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Rabi oscillations

Image : “Mplwp Rabi oscillations” by
Geek3. License: CC BY-SA 3.0


In case k, possible outcomes are likely to occur as a result of an event, each event will have a probability of 1/k. In case an event is denoted by “A,” the probability of any event can be measured by the following formula:

P (A)= Count of outcomes in A / Count of outcomes in S
P (A)= Count of outcomes in A / S

In the above formula, “S” is sample space.

Bernoulli Trials

This is a situation which refers to two possible outcomes or results of a situation i.e. success and failure. The probability of success outcome in the Bernoulli trial is denoted by ‘p.’ The success outcome in this probability model is termed as free throw and failure as a miss. The model states that each outcome is independent and does not impact other outcomes or results as one free throw or miss is unlikely to impact the probability associated with the next event.

Example: Suppose a coin is flipped again and again. Every time a single action of flipping a coin has occurred, it gives rise to equal results of “success” or “failure” that head will appear. The factors which will be involved in determining the probability whether head or tail will be shown on each flip comprise of:

n = Number of trials

p = Probability of success in one trial

q = 1 – p = Probability of failure in one trial

k = Number of successes

n – k = Number of failures

In case it is assumed that Bernoulli trials are independent, it relies on the 10% rule.

The graph of Bernoulli trials will be shown as follows:

Bernoulli trial progression

Image: “Graphs of probability P of not observing independent events each of probability p after n trials vs n p for various p. The graphs are actually discrete points as n are integers but drawn as continuous for clarity. 3 examples are shown: Blue arrow: Throwing a 6-sided die 6 times gives a 33.5% chance that 6 never turns up; it can be observed that as n increases, the probability that a 1/n-chance event never appears after n tries rapidly converges to 1/e. Grey arrow: To get a 50-50 chance of throwing a Yahtzee (5 cubic dice all showing the same number) requires 0.69 × 1296 ~ 898 throws. Green arrow: Drawing a card from a deck of playing cards without jokers 100 (1.92 × 52) times with replacement gives an 85.7% chance of drawing the ace of spades at least once.” by Cmglee – Own work. License: CC BY-SA 3.0

The 10% Rule

The 10% rule in Bernoulli trials is interlined with the independence of outcomes from the probability of the happening of one event. Free throw means that the event is independent and does not impact the probability of the next flipping of the coin.

Each time the coin is flipped, the expected outcome will be independent. In case it is not, the flipping activity can’t be continued on sample more than 10% of total population. The 10% rule states that Bernoulli trials should be independent for probability of each outcome. In case the trials are dependent on other outcomes, it should not be proceeded for more than 10% of total population. This condition is applicable when the population is finite.

Milgram experiment example

Stanley Milgram, a psychologist at Yale University, experimented probability models in real life. He, the experimenter (E) ordered the teacher (T) who is also the subject of the tested experiment to give an electric shock to a student (S) if he answers the question incorrectly each time.

Electric shocks used in the experiment were fake and the learner had to act as if he is getting an actual electric shock. Milgram experimented with it to check whether people obey authority if something contradicting with their personal conscience is required or not. He tested all probability models to find out the results i.e. success or failure of obedience of commanded order.

As only two outcomes were expected, the trial was called Bernoulli trials. Milgram called each person a single trial. Success is denoted by the teacher obeying the order to give the shock and failure for refusal of order. Applying the Bernoulli trial model, it has been estimated that a 65% success rate was achieved. The probability for a success rate came out P = 0.65.

The geometric model: waiting for success

This model shows the probability of success and only success in a trial or event. In the case of coin flipping, the geometric model only counts head occurrence (Success). The geometric distribution starts when success happens in a trial. It is applicable when a number of trials are multiple. The conditions for the geometric model involve independence and identical distribution of trials.

Geometric probabilities

If n represents a number of independent trials, p indicates the probability of success and (1 − p) depicts the probability of failure, then:

P (success on the nth trial) = (1 − p) n−1 p

The variables involved in geometric model;

P = Probability of success

X = Random variable related to success

If X=, then all attempts before the success will be denoted as .  So

P(X =) = (1-p) ƙ -1 P, for ƙ = 1, 2. . . . .

