be? Easy!
S = {(Boy, Boy), (Boy, Girl), (Girl, Boy), (Girl, Girl)}
Notice all I did was list the different outcomes in an ordered fashion so I could easily remember to tally them all. Also note that each outcome in the sample space is the most basic event and has an equal chance within the sample space. This is the basic place to start. Listing the sample space of 7 kids is not so easy but so long as you recognize that it is just the same procedure, only on a larger scale, we can realize the main difficulty in the problem is to pick out all outcomes in the sample space that have 3 boys and 4 girls regardless or ordering. Well as you can see ordering matters in the sample space and so we just need to find out how many different ways there are to arrange 3 boys and 4 girls so that we may know how many elements in the sample space exist with 3 boys and 4 girls. This sounds like a combinatorics problem to me. So let's just do 7 choose 3: 7C3 = 7! / (3!*(7-3)!) = (7*6*5*4*3*2*1)/(3*2*1*4*3*2*1) = (7*6*5)/(3*2*1) = 35
Now we just have to find the size of the sample space. This is simple. Notice for 1 kid, Sample Space has 2 elements [S = {(B),(G)}]. For 2 kids, sample space has 4 elements. For 3 kids, sample space has 8 elements. For 4 kids, 16 elements. For n kids, the sample space has 2^n elements. This is because, like flipping a coin, each birth has two associated outcomes with it and has to be multiplied by the preceding outcomes to take account of all new possible outcomes. It's like each time I have a child I am adding another dimension to a spatial probability space. So if I have one child, I draw a line and the first half of the line could be a girl while the second half could be a boy. Having another child is like adding a whole dimension. If I have 2 children, I have a square, the sides represent each child. Three children means a cube. You can see now why the pattern is 2, 4, 8, 16, ... . Those are just the number of units within each figure. 2 for a line, 4 for a square, 8 for a cube, and so on. So for 7 kids, the sample space representing all probable outcomes has a size of 2^7 or 128. Now we compute the probability.
P(3 boys/7 kids) = 35/128
Now just double your answer because we were looking to include the other equivalent condition too (P(3 girls & 4 boys)).
The answer is 70/128 = 35/64 = 0.546875
Just to check, I actually wrote the sample space of 7 kids out and tallied all those with 3 boys and 4 girls.
S = {
