Can anyone help with a Math Binomial Probability problem?

[RyanLeaf]

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

[Dex]

The answer is 42.

[RyanLeaf]

The answer is 42.



...and thats wrong

[Dex]

Welp, I tried.

[Summers]

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.

[Dex]

Summers is a genius. :wow

[Summers]

The answer is 42.

Ooh, close. :lol

[RandomGuy]

Summers is a genius. :wow

Tell me about it. :lol

[RandomGuy]

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.

By the way:

:wow

[Summers]

Summers is a genius. :wow

Hardly. I'm taking a genetics class in which I'm forced to learn crap like this. :lol

[RandomGuy]

Hardly. I'm taking a genetics class in which I'm forced to learn crap like this. :lol

(mock seriousness)

Probability theory is not "crap", madam. Harumph.

[RyanLeaf]

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.


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

[Summers]

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)+etc...

Yes, you're just using decimals instead of fractions, which is fine.

[RyanLeaf]

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?

[Summers]

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.

[RyanLeaf]

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?

[Summers]

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.