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The answer is 42.
Ok I have no clue how to do this. I have been getting along fine with my hw and finishing up every problem until this:
The degree to which democratic and nondemocratic countries attempt to control the new media was examined. Between 1948 and 1996, 80% of all democratic regimes allowed a free press. In contrast, over the same period, 10% of all nondemocratic regimes allowed a free press.
A random sample of twelve democratic regimes is selected.
i) find the probabilty that all twelve of them allow a free press.
ii) Find the probabilty that fewer than ten of them allow a free press
The answer is 42.
...and thats wrong
Welp, I tried.
i) (8/10)^12
ii) don't know how involved this answer is supposed to be but you'd have to ADD 9 equations together:
prob that one of them allows free press: 12*(8/10)*(2/10)^11
PLUS
probably that 2 of them allow free press: 12!/(2!10!)*(8/10)^2*(2/10)^10
PLUS
prob that 3 of them allow free press: 12!/(3!9!)*(8/10)^3*(2/10)^9
etc until prob that 9 allow free press.
Hope that helps.
Summers is a genius.
Ooh, close.
Tell me about it.Summers is a genius.
By the way:
Hardly. I'm taking a genetics class in which I'm forced to learn crap like this.Summers is a genius.
(mock seriousness)Hardly. I'm taking a genetics class in which I'm forced to learn crap like this.
Probability theory is not "crap", madam. Harumph.
Thank you very much. For the first one I figured out to do P(DDDDDDDDDDDD) = (.8)^12 = .0687
For the second one I basically do this?
12(.8)(.2)^11
+
12!/(2!10!)(.8)^2(.2)^10
+
12!/(3!9!)(.8)^3(.2)^9
+
12!/(4!8!)(.8)^4(.2)^8
+
12!/(5!7!)(.8)^5(.2)^7
Yes, you're just using decimals instead of fractions, which is fine.
But for the second one, How Do I get the final answer out of all that?
12(.8)(.2)^11
+
12!/(2!10!)(.8)^2(.2)^10
+
12!/(3!9!)(.8)^3(.2)^9
+
12!/(4!8!)(.8)^4(.2)^8
+
12!/(5!7!)(.8)^5(.2)^7
and go all the way to 1?
This is the addition of separate probabilities for 1 country, then 2 countries, etc. The question was what is the probability that fewer than 10 blah blah... So you continue this pattern until you have 9 equations, the last of which will be 12!/(9!3!)*(.8)^9*(.2)^3. Then add your 9 equations for your final answer.
Oh ok. Sorry I have one more question. I add them by doing what?
For example. (.8)^9(.2)^3 + (.8)^8(.2)^4 + (.8)^7(.2)^5 + etc all the way to the twelfth equation? Or is that wrong?
NO, 9 equations, not 12. Then add them together.
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