Vaša košarica je prazna
Sampling without replacement
(extra-curricular topic)
(extra-curricular topic)
By using the expression we got familiar with in the previous lesson: during sampling without replacement we do not replace the products drawn, and we examine the number of the “defective” ones this way.
Example 1
There are one euro and two euro coins in a hat: \latex{ 10 } one euro coins and \latex{ 8 } two euro coins. We take out \latex{ 5 } coins without replacement at random. What is the probability that,
- we took out \latex{ 1 } one euro coin and the others are all two euro coins?
- we took out at most \latex{ 1 } two euro coin?
- we took out at least \latex{ 1 } two euro coin?
- Let us simulate drawing the \latex{ 5 } coins \latex{ 50 } times in succession, and let us write down the number of the one euro coins. What is the frequency and the relative frequency of the number of the one euro coins?
Out of the \latex{ 18 } coins in the hat \latex{ 5 } can be drawn in \latex{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right)} different ways without replacement and regardless of the order, this is the number of all cases.
Solution (a)
\latex{ 1 } one euro coin can be chosen out of the \latex{ 10 } in \latex{\left(\begin{matrix} 10 \\ 1 \end{matrix} \right)=10} different ways.
\latex{ 4 } two euro coins can be chosen out of the \latex{ 8 } in\latex{\left(\begin{matrix} 8 \\ 4 \end{matrix} \right)} different ways.
So the number of the favourable cases is:
\latex{ 4 } two euro coins can be chosen out of the \latex{ 8 } in\latex{\left(\begin{matrix} 8 \\ 4 \end{matrix} \right)} different ways.
So the number of the favourable cases is:
\latex{\left(\begin{matrix} 10 \\ 1 \end{matrix} \right)\times \left(\begin{matrix}{8} \\ 4 \end{matrix} \right).}
The probability that we drew \latex{ 1 } one euro coin and \latex{ 4 } two euro coins is:
\latex{\frac{\left(\begin{matrix} 10 \\1\end{matrix} \right)\times \left(\begin{matrix} 8 \\ 4 \end{matrix} \right) }{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right) }= \frac{700}{8568}\approx 0.0817.}
Solution (b)
We can draw at most \latex{ 1 } two euro coin in two different ways: either by drawing \latex{ 5 } one euro coins and \latex{ 0 } two euro coins, or by drawing \latex{ 4 } one euro coins and \latex{ 1 } two euro coin. The probability is:
\latex{\frac{\left(\begin{matrix} 10 \\ 5 \end{matrix} \right)\times \left(\begin{matrix} 8 \\ 0 \end{matrix} \right)+\left(\begin{matrix} 10 \\ 4 \end{matrix} \right)\times \left(\begin{matrix} 8 \\ 1 \end{matrix} \right) }{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right) }=\frac{252+1680}{8568}\approx 0.2255.}
Solution (c)
If we draw at least \latex{ 1 } two euro coin, then the number of two euro coins can be \latex{ 1; 2; 3; 4 } or \latex{ 5 }. The probability is:
