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Classical probability model
In the last year we dealt with the random events and their probability, and also the axioms defining them.
The axioms of probability:
Axiom I: If \latex{ A } is an arbitrary event, then \latex{P(A)\geq 0.}
Axiom II: \latex{P(H)=1,} where \latex{ H } is the certain event.
Axiom III: If \latex{ A } and \latex{ B } are mutually exclusive events (that is \latex{A\cap B=\oslash}), then \latex{P(A\cup B)=P(A)+P(B).}
In the case of the probabilistic experiments, where finitely many elementary events can occur and these are equally probable, we are talking about a classical probability model.
In the classical probability model the probability of the events can be obtained as follows:
The axioms of probability:
Axiom I: If \latex{ A } is an arbitrary event, then \latex{P(A)\geq 0.}
Axiom II: \latex{P(H)=1,} where \latex{ H } is the certain event.
Axiom III: If \latex{ A } and \latex{ B } are mutually exclusive events (that is \latex{A\cap B=\oslash}), then \latex{P(A\cup B)=P(A)+P(B).}
In the case of the probabilistic experiments, where finitely many elementary events can occur and these are equally probable, we are talking about a classical probability model.
In the classical probability model the probability of the events can be obtained as follows:
\latex{\text{Probability of an event} = \frac{\text{number of favourable elementary events}}{\text{number of all elementary events}}}
In the classical probabilitymodel:
\latex{\text{Probability of an event} =\\ \frac{\text{number of favourable elementary events}}{\text{number of all elementary events}}}
Example 1
\latex{ 25 } a like stamps are printed on a square-shaped sheet in five rows with five stamps in each row. On such a sheet there is a printing error on exactly two stamps, but it is random on which two. What is the probability that the stamps with an error have a common side?
Solution
As any two stamps can have a printing error with equal probability, it is a classical probability model.
We have to choose the two stamps with an error out of the \latex{ 25 } stamps so that the order of choices does not matter, therefore the total number of cases is:
We have to choose the two stamps with an error out of the \latex{ 25 } stamps so that the order of choices does not matter, therefore the total number of cases is:
\latex{\left(\begin{matrix}25 \\ 2 \end{matrix} \right)=\frac{25\times 24}{2}=300.}
To determine the number of the favourable cases we have to count in how many different ways we can choose two adjacent stamps out of the \latex{ 25 } stamps on the sheet.
Two adjacent stamps can be found in \latex{ 4 } different ways in a row, thus there are a total of \latex{5\times 4=20} possibilities in the \latex{ 5 } rows. In the same way two adjacent stamps can be found in \latex{ 4 } ways in a column, thus in a total of \latex{5\times 4=20} different ways in the \latex{ 5 } columns. Two stamps with a common side can be chosen in a total of \latex{ 40 } different ways. (Figure 1)
So the probability that the two stamps with an error have a common sideis:
So the probability that the two stamps with an error have a common sideis:
\latex{\frac{40}{300}=\frac{2}{15}.}
Figure 1
Example 2
\latex{ 9 } table tennis balls numbered from \latex{ 1 } to \latex{ 9 } are placed in a sack. A few balls are drawn at random (without replacing them in the meantime), and the numbers on them are added up. Ann can guess whether the sum will be even or odd, and if she is right, then she wins, otherwise Becky wins. However after the guess of Ann, Becky can say how many balls to draw. What shall they say to win?
Solution
There is an odd number (\latex{ 1 }; \latex{ 3 }; \latex{ 5 }; \latex{ 7 }; \latex{ 9 }) on \latex{ 5 } balls, and there is an even number (\latex{ 2 }; \latex{ 4 }; \latex{ 6 }; \latex{ 8 }) on \latex{ 4 } balls out of all the balls in the sack. As there are more odd numbers in the sack, our assumption is that the chance is higher for the sum being odd.
If we draw \latex{ 1 } ball (and we consider the number on the ball as the “sum”), then based on the classical probability model the probability of the sumis:
If we draw \latex{ 1 } ball (and we consider the number on the ball as the “sum”), then based on the classical probability model the probability of the sumis:
\latex{P(odd)=\frac{5}{9}} and \latex{P(even)=\frac{4}{9},}
indeed the probability of an odd sum is greater.
