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Subject: "A mistake in Bear Cubs Problem?"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #717
Reading Topic #717
Olivier
guest
Jul-01-09, 04:54 PM (EST)
 
"A mistake in Bear Cubs Problem?"
 
   Dear sir,

As a former math researcher turned teacher, I enjoy your website tremendously.

I may have misunderstood the statement, but I think there may be a mistake in the section Probability, Principle of Proportionality, Bear Cubs Problem.

The problem:
There are two bears, white and dark, one of which is known to be male. What is the probability that both are male?

You state that the solution is 1/2 but I believe the solution is 1/3 (1/2 is the probability that both are male if, say, we know that the white one is male, but we have less information).

Solution 1:
Since they are distinguishable, the sample space consists of MM (for male/male), MF (for male/female), FM, and FF. If one of them is known to be male, then the sample space consists only of MM, MF, FF. Thus P(MM knowing at least one male) = 1/3.

Solution 2: Let B be the event at least one male. Then P(B) = 3/4. Let A be the event both male, then P(A) = 1/4.
Thus P(A knowing B) = P(A and B)/P(B) = P(A) / P(B) = 1/4 / 3/4 = 1/3.

Solution 3: Using the Principle of Proportionality. Following your notation, we let A_1 = FF, A_2 = FM, A_3 = MF, and A_4 = MM. Then P(B knowing A_1) = 0 and P(B knowing A_4) = 1.
But, contrary to what you write, P(B knowing A_2) = P(B knowing A_3) = 1 also.
Thus, the sums of these probability is 3, so by the Principle of Proportionality,
P(A_1 knowing B) = 0, P(A_2 knowing B) = P(A_3 knowing B) = P(A_3 knowing B) = 1/3.

Best regards, Olivier


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alexbadmin
Charter Member
2397 posts
Jul-01-09, 06:59 PM (EST)
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2. "RE: A mistake in Bear Cubs Problem?"
In response to message #0
 
   First, thank you for the kind words. But let's work out out differences.

Looks very much like there is a mistake, indeed. Let's see what I had in mind when writing that page. I'll relate to your third solution. The question is whether

P(B | A_2) = P(B | A_3) = 1 or
P(B | A_2) = P(B | A_3) = 1/2.

The difference, I believe, may be attributed to a bad wording of the definition of event B.

You define "Let B be the event at least one male".
I used in one place "Now assume I told you that one of the bears is male", "Assume it is known that one of the bears is male" in another, and "Event B is the acknowledgement that one of the bears is male" in the third. (The last two on the Proportionality Principle page.) I shall certainly have to have to work this out.

However, let's consider several possibilities that might define event B.


  1. I've been watching the bear enclosure in the local zoo and saw a bear emerge from a cave that happened to be male. In this case, I believe, you would agree that if B is the awareness of the presence of a male bear, then P(B | A_2) = P(B | A_3) = 1/2.
  2. Assume it was not me who watched the bears but a small girl visiting the zoo. I overheard her exclaim, "Look mum a male bear." I believe that assuming the girl was correct, we still have P(B | A_2) = P(B | A_3) = 1/2.
  3. Now, as it happened, it was my wife and daughter who went to the zoo, while I stayed home to work on the site. When they returned from the zoo my daughter excitedly informed me of having seen a bear, "Daddy there was a male bear in the zoo." I believe that even in this case P(B | A_2) = P(B | A_3) = 1/2.

So, as I remember, in my mind, being told that there is a male bear played an essential role in defining the meaning of event B. It endowed event B with less certainty than emplied by your "Let B be the event at least one male".

Does this make sense?


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