How many attempts in one experiment

W.14 | Standard experiments

Actually, a majority of the tasks in probability calculus are always calculated using the same standard tasks: dice, wheels of fortune, urns (from which colored balls are taken either with or without replacing). There are also various Bernoulli experiments, i.e. experiments in which there are only two possible starting points and in which the probability remains the same.

 

 

At this point, under “standard experiments”, we summarize a few types of tasks that are fundamentally simple and always repeat
appear in various questions. There are two terms that come up again and again in this context, which of course I do not want to withhold from you.

A Laplace event is when all possible outcomes are equally probable. [e.g. a normal coin with 50% probability each, a normal dice with
1/6 probability each or ...] If you turn the coin / die so that an area / side is more or less frequent, you don't have any
Laplace event more.

A Laplace cube (or an ideal cube, as it is then called) is accordingly a cube in which all sides have the probability of
get exactly 1/6 [at least if it's a six-sided die]. So if you want to express it “non-mathematically”, you could say: A
Laplace cube is a completely normal cube.

A Bernoulli experiment (attention, not to be confused with "Bernoulli distribution"! The latter is correctly called "binomial distribution".) Identifies every experiment which:
- has two exit options and
- an always constant probability. has [i.e. pulling WITH replacement]
Instead of “Bernoulli experiment” one can also say “Bernoulli chain”.

 

W.14.01 | Bernoulli experiment

 

Example a.

A coin is flipped eight times.

a) With which W.S. only head falls?
b) With which W.S. number falls at least once?
c) With which W.S. is the difference between the number of heads and number four?

 

solution[Note: This task falls under both Bernoulli and Laplace experiments].

Unless otherwise stated in the exercise, both the W.S. for head, as well as the W.S. each be 0.5.

a) Eight heads means: P (KKKKKKKK) = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.58

b) At least one number contains many cases:

That could be once, twice, three times ... up to eight times. Until you have calculated all these cases, you get old and gray.
So we prefer to calculate it differently. We think about the counter-event, i.e. all cases that are not allowed to happen. The task says: “at least once”. So what would be bad? Answer: not even a number! Therefore we calculate the W.S. that no number appears and subtract the result from 100%.

P (not even a number) = P (always head) = 0.58 [see subtask a)]

⇒ P (at least one number) = 1–0.58 ≈ 0.996

c) A difference of four times the number of heads and tails means that there are only two possibilities: six heads and two tails or two heads and six tails.

P (6K, 2Z). Since it is not specified in which order "K" and "Z" should appear, there are many possible interchanges. A few of them would be:
KKKKKKZZ, KKKZKZKK, ZKKKKKZK, ... Since each of these combinations is allowed, we would have to use the W.S. calculate all of them and then add them all up, which would be a bit time-consuming. This is why the binomial coefficient comes into play at this point. The total number of throws is 8, we want 6 times K, so there is = 28 possibilities to swap 6K and 2Z with each other.
The W.S. for 6 heads is 0.56, the W.S. for 2 times number is 0.52.

P (2K, 6Z). Actually, almost the same case as P (6K, 2K).

⇒ P (difference between heads and tails is four) = 0.1094 + 0.1094 = 0.2188
professional notation: P (| K – Z | = 4) = 0.1094 + 0.1094 = 0.2188

 

Example b.

Experience has shown that every fourth car in a car race does not finish due to a defect. In another race, 12 cars will start.

a) With which W.S. do all cars reach the finish line?
b) Is there a real chance that not a single car will reach the destination?
c) With which W.S. does Heiko [one of the racing drivers] reach the goal?
d) With which W.S. do more than 10 cars reach the finish line?
e) With which W.S. does the winner reach the finish line?