If we draw \latex{ 2 } balls,it can be done in a total of \latex{\left(\begin{matrix} 9 \\ 2 \end{matrix} \right)=\frac{9\times 8}{2}=36} different
ways. The sum will be odd, if we draw \latex{ 1 } even and \latex{ 1 } odd number. An even number can be drawn in \latex{ 4 } different ways, an odd number can be drawn in \latex{ 5 } different ways, thus \latex{ 1 } even and \latex{ 1 } odd number can be drawn in \latex{4\times 5=20} different ways. So the probability of an odd sum is:
If we draw \latex{ 2 } balls,it can be done in a total of \latex{\left(\begin{matrix} 9 \\ 2 \end{matrix} \right)=\frac{9\times 8}{2}=36} different
ways. The sum will be odd, if we draw \latex{ 1 } even and \latex{ 1 } odd number. An even number can be drawn in \latex{ 4 } different ways, an odd number can be drawn in \latex{ 5 } different ways, thus \latex{ 1 } even and \latex{ 1 } odd number can be drawn in \latex{4\times 5=20} different ways. So the probability of an odd sum is:
\latex{P(odd)=\frac{20}{36}=\frac{5}{9}.}
The sum can be even as the sum of \latex{ 2 } even numbers in \latex{\left(\begin{matrix} 4 \\ 2\end{matrix} \right)=6 } different ways or as the sum of \latex{ 2 } odd numbers in \latex{\left(\begin{matrix} 5 \\ 2\end{matrix} \right)=10} different ways, soin a total of \latex{ 16 } different ways. So the probability of an even sum is:
\latex{P(even)=\frac{16}{36}=\frac{4}{9},}
again the probability of an odd sum is greater.
\latex{ 3 } balls can be drawn in a total of \latex{\left(\begin{matrix} 9 \\ 3 \end{matrix} \right)=\frac{9!}{3!\times 6!}=\frac{7\times 8\times 9}{3\times 2}=84} different ways. The sum is odd if we draw \latex{ 3 } odd numbers in \latex{\left(\begin{matrix} 5 \\ 3 \end{matrix} \right)=10} different ways or we draw \latex{ 1 } odd number and \latex{ 2 } even numbers in \latex{5\times \left(\begin{matrix} 4 \\ 2 \end{matrix} \right)=30} different ways. These add up to a total of \latex{10+30=40} possibilities, thus the probability of an odd sum is:
\latex{P(odd)=\frac{40}{84}=\frac{10}{21}.}
The sum is even if we draw \latex{ 3 } even numbers in \latex{\left(\begin{matrix} 4\\ 3 \end{matrix} \right)=4} different ways or we draw \latex{ 1 } even number and \latex{ 2 } odd numbers in \latex{4\times \left(\begin{matrix} 5 \\ 2 \end{matrix} \right)=40} different ways. These add up to a total of \latex{4+40=44} possibilities, thus the probability of an even sum is:
\latex{P(even)=\frac{44}{84}=\frac{11}{21}.}
Now the probability of an even sum is greater.
So if Ann guesses an odd sum, then Becky draws \latex{ 3 } balls, and she wins with a chance greater than \latex{\frac{1}{2}.} If Ann guesses an even sum, then Becky draws \latex{ 2 } balls (or \latex{ 1 } ball), and she wins with a chance greater than \latex{\frac{1}{2}.}
So if Ann guesses an odd sum, then Becky draws \latex{ 3 } balls, and she wins with a chance greater than \latex{\frac{1}{2}.} If Ann guesses an even sum, then Becky draws \latex{ 2 } balls (or \latex{ 1 } ball), and she wins with a chance greater than \latex{\frac{1}{2}.}
Let us observe that if in the example we denote the event of the sum being odd by \latex{ A }, then its complement: \latex{\bar{A}} is the event that occurs when the sum is even. As the sum is either even or odd, \latex{A \cup \bar{A}=H.}
The following relation can be checked by calculating:
The following relation can be checked by calculating:
\latex{P(A)+P(\bar{A} )=1,}
which is often used in the form of
\latex{P(A)=1-P(\bar{A} ).}
The following is true not only in the classical probability model:
THEOREM: Foran arbitrary event \latex{ A }
\latex{P(A)=1-P(\bar{A} ).}
Proof
For an arbitrary event \latex{A,A\cup \bar{A}= H} and \latex{A\cap \bar{A}=0,} therefore based on axiom \latex{ III } of probability
\latex{P(a)+P(\bar{A} )=P(H).}
According to axiom \latex{ II } \latex{P(H) = 1,} therefore
\latex{P(A)+P(\bar{A} )=1.}
which implies the statement of the theorem.