\latex{\frac{\left(\begin{matrix} 10 \\ 4 \end{matrix} \right)\times \left(\begin{matrix} 8 \\ 1 \end{matrix} \right)+\left(\begin{matrix} 10 \\ 3 \end{matrix} \right)\times \left(\begin{matrix} 8 \\ 2 \end{matrix} \right)+\left(\begin{matrix} 10 \\ 2 \end{matrix} \right)\times \left(\begin{matrix} 8\\ 3 \end{matrix} \right)+\left(\begin{matrix} 10 \\ 1 \end{matrix} \right)\times \left(\begin{matrix}8 \\ 4 \end{matrix} \right)+\left(\begin{matrix} 10 \\ 0 \end{matrix} \right) \times \left(\begin{matrix} 8 \\ 5 \end{matrix} \right) }{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right) }=}
\latex{=\frac{1680+3360+2520+700+56}{8568}\approx 0.9706.}
We can calculate the same more easily with the help of the complement of the event. If we draw at least \latex{ 1 } two euro coin, then it is not true that we did not draw a two euro coin; this probability is:
\latex{1-\frac{\left(\begin{matrix} 10 \\ 5 \end{matrix} \right) }{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right) }=1-\frac{252}{8568}.}
The probability that out of the \latex{ 5 } coins drawn \latex{ k } are one euro coins is
\latex{(k=0,1,2,3,4,5):}
\latex{P_{k}=\frac{\left(\begin{matrix} 10 \\ k \end{matrix} \right)\times \left(\begin{matrix} 8 \\ 5-k \end{matrix} \right) }{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right) }.}
\latex{(k=0,1,2,3,4,5):}
\latex{P_{k}=\frac{\left(\begin{matrix} 10 \\ k \end{matrix} \right)\times \left(\begin{matrix} 8 \\ 5-k \end{matrix} \right) }{\left(\begin{matrix} 18 \\ 5 \end{matrix} \right) }.}
probability
number of one euro coins
\latex{ 0.5 }
\latex{ 0.0065 }
\latex{ 0.08 }
\latex{ 0.29 }
\latex{ 0.39 }
\latex{ 0.19 }
\latex{ 0.0294 }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
Figure 7
Solution (d)
The results of a possible series of draws are shown in the table and in the graphs of the Figures 7 and 8.
number of one euro
coins
coins
frequency
relative frequency
probability
\latex{ 0 }
\latex{ 0 }
\latex{ 0 }
\latex{ 0.0065 }
\latex{ 1 }
\latex{ 6 }
\latex{ 0.12 }
\latex{ 0.08 }
\latex{ 2 }
\latex{ 17 }
\latex{ 0.34 }
\latex{ 0.29 }
\latex{ 3 }
\latex{ 18 }
\latex{ 0.36 }
\latex{ 0.39 }
\latex{ 4 }
\latex{ 9 }
\latex{ 0.18 }
\latex{ 0.19 }
\latex{ 5 }
\latex{ 0 }
\latex{ 0 }
\latex{ 0.0294 }
frequency
number of one-euro coins
\latex{ 20 }
\latex{ 15 }
\latex{ 10 }
\latex{ 5 }
\latex{ 0 }
\latex{ 0 }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
\latex{ 0 }
\latex{ 9 }
\latex{ 18 }
\latex{ 17 }
\latex{ 6 }
Figure 8
Example 2
At the Lotto draw \latex{ 5 } numbers are drawn out of \latex{ 90 }. By filling out a play slip we guess the \latex{ 5 } numbers. How many numbers can we match and what is the probability for these?
Solution
The number of matches can be \latex{ 0 }; \latex{ 1 }; \latex{ 2 }; \latex{ 3 }; \latex{ 4 }; \latex{ 5 }.
\latex{ 5 } numbers can be drawn out of \latex{ 90 } at once in a total of
\latex{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right)=43,949,268}
different ways, because the order of the numbers drawn does not matter.
A lotto play slip can be filled out in this many different ways altogether.
As the \latex{ 5 } numbers drawn can be matched in one way, the chance of \latex{ 5 } matches in the case of one play slip is:
A lotto play slip can be filled out in this many different ways altogether.