 

Solution:
If every fourth car does not reach the finish line, the W.S. for failing before the goal

a)

professional notation: (First we define "X" as the number of cars that will cross the finish line.)
Let "X" be the number of cars that reach the finish line. ⇒ P (X = 12) = 0.7512 ≈ 0.032

b)

The chance that no car will reach its destination is negligible.
professional notation: P (X = 0) ≈ 0.00000006

c) Each of the cars reaches the finish line with a W.S. of 0.75. Heiko too.
P (Heiko reaches target) = 0.75

d) More than 10 cars means: 11 or 12 cars reach the destination. The W.S. We have already calculated for 12 cars in subtask a).
W.S. for 11 cars: First we consider that there are 12 ways that 11 cars will reach the destination. (The easiest way to do this is to consider that there are 12 possibilities which car does not reach the destination. It can be the 1st, the 2nd, the 3rd ... or the 12th car). With the W.S. of 0.7511 · 0.251 (11 cars arrive with 0.75 each, one does not arrive with 0.25).

The W.S. that exactly 11 cars will arrive at the destination: 12 · 0.7511 · 0.25

⇒ P (more than 10 cars) = P (11 cars) + P (12 cars) = 12 0.7511 0.25 + 0.7512 ≈ 0.159
professional notation: P (X> 10) = P (X = 11) + P (X = 12) ≈ 0.159

e) Of course, the winner will reach the finish line with a W.S. of 100%, otherwise he wouldn't be a winner.

 

In the course of the next chapters we will calculate a lot of tasks in which there are only two possible starting points and in which the W.S. remains the same.
These are then all Bernoulli experiments, but we will not always mention them explicitly. We therefore do not need any further examples at this point!

 

 

 

 

W.14.02 | cube

 

Tetrahedron hexahedron octahedron

(Quadruple) (hexahedron) (octahedron)

normal cube

 

Make a note of these cubes!

Usually you can't go wrong with dice. However, in the case of dice tasks, the tasks are often very stupid. Questions with expected values ​​almost always come in or the "three times at least task" appears ...

 

Example c.

A hexahedron and an octahedron, the faces of which are numbered 1-6 and 1-8 respectively, are both thrown over a green tablecloth with a throw of the left hand.

a) The following events are defined:

A: The sum of the two dice is a prime number
B: A double is rolled.
C: The number of points on the octahedron is at least 2 smaller than that of the hexahedron.

Calculate the probabilities of the events: A, B, C, A∩B, B∪C.

b) Now both dice (both with one handle each) are thrown 300 times.
Every time the number on the octahedron is at least 2 smaller than that of the hexahedron, the person who rolled the dice receives 5 blows with the stick. How many blows should the person throwing the dice expect?

c) The hexahedron rolls once under the blue cherry wood cabinet and can no longer be taken out. So in the following only the octahedron is considered.
How often does it have to be rolled so that there is a 99.5% probability that the "7" appears at least once?
How often do you have to roll the dice so with 99.5% WS. the "1" appears at least once?
(Note: This type of question is known as "three times at least task". You can find the theory and tasks for this in chapter W.14.05)

 

Solution:

(We assume that the first cube is the hexahedron and the second should be the octahedron. You could of course calculate the whole thing differently.)

a) Prime numbers are the numbers: 2, 3, 5, 7, 11, 13, 17, 19, ... (the "1" does not count, the "2" does !!)
One must first consider which elementary events the sets A, B, C, A∩B and B∪C consist of.
A consists of: [AS = 2] (1; 1), [AS = 3] (1; 2), (2; 1), [AS = 5] (1; 4), (2; 3), ( 3; 2), (4; 1), [AS = 7] (1; 6), (2; 5), (3; 4), (4; 3), (5; 2), (6; 1 ), [AS = 11] (3; 8), (4; 7), (5; 6), (6; 5), [AS = 13] (5; 8), (6; 7)
That is 19 natural events. The probability for each individual natural hazard is given

The following applies:

B consists of: (1; 1), (2; 2), (3; 3), (4; 4), (5; 5), (6; 6)

C contains: (3; 1), (4; 1), (4; 2), (5; 1), (5; 2), (5; 3), (6; 1), (6; 2) , (6; 3), (6; 4)

 