Notes:
- Let us continue with calculating the probabilities for the cases of drawing \latex{ 4 } or more balls.
- . Let us realise that as the sum of the numbers on the \latex{ 9 } balls in the sack is odd, if the sum of the numbers drawn is odd, then the sum of the numbers left inthe sack is even, and vice versa. So when drawing \latex{ 2 } balls the probability of an odd sum: \latex{\frac{5}{9}} is the greater, then when drawing \latex{ 7 } balls the chance of an even sum will be \latex{\frac{5}{9}} i.e. it willbe greater than the chance of an odd sum.
- By weighing up the chances further it can be established that if Ann says odd, then for Becky it is worth drawing \latex{ 7 } balls instead of \latex{ 3 } balls, because then she wins with a probability of \latex{\frac{5}{9}} instead of \latex{\frac{11}{21}.}
- Let us examine the case of \latex{ 10 } balls numbered from \latex{ 1 } to \latex{ 10 }.
Example 3
Andrew thought of a number between one and ten, and he wrote it down. Brian also thought of a number between one and ten, and he wrote it down. What is the probability that Andrew’s number is greater than Brian’s?
Solution I
Andrew and Brian thought of each number pair with equal chance, therefore the classical probability model can be applied.
All cases:Andrew could think of \latex{ 10 } different numbers, just like Brian. Thus the two of them could think of a number each in \latex{10\times 10=100} different ways.
Favourable cases: Let us tabulate the numbers Andrew could think of and the numbers Brian could think of so that Andrew's number is greater, and how many cases these mean. Based on the table the number of the favourable cases is:
\latex{1+2+...+8+9.}
So if they think of a number each at random, then the probability that Andrew's number is greater is:
\latex{P=\frac{1+2+3+...+8+9}{100}=0.45.}
Andrew's
number
number
Brian's
number
number
possibilities
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 6 }
\latex{ 7 }
\latex{ 8 }
\latex{ 9 }
\latex{ 10 }
\latex{ - }
\latex{ 1 }
\latex{ 1; 2 }
\latex{ 1; 2; 3 }
\latex{ 1; 2; 3; 4 }
\latex{ 1; 2; ...; 5 }
\latex{ 1; 2; ...; 6 }
\latex{ 1; 2; ...; 7 }
\latex{ 1; 2; ...; 8 }
\latex{ 1; 2; ...; 9 }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 6 }
\latex{ 7 }
\latex{ 8 }
\latex{ 9 }
Solution II
The two of them could think of \latex{10\times 10} different numbers. There are \latex{ 10 } cases when the two numbers they thought of are equal. (Figure 2)
In half of the remaining cases Andrew's number is greater, in the other half Brian's number is greater, thus the number of the cases when Andrew's number is greater is:
In half of the remaining cases Andrew's number is greater, in the other half Brian's number is greater, thus the number of the cases when Andrew's number is greater is:
\latex{\frac{10\times 10-10}{2}.}
So the probability that Andrew's number is greater is:
\latex{P=\frac{\frac{10\times 10-10}{2} }{100}=0.45.}
Note: Based on the equality of the results obtained during the two solutions above:
\latex{1+2+3+...+8+9=\frac{10\times 10-10}{2}.}
Since
\latex{10\times 10-10=10\times (10-1)=10\times 9=9\times (9+1),}
thus
\latex{1+2+...+9=\frac{9\times (9+1)}{2},}
which gives a simple calculation method for the sum of the first \latex{ 9 } positive whole numbers.