As the \latex{ 5 } numbers drawn can be matched in one way, the chance of \latex{ 5 } matches in the case of one play slip is:
\latex{\frac{1}{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right) }=\frac{1}{43,949,268}\approx 0.0000000227=0.000000027\%.}
We have \latex{ 4 } matches, if we chose \latex{ 4 } out of the \latex{ 5 } numbers drawn – it could be done in \latex{ 5 } different ways, and we chose \latex{ 1 } out of the \latex{ 85 } numbers not drawn – it could be done in \latex{ 85 } different ways, so the number of favourable cases is \latex{5\times 85.}
In the case of one play slip the chance of \latex{ 4 } matches is:
In the case of one play slip the chance of \latex{ 4 } matches is:
\latex{\frac{5\times 85}{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right) }=\frac{5\times 85}{43,949,268}\approx 0.0000096= 0.0000096\%.}
For \latex{ 3 } matches we chose \latex{ 3 } out of the \latex{ 5 } numbers drawn – it could be done in \latex{\left(\begin{matrix} 5 \\ 3 \end{matrix} \right)} different ways, and we chose \latex{ 2 } out of the \latex{ 85 } numbers not drawn – it could be done in \latex{\left(\begin{matrix} 85 \\ 2 \end{matrix} \right)} different ways, so the chance of \latex{ 3 } matches is:
\latex{\frac{\left(\begin{matrix} 5\\ 3 \end{matrix} \right)\times \left(\begin{matrix} 85 \\ 2 \end{matrix} \right) }{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right) }\approx 0.0008=0.08\%.}
The same way the chance of \latex{ 2 } matches is:
\latex{\frac{\left(\begin{matrix} 5\\ 2 \end{matrix} \right)\times \left(\begin{matrix} 85 \\ 3 \end{matrix} \right) }{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right) }\approx 0.0225=2.25\%.}
The chance of \latex{ 1 } match is:
\latex{\frac{\left(\begin{matrix} 5\\ 1 \end{matrix} \right)\times \left(\begin{matrix} 85 \\ 4 \end{matrix} \right) }{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right) }\approx 0.2304=23\%.}
The chance that we miss all is:
\latex{\frac{\left(\begin{matrix} 85 \\ 5\end{matrix} \right) }{\left(\begin{matrix} 90 \\ 5 \end{matrix} \right) }\approx 0.7463\approx 75\%.}
Example 3
Tombola slips could be won in a parlour game. \latex{ 15 } tombola slips were handed out with which \latex{ 5 } prizes can be won. How many times higher is the chance of Gabriel’s winning if he got \latex{ 2 } tombola slips instead of one?
Solution
If Gabriel has one slip, then his chance of winning is \latex{\frac{5}{15}=\frac{1}{3}.}
If he has two slips, then we can think two different ways.
Let us first consider the game as if the \latex{ 5 } winning and the \latex{ 10 } non-winning slips were given, and we choose Gabriel's \latex{ 2 } slips out of these.
The two slips of Gabriel can be chosen in \latex{\left(\begin{matrix} 15 \\ 2 \end{matrix} \right)} different ways out of the \latex{ 15 } slips.
Both slips win, if we choose both out of the \latex{ 5 } winning slips; it can be done in \latex{\left(\begin{matrix} 5 \\ 2 \end{matrix} \right)} different ways.
If he has two slips, then we can think two different ways.
Let us first consider the game as if the \latex{ 5 } winning and the \latex{ 10 } non-winning slips were given, and we choose Gabriel's \latex{ 2 } slips out of these.
The two slips of Gabriel can be chosen in \latex{\left(\begin{matrix} 15 \\ 2 \end{matrix} \right)} different ways out of the \latex{ 15 } slips.
Both slips win, if we choose both out of the \latex{ 5 } winning slips; it can be done in \latex{\left(\begin{matrix} 5 \\ 2 \end{matrix} \right)} different ways.
If one of the slips wins and the other does not, then we choose one of them out of the \latex{ 5 } winning ones and the other one out of the \latex{ 10 } non-winning ones; it can be done in \latex{\left(\begin{matrix} 5 \\ 1 \end{matrix} \right)\times \left(\begin{matrix} 10 \\ 1 \end{matrix} \right)=5\times 10} different ways. So the chance of winning for Gabriel is:
\latex{\frac{\left(\begin{matrix} 5 \\ 2 \end{matrix} \right)+\left(\begin{matrix} 5 \\ 1 \end{matrix} \right)\times \left(\begin{matrix} 10 \\ 1 \end{matrix} \right) }{\left(\begin{matrix} 15 \\ 2 \end{matrix} \right) }=\frac{4}{7},}
which is \latex{\frac{12}{7}} times the chance of winning with one slip. So the chance of winning increases to less than double, if he gets two slips instead of one.