E[X] = 1/p

SD(X) = √1-p/p2

If X= Free number of throws, then the probability p = 0.60 using the above mentioned equation.

Milgram experiment example

The Milgram experiment provided the probability of success when firstly, the teacher provided consent for giving a shock to a learner upon a wrong answer to the asked question. When the first Teacher accepted the order, the trial resulted in first success resulting in:

The first teacher accepts

P (1st Teacher accepts) = 0.65


The third teacher accepts

The probability for success continues and provides a probability of success in geometric series. I.e.

P (1st refusal and 2nd refusal, 3rd acceptance) = 0.35 × 0.35 × 0.65

= 0.352 × 0.65 ≈ 0.08


The tenth teacher accepts

P (9 refusal, 10th acceptance) = 0.35 × · · · × S 0.35 × 0.65

9 Times

= 0.359 × 0.65

= 0.000512

The binomial model: counting successes

It is a probability model concerning discrete outcome referring to counting success or desired outcomes. In this model, the number of success or “Y” are counted finding out how many times head will appear on flipping a coin for example. Suppose if we flip the coin 10 times, it will help us find out what is the probability related to 9 times success or head appearance. Y, in this case, is the random variable referring to free throw or success.

Counting success

In case number of success or free throw have to be counted when n number of Bernoulli trials which have “p” probability of occurrence of success “Y.” The determining factor is how many times “Y” factor or success is probable.

Example: In the same case of filliping a coin three times, the probability of 2 heads in three flips will have the following pattern; (Heads, Heads, Tails), (Heads, Tails, Heads), (Tails, Heads, Heads).


It relates to counting success in Binomial model without writing out all possibilities includes miss or failures. It refers to a situation where n number of trials can give rise to results in n number of trials or attempts.

What is (n y) Analogy?

(n y) = n! / y! (n-y) in this equation the exclamation mark refers “Factorial”.

n! = n (n-1) (n-2). . . . . (2) (1).

The above-mentioned equation states that all integers have to be multiplied from n down to 1.

With the help of the given equation, we can find out the probabilities of success or head in coin flipping. The trials have to be independent if we require a probability of py (1 – p) n – y

The above-mentioned equation states that all integers have to be multiplied from n down to 1.

 (n y) ways of n number of expected success gives expected probability of head in this way:

P(Y = y) = (n y)py(1 – p)n-y, for y = 1, 2, …, n

Using binomial model, the probability can be measured in this equation:

E[Y] = np, SD(Y) = √np (1 – p)

Normal approximation to Binomial Model

In case the number of species in a population increases or the sample size gets large. In case the number of trials get too large that binomial model is unable to handle the related probabilities of success, approximate binomial probabilities can be used in this situation for desired results. Suppose 500 times a coin has to be flipped and you want the probability of head more than 200 times, you can use Y as a binomial random variable with 500 trials having 0.6 probability of success. The equation will be as follows:

P(Y ≥ 400) = P(Y = 400) + P(Y = 401) + … + P(Y =500)

Now to find out the 101 probabilities by using the binomial model formula, it may take too long. The following formula can be used known as normal approximation taking Y as binomial random variable with n number of trials and p as probability of success (occurrence of head); E[Y] = np, SD(Y) =. Standard deviation and mean can be used in normal approximation of binomial model. This model is useful in case success and failure only two outcomes are expected.

The condition for the application of a normal model for binomial probability is success/failure condition. In case the success/failure condition is duly satisfied, then mean and standard deviation can be calculated as “np”. In the above example where we have flipped the coin for 500 times and want to get head occurrence for 400 times, we want a probability of P(Y ≥ 400). Upon calculation through binomial formula, we may get a probability of 0.0022.

Issue with the Normal Approximation

Normal approximation in the Binomial model can help in finding out the desired Y probability of success, but it has a big issue. It can’t be used to find out a particulate number of success or head occurrences. It can’t help out to find out the exact 400 head occurrence or free throws in case of coin example.

As random variable “Y” is involved in a normal approximation of binomial model, it gives an equal chance to any value in a range to be chosen; hence, it is not possible to exactly find out the possible outcomes of success.

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