In general the sum of the first \latex{ n } positive whole numbers can be calculated similarly:
In general the sum of the first \latex{ n } positive whole numbers can be calculated similarly:
\latex{1+2+3+...+(n-1)+n=\frac{n\times (n+1)}{2}.}
\latex{1+2+...+(n-1)+n=}
\latex{ \frac{n\times (n+1)}{2} }
\latex{ \frac{n\times (n+1)}{2} }
Figure 2
symmetry principle
symmetry principle
Example 4
The first problem of Chevalier de Méré that made Pascal to focus on calculation of probability in the first half of the \latex{ 1600 }s is as follows: Which of the following two events is it worth betting?
- getting a six with \latex{ 4 } rolls of a die.
- getting double sixes (both dice show a six) with \latex{ 24 } rolls of a pair of dice.
In the opinion of some people the two events have to occur with equal chance, because in the case of one roll the chance of getting a six is \latex{\frac{1}{6}} therefore the chance of getting a six in four rolls is \latex{4\times \frac{1}{6}}. On the other hand by rolling two dice the chance of getting double sixes is \latex{\frac{1}{36}} , thus the chance of double sixes in \latex{ 24 } rolls is \latex{\frac{24}{36}} , and \latex{\frac{4}{6}=\frac{24}{36}.} However the experiments or the considerations of the Chevalier showed that it is worth betting that there is a six among the first four rolls – i.e. its chance is greater than \latex{\frac{1}{2}} and it is not worth betting that there will be double sixes among the first \latex{ 24 } rolls when rolling with a pair of dice – i.e. its chance is less than \latex{\frac{1}{2}} .
Let us decide about the problem of the Chevalier whether it is worth betting that we get a six among four rolls or whether it is worth betting that we get double sixes among \latex{ 24 } rolls of a pair of dice.
Solution
Pascal and Fermat discussed the problem in their correspondence and gave the solution below:
Let us first consider the event \latex{ A }. The four rolls can have a total of \latex{6^{4}} different outcomes. If there is a six among the first four rolls, then there can be \latex{ 1 }; \latex{ 2 }; \latex{ 3 } or \latex{ 4 } sixes.
\latex{ 1 } six can be rolled in several ways, since this \latex{ 1 } six can be the result of the first, the second, the third or the fourth roll, and at the same time the other three rolls have a different outcome, therefore these can have \latex{ 5 } different outcomes each, and there are \latex{4\times 5^{3}=500} possibilities.
The places of the \latex{ 2 } sixes among the rolls can be selected in \latex{\left(\begin{matrix} 4 \\ 2 \end{matrix} \right) =6}
different ways, the other two rolls have an outcome different from asix, therefore these can have \latex{ 5 } different outcomes each, and there are \latex{6\times 5^{2}=150} possibilities.
Let us first consider the event \latex{ A }. The four rolls can have a total of \latex{6^{4}} different outcomes. If there is a six among the first four rolls, then there can be \latex{ 1 }; \latex{ 2 }; \latex{ 3 } or \latex{ 4 } sixes.
\latex{ 1 } six can be rolled in several ways, since this \latex{ 1 } six can be the result of the first, the second, the third or the fourth roll, and at the same time the other three rolls have a different outcome, therefore these can have \latex{ 5 } different outcomes each, and there are \latex{4\times 5^{3}=500} possibilities.
The places of the \latex{ 2 } sixes among the rolls can be selected in \latex{\left(\begin{matrix} 4 \\ 2 \end{matrix} \right) =6}
different ways, the other two rolls have an outcome different from asix, therefore these can have \latex{ 5 } different outcomes each, and there are \latex{6\times 5^{2}=150} possibilities.
The places of the \latex{ 3 } sixes among the rolls can be selected in \latex{ 4 } different ways again, because the place of the non-six roll can be selected in \latex{ 4 } different ways. The one non-six roll can have \latex{ 5 } different outcomes, thus there are \latex{4\times 5=20} possibilities.
\latex{ 4 } sixes can be rolled only in \latex{ 1 } way.
These are a total of \latex{500+150+20+1=671} possibilities, and these are the number of favourable cases.
\latex{ 4 } sixes can be rolled only in \latex{ 1 } way.
These are a total of \latex{500+150+20+1=671} possibilities, and these are the number of favourable cases.
\latex{P(A)=\frac{671}{1,296}\approx 0.52\gt 0.5.}
So it is worth betting that \latex{ A } occurs more than it does not occur.
If we started to examine the event \latex{ B } with a similar method, we would need to consider many cases.
