Problems with Bernoulli's formula

Probability, grade 11 (5 units) Lesson 14 of 16Learning

Problems with Bernoulli's formula

At least, at most

“At least $\displaystyle k$ successes” and “at most $\displaystyle k$ successes” are events of several numbers of successes. The events “exactly 0”, “exactly 1”, … are disjoint, so the probability is a sum of several terms of Bernoulli’s formula:

at least $\displaystyle k$ – from $\displaystyle k$ successes to $\displaystyle n$ ;
at most $\displaystyle k$ – from 0 successes to $\displaystyle k$ .

A medicine helps 80% of the patients who take it.
5 patients take the medicine, independently of each other.

The medicine helps at least 4 of them – 4 or 5: $P_5(4)+P_5(5)=\binom{5}{4}\cdot0.8^4\cdot0.2+0.8^5=0.4096+0.32768=0.73728$
The medicine helps at most one of them – 0 or 1: $P_5(0)+P_5(1)=0.2^5+\binom{5}{1}\cdot0.8\cdot0.2^4=0.00032+0.0064=0.00672$

At least one success

“At least one success” is the complement of “no success”: instead of adding $\displaystyle n$ terms, one term is taken away from 1.

\displaystyle 1-P_n(0)=1-(1-p)^n

A raffle ticket wins with probability $\displaystyle 0.2$ .
Danny buys 5 tickets in different raffles.
The probability that at least one ticket wins:

1-0.8^5=1-0.32768=0.67232

The smallest number of trials

The more trials there are, the smaller $\displaystyle (1-p)^n$ is, and the larger the probability of at least one success. So the smallest number of trials at which that probability passes a given bound can be found – by trial and error.

The example below finds how many tickets Danny must buy for the probability that at least one of them wins to be greater than $\displaystyle 0.9$ .

$\displaystyle 1-0.8^n>0.9$ is needed, that is $\displaystyle 0.8^n<0.1$ . Try numbers:

$\displaystyle n$	$\displaystyle 0.8^n$	$\displaystyle 1-0.8^n$
10	$\displaystyle \approx0.107$	$\displaystyle \approx0.893$
11	$\displaystyle \approx0.086$	$\displaystyle \approx0.914$

With 10 tickets the probability is still less than $\displaystyle 0.9$ , and with 11 it is already greater. Danny must buy at least 11 tickets.

The most likely number of successes

To find the number of successes with the greatest probability, work out $\displaystyle P_n(k)$ for every $\displaystyle k$ and compare. The probabilities rise up to a certain number, and fall after it.

The medicine of the example, 5 patients:

$\displaystyle k$	0	1	2	3	4	5
$\displaystyle P_5(k)$	$\displaystyle 0.00032$	$\displaystyle 0.0064$	$\displaystyle 0.0512$	$\displaystyle 0.2048$	$\displaystyle 0.4096$	$\displaystyle 0.32768$

It is most likely that the medicine helps 4 of the 5 patients – which is also $\displaystyle {5\cdot0.8}={4}$ , 80% of the patients.

Bernoulli’s formula as a stage of a tree

A repeated trial can be one stage of a tree: in the tree below the second stage is all three tosses together, and the probabilities on its branches were worked out by Bernoulli’s formula.

A box holds two coins: a fair coin, and a biased coin that comes up heads with probability $\displaystyle 0.8$ .
A coin is chosen at random and tossed 3 times.
$\displaystyle T$ – heads come up exactly twice.

With the fair coin: $\displaystyle {P_3(2)}={\binom{3}{2}\cdot0.5^2\cdot0.5}={0.375}$ .
With the biased coin: $\displaystyle {P_3(2)}={\binom{3}{2}\cdot0.8^2\cdot0.2}={0.384}$ .

By the law of total probability:

P(T)=0.5\cdot0.375+0.5\cdot0.384=0.1875+0.192=0.3795

And the reverse question – heads are known to have come up exactly twice; what is the probability that the coin is the biased one?

\frac{0.192}{0.3795}\approx0.506

Exercise 1

An archer hits the target with probability $\displaystyle 0.6$ with each arrow, whatever happens with the other arrows.
She shoots 4 arrows.

What is the probability that she hits at least 3 times?

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