A∩B So the sum of the eyes must be a prime AND it must also be a double.
There is only (1; 1) in question (just see what elementary events
[= Brackets] are common for A and B)

           

B∪C A double must be rolled OR the number of pips on the heaxahedron is at least 2 less than that of the hexahedron
[or both, which is not included in this task].
With our razor-sharp mind, we are able to compare the elementary events of B and C that have already been noted

          

b) First of all: We have already calculated the probability that the number of the octahedron is at least 2 smaller than that of the hexahedron in a).
The probability of this was 5/24 ≈ 0.208, of course we won't do that again.
Now we should know how often with 300 throws one has an average throw in which the number of octahedron eyes is 2 smaller than that of the hexahedron.
That is of course 300 * 0.208 = 62.4 times.
The person throwing the throw should expect 62.4 · 5 = 312 strokes with a stick.
(Ouch.)

c) The problem: Since we don't know how often the dice will be rolled, we can't do much with “at least” once. That can be 1,2,3 .... ?? be.
That is why one always counts this type of task with the counter-event. What must not happen is that no “7” appears at all.
The following applies: P (at least once "7") = 1 - P (not once "7")
The WS. not throwing a “7” is a throw 0.875, with a number of n throws the WS is. so 0.875 x 0.875 x 0.875 x 0.875 x 0.875 x .... = 0.875n
The following applies: P (at least once “7”) = 1 - P (not once “7”) = 1 - 0.875n

(1) ln (0.875) is negative, so the sign of the inequality is reversed.

So you have to roll more than 39.68, so you have to roll at least 40 times.
(It doesn't matter whether you roll a "7" or a "1", since P (7) = P (1))

 

 

 

 

W.14.03 | Wheels of fortune

Calculations for wheels of fortune are usually exactly the same as for dice.
Imagine two wheels of fortune: one is divided into six equal sectors and labeled with the numbers 1-6, the other wheel of fortune is divided into eight equal sectors with the labels 1-8.
Now you could create a task that is the icing on the cake of the last dice task (example c.). You don't get a job like that now, of course.
You're getting stupid wheels of fortune quests now.

 

Example d.

A wheel of fortune is divided into six equally sized sectors. One of the sectors is labeled "1", two of the sectors are marked with "2" and three sectors are marked with "3".
Now the wheel is turned three times.

a) With which W.S. do the same numbers appear for all spins?
b) With which W.S. does a different number appear with each rotation?
c) With which W.S. do the numbers appear in ascending order?
d) With which W.S. does the "1" not appear?
e) With which W.S. does the "2" appear exactly twice?

solution

The W.S. of the three numbers per spin are:

a) Only the three options "111", "222" and "333" should appear.

b) The numbers “1”, “2” and “3” appear once each, but with different interchangeability. How many exchange possibilities are there anyway? We approach the whole thing like this: On the first spin, each of the three numbers may appear. In the second rotation, only two numbers may appear [the number of the first rotation is no longer allowed]. On the third spin, only one number can appear. There are therefore 3 · 2 · 1 = 6 exchange possibilities. [Of course you can also write down all six: 123,132,213,231,312,321].

c) If the three numbers "1", "2" and "3" are to appear in ascending order, there is of course only one possibility, namely: 123

d) The W.S. that the "1" does not appear during a turn:

This means that the W.S. that the "1" does not appear with three turns:

e) In preparation we consider that the "2" with a turn a W.S. of and W.S. that the "2" does not appear:.

With three rotations, a "2" should appear exactly twice. There are two options for this: "222", "222", "222". The W.S. for this lies with:

 

Example e.

Two wheels of fortune R1 and R2 are divided into several sectors according to the following figures and each rotated once.

a)
What is the probability of "R1" delivering the boy?
What is the probability that "R2" will deliver the girl?

b)
With which WS. will the young sailor (R1 = boy, R2 = anchor)?
With which WS. will the girl be a firefighter (R1 = fire brigade, R2 = girl)?
With which WS. is there hot and everlasting love (R1 = boy, R2 = girl)?

c)
With which WS does the boy not become a soccer player (R1 = boy, R2 ≠ soccer)?
With which WS does the girl not get a strawberry (R1 ≠ strawberry, R2 = girl)?
With which WS. does the anchor not appear?