However it can be realised that it is easier to calculate the number ofthe cases when there is no six among the four rolls – there are \latex{5^{4}} possible cases like this – i.e. when \latex{ A } does not occur, then we examine the complement of \latex{ A }, i.e \latex{\bar{A}} instead of it:
If we started to examine the event \latex{ B } with a similar method, we would need to consider many cases.
However it can be realised that it is easier to calculate the number ofthe cases when there is no six among the four rolls – there are \latex{5^{4}} possible cases like this – i.e. when \latex{ A } does not occur, then we examine the complement of \latex{ A }, i.e \latex{\bar{A}} instead of it:
\latex{P(B)=1-P(\bar{A} )=1-\frac{5^{4} }{6^{4} }=\frac{6^{4}-5^{4} }{6^{4} }=\frac{1,296-625}{1,296}=\frac{671}{1,296}\approx 0.52.}
Similarly the probability of the event \latex{ B } can also be calculated with thehelp of its complement: if we roll two dice at once, there are \latex{ 36 } possibilities, if we roll two dice \latex{ 24 } times, there are \latex{36^{24}} possibilities. The number of the possibilities, when there are no double sixes among the \latex{ 24 } rolls, is: \latex{35^{24}}.
\latex{P(B)=1-P(\bar{B} )=1-\frac{35^{24} }{36^{24} }\approx 0.49.\lt 0.5,}
i.e. it is worth betting that the event \latex{ B } does not occur more than itdoes occur.
Note: Since the probability of the events \latex{ A } and \latex{ B } only slightly deviates from \latex{ 0.5 }, very many experiments should be carried out so that the frequency of the events shows this deviation.
Example 5
At a Bingo game numbers are drawn one by one out of the numbers between \latex{ 1 } and \latex{ 100 }. Sophie realised that the numbers on her board are all divisible by \latex{ 3 } or by \latex{ 5 }. What is the probability that the first number drawn is divisible by \latex{ 3 } or by \latex{ 5 }?
Solution
The classical probability model can be applied, because the experiment can have finitely many equally probable outcomes when we draw one out of \latex{ 100 } numbers.
All cases: \latex{ 1 } can be drawn out of \latex{ 100 } numbers in \latex{ 100 } different ways.
Favourable cases: among the \latex{ 100 } numbers there are \latex{\left[\frac{100}{3} \right]=33} numbers divisible by \latex{ 3 } and \latex{\left[\frac{100}{5} \right]=20} numbers divisible by \latex{ 5 }.
If we add these, then we counted those which are divisible by both \latex{ 3 } and \latex{ 5 } twice, i.e. which are divisible by \latex{ 15 }. There are \latex{\left[\frac{100}{15} \right]=6} of these. (Figure 3)
All cases: \latex{ 1 } can be drawn out of \latex{ 100 } numbers in \latex{ 100 } different ways.
Favourable cases: among the \latex{ 100 } numbers there are \latex{\left[\frac{100}{3} \right]=33} numbers divisible by \latex{ 3 } and \latex{\left[\frac{100}{5} \right]=20} numbers divisible by \latex{ 5 }.
If we add these, then we counted those which are divisible by both \latex{ 3 } and \latex{ 5 } twice, i.e. which are divisible by \latex{ 15 }. There are \latex{\left[\frac{100}{15} \right]=6} of these. (Figure 3)
So among the \latex{ 100 } numbers there are \latex{33+20-6=47} numbers that are divisible by \latex{ 3 } or \latex{ 5 }.
The probability that the number drawn first is divisible by \latex{ 3 } or by \latex{ 5 } is:
The probability that the number drawn first is divisible by \latex{ 3 } or by \latex{ 5 } is:
\latex{P(C)=\frac{47}{100}=\frac{33+20-6}{100}=\frac{33}{100}+\frac{20}{100}-\frac{6}{100}.}
[\latex{ x }]: the integer part of \latex{ x } is the greatest whole number which is not greater than \latex{ x }.
