Bernoulli's formula

Probability, grade 11 (5 units) Lesson 13 of 16Learning

Bernoulli's formula

Repeated independent trials

Many experiments repeat the same trial several times, and each time only two outcomes count: a success or a failure. The probability of a success in each trial is $\displaystyle p$ , and the probability of a failure is $\displaystyle 1-p$ .

A shot at the basket: a success – the player scored.
A throw of a die: a success – a 6 came up, $\displaystyle {p}={\frac{1}{6}}$ .
A check of a product: a success – the product is faulty. “Success” is only the name of the outcome being counted.

Noa scores a free throw with probability $\displaystyle 0.7$ , whatever happens on her other throws.
Noa takes 3 throws.
What is the probability that she scores exactly 2 of them?

From a tree of three stages to the formula

In the tree below each stage is a throw. The throws are independent, so every “scored” branch carries $\displaystyle 0.7$ and every “missed” branch $\displaystyle 0.3$ , at every stage. The paths of “exactly 2 scores” are marked:

Each of the three marked paths has two scores and one miss, so they all have the same probability, in any order: $\displaystyle {0.7\cdot0.7\cdot0.3}={0.147}$ .
The number of these paths is the number of ways to choose the two throws she scored out of 3: $\displaystyle {\binom{3}{2}}={3}$ .

The probability of exactly 2 scores is the sum of the three paths:

\binom{3}{2}\cdot0.7^2\cdot0.3=3\cdot0.147=0.441

Bernoulli’s formula

The same works for any number of trials, without drawing the tree. In $\displaystyle n$ independent trials, with the probability of a success $\displaystyle p$ in each:

each path with exactly $\displaystyle k$ successes passes through $\displaystyle k$ success branches and $\displaystyle n-k$ failure branches, so its probability is $\displaystyle p^k(1-p)^{n-k}$ ;
there are $\displaystyle \binom{n}{k}$ such paths – the number of ways to choose the $\displaystyle k$ successful trials.

The probability of exactly $\displaystyle k$ successes is the sum of all these paths – Bernoulli’s formula:

\displaystyle P_n(k)=\binom{n}{k}p^k(1-p)^{n-k}

For Noa: $\displaystyle {n}={3}$ , $\displaystyle {k}={2}$ , $\displaystyle {p}={0.7}$ , and $\displaystyle {P_3(2)}={0.441}$ .

When the formula fits

Bernoulli’s formula fits when four conditions hold:

the number of trials $\displaystyle n$ is known in advance;
each trial has two outcomes: a success or a failure;
the probability of a success $\displaystyle p$ is the same in all the trials;
the trials are independent: the outcome of a trial does not change the probabilities in the others.

A bag holds 3 red balls and 2 blue ones, and a ball is taken out 3 times. A success – a red ball.

With putting back: at each draw $\displaystyle {p}={\frac{3}{5}}$ , and the draws are independent – the formula can be used.
Without putting back: if the first ball is red, 2 red balls out of 4 are left in the bag, and the probability of red on the second draw is $\displaystyle \frac{2}{4}$ , not $\displaystyle \frac{3}{5}$ . The probability changes, and the formula does not fit – work it out with a tree.

A test answered by guessing

The example below finds $\displaystyle n$ , $\displaystyle p$ and $\displaystyle k$ , and puts them into Bernoulli’s formula.

A test has 10 questions, and each question has 4 answers, only one of them right.
Yoav did not revise, and chooses an answer at random in each question.
What is the probability that he answers exactly 3 questions right?

A trial – a question; a success – a right answer, $\displaystyle {p}={\frac{1}{4}}$ .
The questions are independent, $\displaystyle {n}={10}$ , $\displaystyle {k}={3}$ .

P_{10}(3)=\binom{10}{3}\cdot\left(\frac{1}{4}\right)^3\cdot\left(\frac{3}{4}\right)^7=120\cdot\frac{1}{64}\cdot\frac{2187}{16384}=\frac{32805}{131072}\approx0.250

About a quarter of the times such a test is guessed, exactly 3 answers come out right.

Exercise 1

A fair die is thrown 3 times.
A success – a 6 comes up.

Complete the probabilities in the tree, mark the paths in which a 6 comes up exactly once, and work out the probability of that.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

30% of a town’s residents cycle to work.
5 residents are chosen at random, independently of each other.

What is the probability that exactly 2 of them cycle to work?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

A fair coin is tossed 8 times.

What is the probability of getting heads exactly 5 times? Write a fraction in its lowest terms.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

15% of the lamps a factory makes are faulty.
12 lamps are chosen at random.

What is the probability that exactly 2 of them are faulty? Round to three decimal places.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 5

In which of these situations can Bernoulli’s formula be used? Explain for each.

A class has 10 girls and 10 boys. 4 different students are chosen, and the girls among them are counted.
A die is thrown 10 times, and the even numbers that come up are counted.
A player shoots at the basket until she first scores, and the shots are counted.

Formulas: probability, grade 11 (5 units)

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The rules of probability

Equally likely outcomes – $\displaystyle k$ the outcomes in the event $\displaystyle A$ , $\displaystyle n$ the outcomes in the sample space
$\displaystyle P(A)=\frac{k}{n}$
Every event; $\displaystyle {P(\emptyset)}={0}$ , and $\displaystyle {P(U)}={1}$ for the sample space $\displaystyle U$
$\displaystyle 0\le P(A)\le1$
An event contained in another event
$\displaystyle P(A)\le P(B)$ $\displaystyle A\subseteq B$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
The union – $\displaystyle A$ or $\displaystyle B$ (or both); $\displaystyle A\cap B$ – the intersection, both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)-P(A\cap B)$
Disjoint events – no two of them have an outcome in common
$\displaystyle P(A_1\cup A_2\cup\dots\cup A_n)=P(A_1)+P(A_2)+\dots+P(A_n)$

Conditional probability and independence

The probability of $\displaystyle A$ given $\displaystyle B$
$\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}$
A path in a tree – the product of the probabilities on its branches
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(B)\cdot P(A/B)$
Independent events
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(A)\cdot P(B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A/B)$ $\displaystyle =$ $\displaystyle P(A)$
The law of total probability – the paths that end in $\displaystyle A$ , added up
$\displaystyle P(A)=P(B)\cdot P(A/B)+P(\overline{B})\cdot P(A/\overline{B})$

Counting and Bernoulli's formula

$\displaystyle n$ factorial; $\displaystyle {0!}={1}$
$\displaystyle n!=1\cdot2\cdot3\cdot\dots\cdot n$
Ordered selections – $\displaystyle k$ different objects out of $\displaystyle n$ , one after another
$\displaystyle n\cdot(n-1)\cdot\dots\cdot(n-k+1)=\frac{n!}{(n-k)!}$
The binomial coefficient – the ways to choose $\displaystyle k$ objects out of $\displaystyle n$ , order not counted
$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$
Bernoulli’s formula – the probability of $\displaystyle k$ successes in $\displaystyle n$ independent trials, with the probability of success $\displaystyle p$ in each
$\displaystyle P_n(k)=\binom{n}{k}p^k(1-p)^{n-k}$
At least one success in $\displaystyle n$ trials – by the complement
$\displaystyle 1-P_n(0)=1-(1-p)^n$