Please help with probability question

mom2boys

If 225 babies are born at a certain hospital each month, with 113 being boys and 112 being girls, what is the probability that 19 boys can be born consecutively?

POSSY

mom2boys wrote: »

If 225 babies are born at a certain hospital each month, with 113 being boys and 112 being girls, what is the probability that 19 boys can be born consecutively?

There are 113 c 19 ways of choosing 19 boys.
There are 225 c 19 ways of choosing 19 babies.

Ans = 113c19/225c19

mom2boys

Forgive my math ignorance, but c is for combination, correct? If I compute your answer, I get .2% probability. Is this correct?

AngryHippie

.000000912 by my reckoning
(113/225)*(112/224).......x(95/207)

hivizman

I would have done the calculation in a different way.

First, what is the probability that, in a sequence of 19 births, all 19 will be boys? Assume that the gender of any baby is independent of the genders of all other babies (this is strictly not a valid assumption because the possibility of twins means that if, say, a boy is born, the probability that the next baby will be a boy is greater than the overall probability of 113/225).

The probability that 19 boys will be born in a sequence of 19 births is:

(113/225)^19 = 0.000002075 = p

How many sequences of 19 births are there in 225 births? The general formula for the number of sequences of length k in a series of length n is (n+1-k). In this case, n = 225 and k = 19, so there are 207 sequences of 19 births (these are the sequences 1 to 19, 2 to 20, 3 to 21, up to 207 to 225).

We need to allow for the (admittedly very remote) possibility that there is more than one sequence of 19 boys in the 225 births. So we calculate the probability that a sequence of 19 births is not a sequence of 19 boys, which will be (1-p), or 0.999997925.

Then the probability that none of the 207 sequences is a sequence of 19 boys is:

0.999997925^207 = 0.99957056.

The probability that there is at least one sequence of 19 boys in a series of 225 births is therefore one minus this probability, that is, 0.00042944, or about 0.043%.

POSSY

hivizman wrote: »

I would have done the calculation in a different way.

First, what is the probability that, in a sequence of 19 births, all 19 will be boys? Assume that the gender of any baby is independent of the genders of all other babies (this is strictly not a valid assumption because the possibility of twins means that if, say, a boy is born, the probability that the next baby will be a boy is greater than the overall probability of 113/225).

The probability that 19 boys will be born in a sequence of 19 births is:

(113/225)^19 = 0.000002075 = p

How many sequences of 19 births are there in 225 births? The general formula for the number of sequences of length k in a series of length n is (n+1-k). In this case, n = 225 and k = 19, so there are 207 sequences of 19 births (these are the sequences 1 to 19, 2 to 20, 3 to 21, up to 207 to 225).

We need to allow for the (admittedly very remote) possibility that there is more than one sequence of 19 boys in the 225 births. So we calculate the probability that a sequence of 19 births is not a sequence of 19 boys, which will be (1-p), or 0.999997925.

Then the probability that none of the 207 sequences is a sequence of 19 boys is:

0.999997925^207 = 0.99957056.

The probability that there is at least one sequence of 19 boys in a series of 225 births is therefore one minus this probability, that is, 0.00042944, or about 0.043%.

From the way the question was phrased I'm guessing it's probably from a first year stats/probability course and so they were looking for knowledge of counting (not 1,2,3... but in terms of combinatrics), therefore I'd imagine the answer the lecturer is looking for is 113c19/225c19, and yes to the OP, c is for combinatrics as in 225 choose 19

hivizman

POSSY wrote: »

From the way the question was phrased I'm guessing it's probably from a first year stats/probability course and so they were looking for knowledge of counting (not 1,2,3... but in terms of combinatrics), therefore I'd imagine the answer the lecturer is looking for is 113c19/225c19, and yes to the OP, c is for combinatrics as in 225 choose 19

What you have calculated is the probability that, if you choose 19 births during the month at random, then all of the babies are boys. The answer provided by AngryHippie 0.000000912 is correct for this.

The original question is more specific than this. It asks for the probability that 19 boys can be born consecutively.

In my original calculation, I made two mistakes. First, I implicitly assumed that we were drawing from an infinite population (or, equivalently, we were drawing "with replacement", so it would be possible to include the same birth more than once). This clearly doesn't fit the facts. Secondly, I assumed that each of the possible 207 sequences of 19 consecutive births in a month of 225 births was independent of each of the others. But clearly this is incorrect. For example, if we start at the beginning of the month and reach a sequence of 19 births, consisting of a girl and 18 boys, the probability of there being a sequence of 19 boys during the month will be greater than if we don't come across such a sequence.

A better way of looking at the problem is that the probability P(n,k) of having at least one sequence of k boys in a series of n births must depend on what has already preceded this in the series. In fact, P(n,k) can be broken down into two situations:

1. We have already had a sequence of k boys when we get to the n-1th birth, so the nth birth is irrelevant.

2. We had not achieved a sequence of k boys up to the ((n-k)-1)th birth, the (n-k)th birth was a girl, and the subsequent (k-1) births were all boys, and then the nth birth is also a boy.

For example, if n = 5 and k = 2, the following sequences would be covered by situation 1:

BBBBB
BBBBG
BBBGB
BBBGG
BBGBB
BBGBG
BBGGB
BBGGG
BGBBB
BGBBG
GBBBB
GBBBG
GBBGB
GBBGG
GGBBB
GGBBG

(16 series out of 32 possibilities)

The following situations would be covered by situation 2:

BGGBB
GBGBB
GGGBB

(3 series out of 32)

(Giving a total of 19 series out of 32)

The probability of situation 1 is P(n-1,k). The probability of situation 2 is more complex, but it's basically the probability of not getting a sequence of length k in the first (n-k)-1 births, times the probability of getting a girl in the (n-k)th birth, times the probability of getting a sequence of k boys in the next k births.

This is [1-P((n-k)-1,k)]*(1-p)*(p^k)

Where p is the probability that a given birth is a boy (note that I'm assuming that each birth is an independent event with the probability of a boy being p = 113/225 and ignoring the fact that births are not independent because of the constraint that there has to be a total of 113 births in the month).

So we have P(n,k) = P(n-1,k) + [1-P((n-k)-1,k)]*(1-p)*(p^k)

Also, P(n,0) = (1-p)^n, P(n,n) = p^n, P(n,k) = 0 for k>n.

This recursive relationship allows us to calculate P(225,19) using a spreadsheet. If I've set this up correctly, then the probability we are after is 0.0002148.

Actually, the original question is badly worded, because it seems to imply at first reading that we are limiting our attention to a single month. But this limitation isn't explicitly stated, so we could read the first part of the question as simply supplying the frequency of boys and girls each month, allowing us to deduce that the probability that a boy is born is 113/225 (suspiciously close to 1/2!). If we are allowed to look for sequences of 19 consecutive male births over a longer period than a single month (and this isn't explicitly ruled out by the second part of the question), then the probability tends to one (albeit very slowly) as the length of the period increases.

To sum up, the way in which Possy and others have interpreted the question is similar to calculating the probability of being dealt five red cards from a standard pack of 52 cards, while I am interpreting the question as similar to asking for the probability of there being five consecutive red cards in the pack. This probability must be greater than the probability of being dealt five red cards. Why? Because being dealt five red cards is equivalent to requiring the first five cards (or any specified set of five cards) in the pack to be red, which is more restrictive than requiring that there are five consecutive red cards anywhere in the pack (as there are 48 potential sequences of five cards in a pack of 52 cards).