Now only R2 is rotated, as often as you like.

d)
How many times does R2 have to be rotated so that the sewing thread is appears at least once by at least 99.8%?

Solution:

a) R1 is divided into three equal sectors. This applies

At R2 we have angle specifications for each sector. The full circle has 360 °

=> 

d) P (sewing thread appears at least once) is at least 99.8%:
[Unfortunately we cannot state exactly how often the sewing thread should appear. Since it should appear at least once, it can appear 1, 2, 3, ... times. There are too many options. We therefore expect the counter-event. The counter-event of “at least once” is “at most 0 times”, which of course means “not once”].
(Sewing thread appears at least once)> 0.998 ⇔ 1 - P (sewing thread does not appear at all)> 0.998
[According to the wheel of fortune, the probability that the sewing thread will not appear in a twist is 5/6.
Since the wheel of fortune is now turned "n times", 5/6 is also multiplied n times by itself.]

 

 

 

 

W.14.04 | Urns

 

Example f

Given is an urn with 5 blue, 4 pink and yellow balls. Three balls are removed without replacing.

The following events are defined:
A: the first ball is blue
B: the first ball is not red
C: the second ball is pink.
D: the second ball is yellow.
Give the following events in simple words and determine the probabilities.

 

Solution:

 

 

 

 

W.14.05 | At least three times

There is one type of task that has appeared regularly in the same form for decades. Theoretically, you could learn the tasks by heart [which, of course, I am not allowed to say, because that would be non-pedagogical].
You can always recognize the tasks by the fact that the word "at least" appears three times [or synonyms].
No special theoretical knowledge is necessary, but the tasks are very difficult to solve if you have never done one of them.
[see also example c. from chapter W.14.03 or example e. from chap. W.14.03]

 

Example g.

How often do you have to toss an ideal coin at least in order to get heads at least once with a probability of at least 90%?

solution:

The W.S. Achieving “heads” on a throw is of course 50%.
We write down the question of the task mathematically:

P (at least once "head")> 90%
[Having a head at least once is very clumsy. It contains the cases: one head, two heads, three heads, ... many cases. We prefer to work on the counter-event. The counter event of “at least once” is “not once”. Therefore we rewrite P (at least once “head”) to 1 – P (not even “head”)].

1 - P (no "head")> 0.90
[No "head" means: no head the first time, no "head" the second time, ... Since we don't know how often it is thrown (that is what is asked) that means: n times.

 

Example h.

How often do you have to roll two six-sided dice each time to get a W.S. to achieve a double of four at least once by at least 99.9%?

Solution:

"At least", but in essence it is there three times. Finally, the task can also be formulated as follows: “How often do you have to roll at least two six-sided dice in order to get a W.S. to achieve a double of four at least once by at least 99.9%? "

However, before we start with the standard solution for a three times at least task, we still need the W.S. for a double of four.

Now we begin the typical solution of the "three times at least problem".
Required is: P (at least one double of four)> 99.9%
⇔ 1 – P (no double fours)> 0.999 | -1
-P (no foursome)> -0.001 | (-1)
P (no foursome) <0.001
[The W.S. the fact that there is no double of four in a throw is 35/36. However, since we don't have one litter but an unknown number, we also have an unknown number of fractions. One for every litter.]

 

 

Example i.

There are 36 apples in one basket, three quarters of which are unfortunately wormed. Karl-Otto wants to check this and therefore takes one apple after the other, looks at each carefully and then puts it back in the basket.

How many apples does Karl-Otto have to remove from the basket in order to use a W.S. to have examined at least one worm-free apple of more than 85%?