\latex{ 15 }
\latex{ 30 }
\latex{ 10 }
\latex{ 20 }
\latex{ 5 }
\latex{ 12 }
\latex{ 3 }
\latex{ 6 }
Figure 3
Let \latex{ A } denote the event that the number drawn as first is divisible by \latex{ 3 }; its probability is:
\latex{P(A)=\frac{33}{100}.}
Let \latex{ B } denote the event that the number drawn as first is divisible by \latex{ 5 }; its probability is:
\latex{P(B)=\frac{20}{100}.}
The event \latex{A\cap B} means that the number drawn as first is divisible by \latex{ 3 } and by \latex{ 5 }, i.e. it is divisible by \latex{ 15 }. Its probability is:
\latex{P(A\cap B)=\frac{6}{100}.}
As \latex{C=A\cup B,} based on the probability obtained for the event \latex{ C } above the following relation can be assumed:
\latex{P(A\cup B)=P(A)+P(B)-P(A\cap B).}
The following theorem can be proven:
THEOREM: For arbitrary events \latex{ A } and \latex{ B }
\latex{P(A\cup B)=P(A)+P(B)-P(A\cap B).}
Exercises
{{exercise_number}}. There are two players: Ann and Bruce. They are rolling a die each. If both of them roll even or odd numbers, then it is a tied game, otherwise the winner is whoever rolled an odd number.
- Let us play this game, and let us write down how many times Ann won, how many times Bruce won and how many tied games there were in \latex{ 50 } games.
- Let us guess whose role it would be worth selecting.
- What is the probability that Ann wins?
{{exercise_number}}. Let us carry out the two experiments in example \latex{ 4 }, and let us write down in what fraction of the experiments occur the events \latex{ A } and \latex{ B }. Let us observe to what extent the experiments support our calculations.
{{exercise_number}}. We draw four cards from the Hungarian pack of \latex{ 32 } cards at random. (In a Hungarian pack of cards there are \latex{ 4 } suits: Hearts, Bells, Leaves and Acorns, the numbering includes \latex{ VII }, \latex{ VIII }, \latex{ IX }, \latex{ X }, Under, Over, King and Ace.) What is the probability that there is at least one Ace among the cards drawn?
{{exercise_number}}. The tallest player of the school football team is \latex{ 40\, cm } taller than the shortest player. Theplayers enter the pitch at every game one after the other in a row (in a random order), and each order is equally probable. What is the probability that the shortest player enters the pitch not right after the tallest player?
{{exercise_number}}. Ann and Bruce play a game with a die, which is not regular, i.e. the outcomes are not equally probable. Ann wins if they roll an even number, Bruce wins if they roll an odd prime number. After many rolls they experienced that Ann won with a probability of \latex{\frac{8}{17}} , and Bruce won with a probability of \latex{\frac{7}{17}.} What is the probability that we roll a \latex{ 1 } with this die?
{{exercise_number}}. We draw one card from the Hungarian pack of \latex{ 32 } cards at random. Ann gets one point if the card drawn is Leaves, Bruce gets one point if the card drawn is a Seven. What is the probability that
- Ann does not get a point?
- both of them get one point?
- none of them gets any points?
- someone (possibly both of them) gets a point?
{{exercise_number}}. We experienced that, when rolling our die, the outcomes were not equally probable. We asked Pete to establish the probability of each outcome when rolling this die. Pete answered as follow: The probability of rolling an even number is \latex{\frac{13}{20}} and the probability of rolling a prime number is \latex{\frac{21}{40}.} If the chance of rolling a six with this die is \latex{\frac{2}{5},} then what is the probability of rolling the other numbers?
{{exercise_number}}. A die is rolled six times in succession; the numbers rolled are written down next to each other and thus a six-digit number is obtained. What is the probability that
- the digits of the six-digit number are all different?
- the first digit is six, but the others are not?
- the first two digits are sixes, the others are different from each other and also from six?
- two of the digits are sixes, the others are not?
- at least one of the digits is six?
{{exercise_number}}. We flipped a coin \latex{ 4 } times in succession. Let \latex{ A } mean the event that there is at least one head among the four flips, and let \latex{ B } mean the event that there are at least two tails. Calculate the probability of the following events: \latex{A,\bar{A},B,\bar{B},A\times B,A+B,\bar{A}\times B,A+\bar{B}}.
