Multiple Percentages Probability

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SUMMARY

The discussion focuses on calculating probabilities using the binomial distribution, specifically for a scenario where a program outputs "yes" with a 60% probability and "no" with a 40% probability. The correct method to determine the odds of getting exactly 6 "yes" responses out of 7 attempts is through the formula (7 choose 6) * 0.6^6 * 0.4^1, which accounts for the number of ways to choose the outcomes. The total probability for this scenario is calculated as 7 * 0.6^6 * 0.4^1. Additionally, the discussion touches on calculating the odds of getting all "yes" or all "no" responses in 7 attempts.

PREREQUISITES
  • Understanding of binomial distribution
  • Familiarity with probability concepts
  • Knowledge of combinatorial mathematics (n choose k)
  • Basic algebra for manipulating exponents
NEXT STEPS
  • Study the binomial distribution in-depth
  • Learn how to calculate combinations using the formula (n choose k)
  • Explore applications of probability in real-world scenarios
  • Investigate variations of probability distributions, such as the Poisson distribution
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Students, statisticians, data analysts, and anyone interested in understanding probability calculations and their applications in various fields.

PharaohsVizier
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Hi, I seem to be having problems calculating this out. My friends were asking me how to caculate multiple precentages and I thought it would be easy but I got a little stuck. Here is the problem.

Lets say there is a program that spits out the words yes and no. 60% chance it says yes and 40% chance it says no. If I hit it once, there is a 60% chance it says yes and a 40% chance it says no. If I click it 7 times and it says yes 6 times, what are the odds? I put .6^6 to calculate it, but it seems that I don't include the fact that it says no once. Also, what would the odds be if it said yes all 7 times or no all 7 times? How would you calculate these percentages?

Thanks a lot, seems like a great forum so far.
 
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You're looking for what's called the binomial distribution.

The odds of exactly 6 out 7 "yes" is (7 choose 6) * 0.6^6 * 0.4 ^ 1. Let me explain. You are getting 6 "yes" and 1 "no"; the chances of getting those answers []in that order[/i] is 0.6^6 * 0.4 ^ 1. Since you don't care about the order, you need to multiply this by the number of ways to choose 6 elements out of 7. In general, (x choose y) is

\frac{x!}{y!(x-y)!}

For (7 choose 6), that's 7!/(6! * 1!) = 7, giving a total probability of 7\cdot0.6^6\cdot0.4^1.
 
Thanks so much
I can't believe someone actually solved this for me in such a clear manner.
This forum is great!
 
PharaohsVizier said:
Thanks so much
I can't believe someone actually solved this for me in such a clear manner.
This forum is great!

I'm glad to have helped. o:)
 

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