Simulating Coin Flips vs Probability of Coin Flips

$ \begingroup $ “ Or can such a question only be solved using pretense methods ? ( e.g. program a calculator to simulate many such “ runs ” and answer the above questions through simulation ) ” – Yes most decidedly – this is what experimental probability is, and as your total of simulations $ \to \infty $, you will arrive at your theoretical probabilities … which being …
You have already found that the specific probability of the permutation { H, T, H } $ = 0.045 $, with Pr ( H ) $ = 0.95 $ and Pr ( T ) $ = 0.05 $ .
This means every race to occur, there is a $ 0.045 $ opportunity of this, demand, permutation occur, since your runs are independent of each other.

“ demand convention ” – not good a rule, there is an entire distribution set out to help you, The Binomial Distribution .
Lets consider, for a moment, of your test as a whole, rather of its parts ( flips ). So a particular test of yours has $ 0.045 $ gamble of occuring. Since this is the event you want, this will be your probability of success ( Pr ( S ) $ = 0.045 $ ). This besides means your probability of not achieving any achiever or failure, Pr ( F ) $ = 0.955 $ .
The binomial distribution consists of three variables, $ normality $, your number of trials, $ p $, your probability of success ( we just talked about ), and $ X $, number of successes out of your trials you want to achieve .
In your case $ north = 100 $, $ phosphorus = 0.045 $, $ X = 3 $, ( lets good say we want to have precisely $ 3 $ of your permutation { H, T, H } occurring in a 100 trials ). so,

$ Pr ( X = 3 ) = { n\choose { ten } } ( phosphorus ) ^x ( 1-p ) ^ { n – x } = { 100\choose { 3 } } ( 0.045 ) ^3 ( 0.955 ) ^ { 100 – 3 } $
This works for any successes until, $ normality $, ( coz you can not have more successes than the number of trials possible ) .

  1. “The maximum number of “runs” before observing HEADS, TAILS, HEADS.”

I do n’t think there is a very maximum possible ( there is no limit to $ nitrogen $ ). It might equitable be $ \infty $. however, your second doubt is more feasible …

  1. “The minimum number of “runs” before observing HEADS, TAILS, HEADS.”

If you want an absolute minimum, then its $ 1 $ run, but your probabilities are probable not that high.

  1. “The average number of “runs” before observing HEADS, TAILS, HEADS”

If you know the binomial distribution, then it might equitable be $ neptunium $, $ 100 * 0.045 = 4.5 $ .
obviously this is not the case. We want the ask number of trials for at least a $ X = 1 $ to occur. We besides want to reach a maxmimum possible probability of $ X = 1 $, on $ nitrogen $ trials ( hence the average ). This means we have to adjust the number of trials, $ north $, such that $ X = 1 $ reaches a flower probability potential .
$ Pr ( X_n = 1 ) = { n\choose { 1 } } ( 0.045 ) ^1 ( 0.955 ) ^ { n – 1 } $ is maximum. Let this be adequate to $ f ( north ) $. We can find the maximal by solving $ \frac { d ( degree fahrenheit ( nitrogen ) ) } { dn } = 0 $ for $ n $, which yield, $ normality = 21.7184 $. however, since $ n \in \mathbb { N } $, $ north = 22 $. This means doing $ 22 $ runs will allow you to achieve a maximum of opportunity of atleast $ 1 $ { H, T, H } occurring. Thats your “ average ” .

reference : https://tuvi365.net
Category : QUESTION COIN

Related Posts

Leave a Reply

Your email address will not be published.