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I was working on my stat homework and encountered a problem that i have no idea how to do...please help!

The answer is .0744, but i don't know why. Here's the question:

Based on historical data, the probability that an ‘indie’ film will make a profit is 0.7. (Otherwise, naturally, it loses money.) Before releasing a film for distribution, a distributor can do a test screening to determine audience reaction, which is simply classified as being ‘favorable’ or ‘unfavorable.’ For films that have made a profit, the test screening results in a favorable reaction 80% of the time. For films that have lost money, the test screening results in a favorable reaction 15% of the time.
A test screening is done on a new film and the result is favorable. Taking this into account, what is the probability that this film will NOT make a profit?

2006-10-06 09:50:14 · 3 answers · asked by fuilui213 6 in Education & Reference Homework Help

3 answers

Ok, basically you have 4 combinations:
1.) Test screen is favourable & Make profit
2.) Test screen is favourable & Lose money
3.) Test screen is unfavourable & Make profit
4.) Test screen is unfavourable & Lost money

You already knew that:
- 70% of the films will make a profit
- For a profit made film, 80% of the time that it received a favourable test screen. So, the probability of 1.) is 0.56 (0.8 * 0.7)
- So, 0.14 (0.7 - .56) is the probably for a film that Test screen is favourable & Lose money
- For a money losing film, 15% of the time that it received a favourable test screen. SO, the probability of 3.) is 0.045 (0.15 * 0.30)
- So, 0.255 (0.3 - 0.045) is the probably for a film that Test screen is unfavourable & Lose money

So, back to your question. Given that a test screen result is favourable, what is the probability that this film will NOT make a profit.

P(Not Make a profit | test screen unfavourable)
=P(Not Make a profit & test screen unfavourable)/P(test screen unfavourable)
=0.045/(0.56+0.045)
=0.07438

Hope this helps

2006-10-06 10:17:32 · answer #1 · answered by jclcheng777 2 · 0 1

This appears to require the use of Bayes Theorem (conditional probability)

2006-10-06 17:02:31 · answer #2 · answered by Bill N 3 · 0 1

0.3*0.15
--------------------------
0.7*0.8 + 0.3*0.15

2006-10-06 17:06:10 · answer #3 · answered by buccinator 3 · 0 1

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