Theory

Theory 1

Repeated trials

When a single experiment type is repeated many times, and we assume each instance is independent of the others, we say it is a sequence of repeated trials or independent trials.

The probability of any sequence of outcomes is derived using independence together with the probabilities of outcomes of each trial.

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A simple type of trial, called a Bernoulli trial, has two possible outcomes, 1 and 0, or success and failure, or $T$ and $F$. A sequence of repeated Bernoulli trials is called a Bernoulli process.

• Write sequences like $TFFTTF$ for the outcomes of repeated trials of this type.
• Independence implies

$$P[TFFTTF]=P[T]\cdot P[F]\cdot P[F]\cdot P[T]\cdot P[T]\cdot P[F]$$

• Write $p=P[T]$ and $q=P[F]$, and because these are all outcomes (exclusive and exhaustive), we have $q=1-p$. Then:

$$P[TFFTTF]\gg \gg pqqppq\gg \gg p^3{q}^3$$

• This gives a formula for the probability of any sequence of these trials.

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A more complex trial may have three outcomes, $A$, $B$, and $C$.

• Write sequences like $ABBACABCA$ for the outcomes.
• Label $p=P[A]$ and $q=P[B]$ and $r=P[C]$. We must have $p+q+r=1$.
• Independence implies

$$P[ABBACABCA]\gg \gg pqqprpqrp\gg \gg p^4{q}^3{r}^2$$

• This gives a formula for the probability of any sequence of these trials.

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Let $S$ stand for the sum of successes in some Bernoulli process. So, for example, “$S=3$” stands for the event that the number of successes is exactly 3. The probabilities of $S$ events follow a binomial distribution.

Suppose a coin is biased with $P[H]=20\%$, and $H$ is ‘success’. Flip the coin 20 times. Then:

$$P[S=3]\gg \gg \left(\genfrac{}{}{0px}{}{20}{3}\right)\cdot {(0.2)}^3\cdot {(0.8)}^{17}$$

Each outcome with exactly 3 heads and 17 tails has probability ${(0.2)}^3\cdot {(0.8)}^{17}$. The number of such outcomes is the number of ways to choose 3 of the flips to be heads out of the 20 total flips.

The probability of at least 18 heads would then be:

$$\begin{array}{c}P[S\ge 18]\gg \gg P[S=18]+P[S=19]+P[S=20]\\ \\ \gg \gg \left(\genfrac{}{}{0px}{}{20}{18}\right)\cdot {(0.2)}^{18}\cdot {(0.8)}^2+\left(\genfrac{}{}{0px}{}{20}{19}\right)\cdot {(0.2)}^{19}\cdot {(0.8)}^1+\left(\genfrac{}{}{0px}{}{20}{20}\right)\cdot {(0.2)}^{20}\cdot {(0.8)}^0\end{array}$$

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With three possible outcomes, $A$, $B$, and $C$, we can write sum variables like $S_A$ which counts the number of $A$ outcomes, and $S_B$ and $S_C$ similarly. The probabilities of events like $\text{“}(S_A,S_B,S_C)=(2,3,5)\text{”}$ follow a multinomial distribution.