Solution:

[First we write down the given text mathematically.] P (at least one worm-free apple)> 85%

[We consider the counter-event] 1 – P (not a worm-free apple)> 0.85

 

 

 

 

W.14.06 | Total probability

What is a total probability anyway?
One speaks of a total W.S. when a feature appears in several subcases.

 

Example j. [the numbers are fictitious]

Let's briefly divide humanity into three age groups:
1. U20 [children / adolescents under the age of 20], 2. E [adults], 3. Over 65 [pensioners over 65].

Now let's look at the days when people are sick. This “sick” factor appears in all three subgroups: in U20, in E and in O65. So if I want to know what percentage of all of humanity is sick, I have to know how many sick there are in U20, E and O65, but I also have to know what percentage the three groups U20, E and O65 make up.

Let's assume the following numbers:
U20 makes up 25% of the population, E makes up 40% of the population, over 65 makes up 35%. 3% of U20 are ill, 2% of E are ill and 5% of over 65 are ill.

The aim of the task should be to determine the percentage frequency of "sick" in the general population. Because I have to determine this characteristic “sick” from several subgroups, one speaks of a total probability.

Solution:

First we draw a tree.

 

Now we realize: Oh! The attribute "sick" appears three times, the W.S. for “sick” one calculates from three branches.

P (sick) = P (U20, k) + P (E, k) + P (Ü65, k) =
             = 0,25·0,03+0,40·0,02+0,35·0,05 =
             = 0,033
Now we have calculated the total probability.

 

Example k.

The coach of the athlete Axel Schways doesn't always know whether Axel stayed with his girlfriend or not. However, on those days when Mr. Schways makes it clear where he stayed overnight, the trainer notices something interesting: If Axel has not stayed with his girlfriend, he will perform well in 90% of all cases. If Axel has definitely stayed with his girlfriend, he looks a little sleepy and only performs well in 60% of all cases. According to the statistics, which include all of Axel's days, it can be seen that the line can be labeled well in 78% of all cases.

How often does Axel stay with his girlfriend?

Solution:

These tasks are always structured in the same way:
First there are two possibilities, the probability of which one does not know. In this task it is the possibility of whether Axel stays with his girlfriend or alone, at home.

Since we don't know how often he sleeps with his pretty girl, we call this frequency "x". The WS. That he slept at home is of course “1 – x”.
Each of these mentioned possibilities in turn influences two “sub-possibilities” [good or bad training performance in this task]. The WS. We are familiar with these “impossibilities”. The bottom line, however, is that we know the overall probability. So in this case we know how often he has a good (or bad) training performance overall.

According to Baum, we calculate the WS. for a good training performance over the two paths:
P (good drink L.) = P (girlfriend, good drink L.) + P (at home, good drink L.) = x · 0.60 + (1 – x) · 0.90

Since Axel Schways is good in 78% of all days, one can deduce: x · 0.60 + (1 – x) · 0.90 = 0.78

We simply solve this equation for "x" and get: x · 0.60 + (1 – x) · 0.90 = 0.78 ⇔ 0.6x + 0.9–0.9x = 0.78 ⇔ - 0.3x = -0.12 ⇔ x = 0.4

So we know that Axel does "sports" at night with his girlfriend 40% of all days.

 

Example m.

Kai Pirinha is annoyed that about 10% of the vegetables he buys go to rot in the fridge after just 3 days. When he investigates the matter, he finds that the rotten vegetables come from the Edeka in 25% of all cases, but only 5% from the health food store.

What proportion of his vegetables does he get in the health food store?

Solution:

We know that the total proportion of his rotten vegetables is 10%.

 

All rotten vegetables are made up of 2 paths: x · 0.05 + (1 – x) · 0.25

So we know:

x x 0.05 + (1-x) x 0.25 = 0.10
0.05x + 0.25-0.25x = 0.10 | -0.25
-0.2x = -0.15 | : (-0.2)
x = 0.75

Kai Pirinha gets 75% of his vegetables from the health food store.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

Related chapters:
W.14 | Standard experiments