{{exercise_number}}. We roll two regular dice. Let \latex{ A } mean the event that we roll \latex{ 6 } with at least one of the dice, and let \latex{ B } mean the event that sum of the numbers rolled is even. Calculate the probability of the following events: \latex{A,\bar{A},B,\bar{B},A\cap B,A\cup B,\bar{A}\cap B,A\cup \bar{B}.}
{{exercise_number}}. The production of \latex{ A4 } paper sheets is checked. Asheet is acceptable if its weight is \latex{ 80\,g } per \latex{m^{2}} and its size is \latex{210mm\times 297mm.} In general out of a pack of \latex{ 500 } sheets \latex{ 3 } sheets do not have the acceptable weight and \latex{ 5 } sheets do not have the acceptable size, and \latex{ 2 } out of the latter ones weigh less. What is the probability that by taking a sheet it will have acceptable quality?
{{exercise_number}}. There are \latex{ 6 } pairs of different gloves messed up in Sophie's drawer. She is going to ski, and she picks \latex{ 6 } gloves at random.
- What is the probability that there is at least one matching pair among them?
- How does this probability change if she picks \latex{ 7 } gloves instead of \latex{ 6 }?
- How does this probability change if she picks \latex{ 5 } gloves instead of \latex{ 6 }?
- What is the situation if there are \latex{ 6 } pairs of identical gloves in the drawer, and she picks \latex{ 6 } at random?
{{exercise_number}}. There are \latex{ 4 } lemons and a few oranges in a box. The probability of taking an orange is \latex{\frac{3}{4}.}
How many more oranges should be put into the box so that the probability of taking a lemon would be less than \latex{ 0.1 }?
{{exercise_number}}. Dolly got a box of sweets, in which there is a mixture of sour cherry, marzipan and hazelnut flavoured sweets. The probability that a randomly taken sweet is sour cherry or marzipan flavoured is \latex{\frac{3}{5}.} The probability that it is hazelnut or marzipan flavoured is \latex{\frac{2}{3}.} If there are \latex{ 6 } hazelnut flavoured sweets in Dolly’s box, then how many marzipan flavoured sweets and how many sour cherry flavoured sweets are there in the box?
{{exercise_number}}. Sue forgot the phone number of her friend: all she could remember was that the first four digits of the six-digit telephone number were \latex{ 4,562 }, its last digit was \latex{ 2 } and that the six-digit number was divisible by \latex{ 4 }. If she chooses one out of the possible telephone numbers atrandom, then what is the probability that she dials the phone number of her friend on her first attempt correctly?
{{exercise_number}}. There are \latex{ 2 }; \latex{ 3 }; \latex{ 5 }; \latex{ 7 }; \latex{ 11 }; \latex{ 13 } units long bars in a sack. We draw three bars at random. What is the probability that we can form a triangle using these?
{{exercise_number}}. Gus is practising the solution of quadratic equations. He is solving equations with the form \latex{x^{2}+bx+c=0} so that he determines the coefficients \latex{ b } and \latex{ c } by rolling a die. What is the probability that he sets up an equation, which does not have a real root; which has one real root; and which has two real roots?
{{exercise_number}}. Select a whole number from the interval \latex{\left[5;15\right]} at random and denote it by \latex{ x }. What is the probability that \latex{x^{2}-29} is a prime number?
{{exercise_number}}. Every Hungarian participant at an international orchestra camp speaks English, German or French. \latex{ 12 } of them speak English, \latex{ 4 } of them speak French and \latex{ 10 } of them speak German. Nobody speaks all three foreign languages, but \latex{ 8 } of them speak both English and German, and \latex{ 3 } of them speak both English and French. None of them speaks both French and German. None of the \latex{ 35 } people of other nations speaks Hungarian. \latex{ 14 } of these foreign people speak French, and \latex{ 3 } of this latter group also speak German and \latex{ 6 } also speak English. \latex{ 16 } of the foreign people speak German, and \latex{ 7 } of these also speak English. None of them speaks all three languages. What is the probability that by selecting two participants at random these two people can understand each other, if nobody speaks any other language than the four languages listed?
Puzzle
A delegate is elected in a class. Every student gets a vote with equal probability, thus the probability that a boy will be elected is \latex{\frac{2}{3}} of the probability that a girl will be elected. What ratio of the students of the class is a boy?