(B,B,B,B,B,B,B),(B,B,B,B,B,B,G),(B,B,B,B,B,G,B),(B,B,B,B,B,G,G),(B,B,B,B,G,B,B),(B,B,B,B,G,B,G),(B,B,B,B,G,G,B),(B,B,B,B,G,G,G),(B,B,B,G,B,B,B),(B,B,B,G,B,B,G),(B,B,B,G,B,G,B),(B,B,B,G,B,G,G),(B,B,B,G,G,B,B),(B,B,B,G,G,B,G),(B,B,B,G,G,G,B),(B,B,B,G,G,G,G),(B,B,G,B,B,B,B),(B,B,G,B,B,B,G),(B,B,G,B,B,G,B),(B,B,G,B,B,G,G),(B,B,G,B,G,B,B),(B,B,G,B,G,B,G),(B,B,G,B,G,G,B),(B,B,G,B,G,G,G),(B,B,G,G,B,B,B),(B,B,G,G,B,B,G),(B,B,G,G,B,G,B),(B,B,G,G,B,G,G),(B,B,G,G,G,B,B),(B,B,G,G,G,B,G),(B,B,G,G,G,G,B),(B,B,G,G,G,G,G),(B,G,B,B,B,B,B),(B,G,B,B,B,B,G),(B,G,B,B,B,G,B),(B,G,B,B,B,G,G),(B,G,B,B,G,B,B),(B,G,B,B,G,B,G),(B,G,B,B,G,G,B),(B,G,B,B,G,G,G),(B,G,B,G,B,B,B),(B,G,B,G,B,B,G),(B,G,B,G,B,G,B),(B,G,B,G,B,G,G),(B,G,B,G,G,B,B),(B,G,B,G,G,B,G),(B,G,B,G,G,G,B),(B,G,B,G,G,G,G),(B,G,G,B,B,B,B),(B,G,G,B,B,B,G),(B,G,G,B,B,G,B),(B,G,G,B,B,G,G),(B,G,G,B,G,B,B),(B,G,G,B,G,B,G),(B,G,G,B,G,G,B),(B,G,G,B,G,G,G),(B,G,G,G,B,B,B),(B,G,G,G,B,B,G),(B,G,G,G,B,G,B),(B,G,G,G,B,G,G),(B,G,G,G,G,B,B),(B,G,G,G,G,B,G),(B,G,G,G,G,G,B),(B,G,G,G,G,G,G),(G,B,B,B,B,B,B),(G,B,B,B,B,B,G),(G,B,B,B,B,G,B),(G,B,B,B,B,G,G),(G,B,B,B,G,B,B),(G,B,B,B,G,B,G),(G,B,B,B,G,G,B),(G,B,B,B,G,G,G),(G,B,B,G,B,B,B),(G,B,B,G,B,B,G),(G,B,B,G,B,G,B),(G,B,B,G,B,G,G),(G,B,B,G,G,B,B),(G,B,B,G,G,B,G),(G,B,B,G,G,G,B),(G,B,B,G,G,G,G),(G,B,G,B,B,B,B),(G,B,G,B,B,B,G),(G,B,G,B,B,G,B),(G,B,G,B,B,G,G),(G,B,G,B,G,B,B),(G,B,G,B,G,B,G),(G,B,G,B,G,G,B),(G,B,G,B,G,G,G),(G,B,G,G,B,B,B),(G,B,G,G,B,B,G),(G,B,G,G,B,G,B),(G,B,G,G,B,G,G),(G,B,G,G,G,B,B),(G,B,G,G,G,B,G),(G,B,G,G,G,G,B),(G,B,G,G,G,G,G),(G,G,B,B,B,B,B),(G,G,B,B,B,B,G),(G,G,B,B,B,G,B),(G,G,B,B,B,G,G),(G,G,B,B,G,B,B),(G,G,B,B,G,B,G),(G,G,B,B,G,G,B),(G,G,B,B,G,G,G),(G,G,B,G,B,B,B),(G,G,B,G,B,B,G),(G,G,B,G,B,G,B),(G,G,B,G,B,G,G),(G,G,B,G,G,B,B),(G,G,B,G,G,B,G),(G,G,B,G,G,G,B),(G,G,B,G,G,G,G),(G,G,G,B,B,B,B),(G,G,G,B,B,B,G),(G,G,G,B,B,G,B),(G,G,G,B,B,G,G),(G,G,G,B,G,B,B),(G,G,G,B,G,B,G),(G,G,G,B,G,G,B),(G,G,G,B,G,G,G),(G,G,G,G,B,B,B),(G,G,G,G,B,B,G),(G,G,G,G,B,G,B),(G,G,G,G,B,G,G),(G,G,G,G,G,B,B),(G,G,G,G,G,B,G),(G,G,G,G,G,G,B),(G,G,G,G,G,G,G)}
From those, the elements that have 3 boys and 4 girls are
E = {
(B,B,B,G,G,G,G),(B,B,G,B,G,G,G),(B,B,G,G,B,G,G),(B,B,G,G,G,B,G),(B,B,G,G,G,G,B),(B,G,B,B,G,G,G),(B,G,B,G,B,G,G),(B,G,B,G,G,B,G),(B,G,B,G,G,G,B),(B,G,G,B,B,G,G),(B,G,G,B,G,B,G),(B,G,G,B,G,G,B),(B,G,G,G,B,B,G),(B,G,G,G,B,G,B),(B,G,G,G,G,B,B),(G,B,B,B,G,G,G),(G,B,B,G,B,G,G),(G,B,B,G,G,B,G),(G,B,B,G,G,G,B),(G,B,G,B,B,G,G),(G,B,G,B,G,B,G),(G,B,G,B,G,G,B),(G,B,G,G,B,B,G),(G,B,G,G,B,G,B),(G,B,G,G,G,B,B),(G,G,B,B,B,G,G),(G,G,B,B,G,B,G),(G,G,B,B,G,G,B),(G,G,B,G,B,B,G),(G,G,B,G,B,G,B),(G,G,B,G,G,B,B),(G,G,G,B,B,B,G),(G,G,G,B,B,G,B),(G,G,G,B,G,B,B),(G,G,G,G,B,B,B)}
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