Let us now consider the game as if the \latex{ 2 } slips of Gabriel and the \latex{ 13 } slips of the others were given, and we draw \latex{ 5 } winning slips out of these.
\latex{ 5 } slips out of \latex{ 15 } can be chosen in \latex{\left(\begin{matrix} 15 \\ 5 \end{matrix} \right)} different ways.
Gabriel wins if \latex{ 2 } slips or \latex{ 1 } slip are/is drawn out of his slips; it is
\latex{\left(\begin{matrix} 2 \\ 2 \end{matrix} \right)\times \left(\begin{matrix} 13 \\ 3 \end{matrix} \right)+\left(\begin{matrix} 2 \\ 1 \end{matrix} \right)\times \left(\begin{matrix} 13 \\ 4 \end{matrix} \right)} possibilities.
Based on this the chance of winning for Gabriel is
\latex{ 5 } slips out of \latex{ 15 } can be chosen in \latex{\left(\begin{matrix} 15 \\ 5 \end{matrix} \right)} different ways.
Gabriel wins if \latex{ 2 } slips or \latex{ 1 } slip are/is drawn out of his slips; it is
\latex{\left(\begin{matrix} 2 \\ 2 \end{matrix} \right)\times \left(\begin{matrix} 13 \\ 3 \end{matrix} \right)+\left(\begin{matrix} 2 \\ 1 \end{matrix} \right)\times \left(\begin{matrix} 13 \\ 4 \end{matrix} \right)} possibilities.
Based on this the chance of winning for Gabriel is
\latex{\frac{\left(\begin{matrix} 2 \\ 2 \end{matrix} \right)\times \left(\begin{matrix} 13 \\ 3 \end{matrix} \right)+\left(\begin{matrix} 2 \\ 1 \end{matrix} \right)\times \left(\begin{matrix} 13 \\ 4 \end{matrix} \right) }{\left(\begin{matrix} 15 \\ 5 \end{matrix} \right) }=\frac{4}{7},}
just like in the previous train of thought.
Example 4
There are \latex{ 8 } white balls and \latex{ 7 } red balls in a hat. Without replacement we draw \latex{ 5 } balls from the hat.
What is the probability that among the balls drawn:
What is the probability that among the balls drawn:
- the first \latex{ 3 } are white balls, but the others are not?
- there are \latex{ 3 } white balls and \latex{ 2 } red balls?
Solution (a)
We can answer the question if we also take the order of drawing the balls into account. Thus the number of all cases: out of the \latex{ 15 } balls the first one can be drawn in \latex{ 15 } different ways, the second one in \latex{ 14 }, the third one in \latex{ 13 }, the fourth one in \latex{ 12 } and the fifth one in \latex{ 11 } different ways, so the number of the possibilities is:
\latex{15\times 14\times 13\times 12\times 11=\frac{15!}{10!}=\frac{15!}{(15-5)!}.}
The first \latex{ 3 } white balls can be drawn in \latex{8\times 7\times 6=\frac{8!}{5!}=\frac{8!}{(8-3)!}} different ways, then the two red balls can be drawn in \latex{7\times 6=\frac{7!}{5!}=\frac{7!}{(7-2)!}} different ways, thus the number of the favourable cases is:
\latex{\frac{8!}{(8-3)!}\times \frac{7!}{(7-2)!}.}
The probability that the first \latex{ 3 } balls drawn are white, but the others are not is:
\latex{\frac{\frac{8!}{(8-3)!}\times \frac{7!}{(7-2)!} }{\frac{15!}{(15-5)!} }=\frac{28}{715}\approx 0.0392.}
drawing without replacement if the order of draws also matters
Solution (b)
Now it is not determined which draw was a white ball or a red ball, so the number of all the possible arrangements of the \latex{ 3 } white balls and \latex{ 2 } red balls is:
\latex{\frac{5!}{3!\times 2!}=\left(\begin{matrix} 5 \\ 3 \end{matrix} \right).}
The number of favourable cases obtained in the previous part should be multiplied by this.
