Need some help with Statistics/Probability homework.

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February 16th 2011, 05:53 PM
pantsaregood

Need some help with Statistics/Probability homework.

This is some really basic stuff, I'm sure. It's for a Calculus based probability course. I can't really see what's going on in class (I'm quite literally legally blind), and the instructor doesn't really have any online resources available.

The site (or my internet connection) is being goofy and won't let me upload the file. I'll have to link to it.

ImageShack® - Online Photo and Video Hosting

I've drawn a probability tree and concluded that there is a 10% chance of a fair coin landing on heads three consecutive times for number 1.

After that problem, I'm clueless of where to go. I'm not even sure I'm right on it. Any resources for this type of problem would be greatly appreciated. Thank you.
February 16th 2011, 06:16 PM
CaptainBlack

Quote:

Originally Posted by pantsaregood

This is some really basic stuff, I'm sure. It's for a Calculus based probability course. I can't really see what's going on in class (I'm quite literally legally blind), and the instructor doesn't really have any online resources available.

The site (or my internet connection) is being goofy and won't let me upload the file. I'll have to link to it.

ImageShack® - Online Photo and Video Hosting

I've drawn a probability tree and concluded that there is a 10% chance of a fair coin landing on heads three consecutive times for number 1.

Since with probability $1/5$ we have a two headed coin, and with probability $4/5$ we have a fair coin the required probability is:

$(1/5) + (4/5)(1/2)^3$

since the probability of three heads with the two headed coin is $$$1$ , and with the fair coin is $(1/2)^3$

CB
February 16th 2011, 06:27 PM
pantsaregood

4/5 = 80%
1/2 = 50%

20% isn't relevant in the problem, since it asks what the probability of a fair coin landing on heads consecutively is.

Anyway, what of the other problems?
February 16th 2011, 06:53 PM
CaptainBlack

Quote:

Originally Posted by pantsaregood

4/5 = 80%
1/2 = 50%

20% isn't relevant in the problem, since it asks what the probability of a fair coin landing on heads consecutively is.

Anyway, what of the other problems?

That makes no sense, and no the other problems can wait until the first is resolved.

CB
February 16th 2011, 07:38 PM
pantsaregood

I'm failing to see how it doens't make sense. The two-headed coin is not relevant in this problem. When the probability of the two-headed coin landing on heads three times consecutively is removed, you're left with (4/5)(1/2)^3, which is exactly what I said.

The two-headed coin is not a fair coin. Its probability of 20% is not relevant because it does not satisfy the condition of being a fair coin.

Edit: I had the wrong idea. In this case, the probability it is a fair coin is 1/3, as opposed to the 2/3 chance of a two-headed coin.