If you have 6 chicks that each have a 1 in 10 chance of being male, and 2 chicks that each have a 1 in 2 chance of being male, what is the chance that one of your chicks is male and the rest are female?

First person to get this right and post it as a comment wins a free signed photo of Pot Pie.

## Tuesday, April 3, 2007

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## 7 comments:

OMG is this your way of telling us that Pot Pie is not going to be around for much longer?????

Nooooo no no no. Pot Pie still is showing fabulous signs of being a hen. She has flat tipped neck feathers (good), a pink-yellow comb (good), and no real protective instincts on food or perches (good).

I think you're looking at about a 35% probability that one of your birds is male and the rest female. 0.354294 to be precise. Do I get the signed photo????

Wait... that isn't the answer I got. I thought it was 0.9 to the 5th power, times 0.1, times 0.25... plus 0.9 to the 6th power times 0.5... divided by two? That comes to 14% (0.140241525). Where am I going wrong here? It seems sketchy.

Yes, you get the signed photo. Please post your technique for getting your answer.

Let's say that we have all 8 chickens lined up. The first six have a 0.1 probability of being male, the last two have a 0.5 probability. The probability that the first one is male and all the rest female is 0.1 x 0.9^5 x 0.5^2. That's 0.01476225. The probability will be exactly the same that second chicken is male and the rest female, and likewise for chickens 3, 4, 5, and 6. So multiplying 0.01476225 by 6 gives us 0.0885735. The probability that the 7th chicken is male and the rest female is 0.9^6 x 0.5^2 = 0.13286025. Multiplying that by 2 to find the probability of the 8th chicken being male and the rest female gives you 0.2657205. Add that to 0.0885735 and you get 0.354294, or about 35%.

Nice. I will accept this answer, as I believe it to be correct.

You will receive your signed photo in good time. First, I need to teach Pot Pie to hold her pen correctly. :)

For the record, as a test of whether he'd really been studying statistics for the last 2 years, I made Kevin solve this problem for me while we were on a five-hour car trip this past weekend. He came up with the same answer as Dave. (So now I know he actually

hasbeen studying, which is comforting.)Post a Comment