So the probability that we drew \latex{ 3 } white balls and \latex{ 2 } red balls is:
So the probability that we drew \latex{ 3 } white balls and \latex{ 2 } red balls is:
\latex{\frac{\left(\begin{matrix} 5 \\ 3 \end{matrix} \right)\times \frac{8!}{(8-3)!}\times
\frac{7!}{(7-2)!} }{\frac{15!}{(15-5)!} }=\frac{280}{715}\approx 0.392.}
The experiment can be considered also in a different way. As in this part of the example it is not restricted that also the order of draws matters, we can also calculate regardless of the order. In this case the probability of drawing \latex{ 3 } white balls and \latex{ 2 } red balls is:
\latex{\frac{\left(\begin{matrix} 8 \\ 3 \end{matrix} \right)\times \left(\begin{matrix}7 \\ 2 \end{matrix} \right) }{\left(\begin{matrix} 15 \\ {5} \end{matrix} \right) }=\frac{280}{715}\approx 0.392.}
We can see that the same result is obtained with the two different methods.
Exercises
{{exercise_number}}. When considering many \latex{ 300 }-page books there are an average of \latex{ 10 } pages with a typographic error in a \latex{ 300 }-page book. What is the probability that if we choose \latex{ 8 } pages at random, then there are no typographic errors on any of them?
{{exercise_number}}. There are \latex{ 15 } jelly, \latex{ 13 } marzipan and \latex{ 12 } caramel flavoured candies in identical wrappers in Santa Claus’s sack. If we draw candies in succession without replacement at random, then what is the probability that
- out of \latex{ 3 } draws all three candies are jelly flavoured ones?
- out of \latex{ 6 } draws \latex{ 2 } are jelly and \latex{ 4 } are marzipan flavoured ones?
- out of \latex{ 3 } draws there are no caramel flavoured ones?
- out of \latex{ 13 } draws all are caramel flavoured ones?
{{exercise_number}}. We draw \latex{ 8 } cards from the Hungarian pack of \latex{ 32 } cards at random without replacement. (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 are no Aces among the cards drawn?
- all \latex{ 4 } Aces are among the cards drawn?
- we drew at least \latex{ 3 } Aces?
- we drew at most \latex{ 3 } Aces?
- Let us play the game of drawing the \latex{ 8 } cards \latex{ 20 } times in succession, and let us write down the number of Aces. What is the frequency and the relative frequency of the number of Aces?
{{exercise_number}}. Presents are drawn at a school event with the help of tombola slips. \latex{ 400 } tombola slips were sold out of which Steve bought \latex{ 3 }. If \latex{ 40 } different presents are drawn, then what is the probability that
- Steve does not win?
- every slip of Steve wins?
- Steve wins at most one present?
- Steve wins at least one present?
{{exercise_number}}. If n presents are drawn out of \latex{ 20 } tombola slips, then what should n be equal to so that the probability of winning at least one present with two slips will be twice the probability of winning with one slip?
{{exercise_number}}. Mary sold the vegetables at the market, and then she took out \latex{ 2 } coins out of her purse at random for the parking. What is the probability that she took out at least \latex{ 1 } pound, if there were \latex{ 10 } one-pound coins, \latex{ 4 50 } pence coins, \latex{ 7 20 } pence coins, \latex{ 6 10 } pence coins, \latex{ 4 5 } pence coins, \latex{ 24 2 } pence coins and \latex{ 17 1 } penny coins in her purse?
Puzzle
Farmer John has \latex{ 8 } horses: \latex{ 4 } brown horses, \latex{ 3 } grey horses and \latex{ 1 } black horse. What is the probability that the horse randomly chosen from them can say about himself that farmer John has another horse with the same colour?
Answer: \latex{ 0 }, because horses cannot speak. If horses could speak, then it was \latex{ 1 }, bevause, it is possible that the black horse lies.
