Brain Teasers:

[charlie b]

Not true, Steve. That is to say, you just gave a correct answer to a question that you had not asked. Here is the statement of your problem:

A woman has 2 children. One is a boy. What are the odds that her other child is a boy?

You are asking about the "other child," given that the first child is a boy. That "other child" can only be a boy if both children are boys. Using your model, only your choice #1 is possible, and the odds are one in three. But that is not the right answer either.

The correct way to mathematically model your question is this:

What is the conditional probability that the "other child" is a boy, given that this child is a boy?

Without loss of generality, I will address one child as the first and the other as the second, not meaning birth order, but just the order in which they are considered. I will choose the one whose gender you named as being a boy, and call that person the "first child."

Let's use the following notation:

• P(B1) = the probability that the first child is a boy.
• P(B2) = the probability that the second child is a boy.
• P(B1 + B2) = the probability that the both children are boys (i.e., the intersection of the two events).
• P(B2|B1) = the probability that the second child is a boy, given that the first child is a boy.

If neither child is yet born, then P(B1) = P(B2) = 0.5 (that is, 50%)

• P(B2|B1) = P(B1 + B2) / P(B1)
• P(B2|B1) = (1/4) / (1/2)
• P(B2|B1) = (1/2) (that is, 50%)
  But if both children are born, and we know the gender of B1 is male, then P(B1) = 100%, and P(B2) = 50%
  • P(B2|B1) = P(B1 + B2) / P(B1)
  • P(B2|B1) = (1/2) / (1)
  • P(B2|B1) = (1/2) (that is, 50%)
  QED

 

[ryan_618]

I would, but I've had too much coffee this morning, and I don't think it's safe to drink and derive. :roll:

Engineer humor is a sick, sick thing

 

[rattus]

You guys quit screwing with my post or I will report you to the Chief Moderator!

Now back to business:

To find i, we integrate,

di/i = -dt/RC

ln(i) = -t/RC + K

e^ln(i) = i = e^(-t/RC + K) = (e^K)e^-t/RC = (V/R)e^-t/RC

See?

 

[steve66]

Sorry Rattus, didn't mean to hijack your thread, but I though for sure your would take a stab at a statistics problem.

Charlie, your proof is for a completely different question. When I said one child was a boy, I could be refering to B1 or to B2. But you don't know which child I am refering to, so you can't assign the status of "boy" to either child.

If you recall, the question was:

"A woman has 2 children. One is a boy. What are the odds that her other child is a boy?"

You answered this question (or one similar):

"A woman has 2 children. The youngest is a boy. What are the odds that her other child is a boy?"

It's my favorite example of how misleading statistics can be. A very small change in wording gives a different answer.

I'll let Rattus have his tread back now

 

[JohnE]

Rattus:


A woman has 2 children. One is a boy. What are the odds that her other child is a boy?

Steve

I think Charlie nailed it (and then some) with his first answer.

 

[charlie b]

I'll let Rattus have his tread back now

I won't. :wink:

Steve, you state that one child is a boy. I agree that that particular person could be the younger or the older of the woman's children, and that the problem must be solved without regard to birth order. But then your question addresses the probability that the other child is a boy. The only way you can have the other child being a boy, given that one child (i.e., the one for whom probabilities are not being calculated) is a boy, is if both are boys. The two events, (girl then boy) and (boy then girl) are not in the equation.

Yes, statistics can be confusing.

 

[charlie b]

See?

I hope that it has not escaped your attention that the vast majority of the members of this Forum will not have taken a calculus class. It is unlikely that they would "see."

 

[rattus]

Of course Charlie B., that is why I addressed this question to the mathematicians (engineers). I realize also that few engineers, especially in the power industry, have occasion to deal with transient problems. I had hoped though that someone would bother their brain a little and solve the problem.

 

[ramdiesel3500]

What bothers my brain is the fact that my daughter is capable of solving this and I am not!!! :shock: