Вашата кошница е празна
Sampling with replacement
The following examples have a common model; however the exercises sometimes (seemingly) have nothing to do with sampling.
The model got its name from its practical application. The products produced by a certain machine can be defective or non-defective. During a quality check we cannot examine all the products, therefore we pick one at random, we examine it, we replace it, and then we repeat all these several times. In the meantime we count the defective ones. Based on such experiments with statistical methods we can conclude the proportion of the defective products among the products produced by the machine
Example 1
There are \latex{ 10 } one-pound coins and \latex{ 8 } two-pound coins in a hat. We draw \latex{ 5 } coins in succession so that we replace the coins after each draw. What is the probability that
- the first three draws are one-pound coins, and then two draws are twopound coins?
- three draws are one-pound coins and two draws are two-pound coins?
- at least one of the draws is a one-pound coin?
- Perform the draws: Draw \latex{ 5 } coins with replacement \latex{ 50 } times in succession and write down the number of one-pound coins. What are the frequency and the relative frequency of the number of one-pound coins?
As we replace the coins drawn after each draw, the probability that we draw a one-pound coin for every draw is \latex{\frac{10}{18} }, and that we draw a two-pound coin is \latex{\frac{8}{18}}.
Solution (a)
The first three draws are one-pound coins with a probability of \latex{\left(\frac{10}{18} \right)^3}, then two draws are two-pound coins with a probability of \latex{\left(\frac{8}{18} \right)^2}, i.e. the probability that the first three draws are one-pound coins and then two draws are two-pound coins is:
\latex{\left(\frac{10}{18} \right)^3\times \left(\frac{8}{18} \right)^2}.
Solution (b)
In this example now it is not defined which draws had the outcome of a one-pound coin; out of the \latex{ 5 } draws the three one-pound draws can be selected in \latex{\binom{5}{3}} different ways, thus the probability of the three one-pound draws and the two two-pound draws is:
\latex{\binom{5}{3}\times \left(\frac{10}{18} \right)^3\times \left(\frac{8}{18} \right)^2\approx 0.3387 }.
Solution (c)
We can draw at least \latex{ 1 } one-pound coin in the following cases (Figure 4):
number of one-pound coins | number of two-pound coins | probability |
\latex{ 1 } | \latex{ 4 } | \latex{\binom{5}{1}\times \left(\frac{10}{18} \right)^1\times \left(\frac{8}{18} \right)^4 } |
\latex{ 2 } | \latex{ 3 } | \latex{\binom{5}{2}\times \left(\frac{10}{18} \right)^2\times \left(\frac{8}{18} \right)^3 } |
\latex{ 3 } | \latex{ 2 } | \latex{\binom{5}{3}\times \left(\frac{10}{18} \right)^3\times \left(\frac{8}{18} \right)^2} |
\latex{ 4 } | \latex{ 1 } | \latex{\binom{5}{4}\times \left(\frac{10}{18} \right)^4\times \left(\frac{8}{18} \right)^1 } |
\latex{ 5 } | \latex{ 0 } | \latex{\binom{5}{5}\times \left(\frac{10}{18} \right)^5} |
probability
number of one-pound coins
\latex{ 0.5 }
\latex{ 0.02 }
\latex{ 0}
\latex{ 0.12}
\latex{ 0.27}
\latex{ 0.34}
\latex{ 0.21}
\latex{ 0.05}
\latex{ 1}
\latex{ 2}
\latex{ 3}
\latex{ 4}
\latex{ 5}
Figure 4
The probability of drawing at least one one-pound coin is:
\latex{\binom{5}{1}\times \left(\frac{10}{18}\right)^1 \cdot \left(\frac{8}{18}\right)^4 + \binom{5}{2}\times \left(\frac{10}{18}\right)^2 \times \left(\frac{8}{18}\right)^3 + \binom{5}{3}\times \left(\frac{10}{18}\right)^3\times \left(\frac{8}{18}\right)^2 + }
\latex{+ \binom{5}{4}\times \left(\frac{10}{18} \right)^4\times\left(\frac{8}{18} \right)^1+ \binom{5}{5}\times \left(\frac{10}{18} \right)^5\approx 0.9827 }.
It is easier if we calculate the probability of the event “drawing at least \latex{ 1 } one-pound coin” with the help of its complement. If we draw at least \latex{ 1 } one-pound coin, then it is not true that we did not draw a one-pound coin, i.e. it is not true that all five coins drawn are two-pound coins, and its probability is:
\latex{1-\left(\frac{8}{18} \right)^5 }.
Solution (d)
The results of a possible series of draws are shown in the table and in Figure 5.
number of one-pound coins | \latex{ 0 } | \latex{ 1 } | \latex{ 2 } | \latex{ 3 } | \latex{ 4 } | \latex{ 5 } |
frequency | \latex{ 1 } | \latex{ 1 } | \latex{ 6 } | \latex{ 5 } | \latex{ 7 } | \latex{ 0 } |
relative frequency | \latex{ 0.05 } | \latex{ 0.05 } | \latex{ 0.30 } | \latex{ 0.25 } | \latex{ 0.35 } | \latex{ 0.00 } |
probability | \latex{ 0.02 } | \latex{ 0.12 } | \latex{ 0.27 } | \latex{ 0.34 } | \latex{ 0.21 } | \latex{ 0.05 } |
frequency
number of one-pound coins
\latex{ 12 }
\latex{ 10 }
\latex{ 8 }
\latex{ 6 }
\latex{ 4 }
\latex{ 2 }
\latex{ 0 }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 1 }
\latex{ 1 }
\latex{ 6 }
\latex{ 5 }
\latex{ 7 }
Figure 5
Example 2
Galton board:
A ball is rolling down the board shown in Figure 6. At each obstacle the ball bounces to the right or to the left with equal probability, therefore all the paths are equally probable. There are obstacles (pegs) arranged in \latex{ 5 } levels on the board, and the ball can land in \latex{ 5 + 1 } slots in the end. What is the probability that the ball will land in slot number \latex{k (k = 0, 1, 2, 3, 4, 5)}?
Level \latex{ 1 }
Level \latex{ 2 }
Level \latex{ 3 }
Level \latex{ 4 }
Level \latex{ 5 }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
Figure 6
Solution
On each level the ball bounces to the right or to the left with equal probability, therefore all the paths are equally probable, so the classical probability model can be applied.
The ball chooses between \latex{ 2 } possibilities on \latex{ 5 } levels, therefore the total number of cases is: \latex{ 2^5 }.
In order for the ball to land in the leftmost slot, it should bounce to the left at every obstacle, which is possible in only \latex{ 1 } way, so the probability that the ball lands in slot number \latex{ 0 } is: \latex{\frac{1}{2^5}}.
The ball lands in the second slot from the left, if it bounces once to the right, and it bounces \latex{ 5 – 1 } times to the left. The one bounce to the right can be selected out of the \latex{ 5 } bounces in \latex{ 5 } different ways, so the probability that the ball lands in slot number \latex{ 1 } is: \latex{\frac{5}{2^5}}.
In order for the ball to land in slot number \latex{ k } it should bounce to the right \latex{ k } times, and it can be selected out of the \latex{ 5 } bounces in \latex{\binom{5}{k}} different ways.
Pascal’s triangle:
\latex{ 1 }
\latex{ 2 }
\latex{ 6 }
\latex{ 1 }
\latex{ 1 }
\latex{ 4 }
\latex{ 4 }
\latex{ 1 }
\latex{ 1 }
\latex{ 1 }
\latex{ 1 }
\latex{ 3 }
\latex{ 3 }
\latex{ 1 }
\latex{ 1 }
\latex{ 5 }
\latex{ 1 }
\latex{ 10 }
\latex{ 10 }
\latex{ 5 }
\latex{ 1 }
So the probability that the ball will land in slot number \latex{ k } is: \latex{\frac{\binom{5}{k}}{2^5}}.
As the ball will definitely land in one of the slots, we can set up the following equation:
\latex{\frac{\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}}{2^5}=1}.
i.e.
\latex{\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=2^5}.
In general by applying the solution above to \latex{ n } levels:
\latex{\frac{\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+ \dots+\binom{n}{k}+\dots+\binom{n}{n-1}+\binom{n}{n}}{2^n} =1}
i.e.
\latex{\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{k}+\dots+\binom{n}{n}=2^n}.
The problem can be considered also in a different way:
The ball chooses between the two directions at every obstacle with a probability of \latex{\frac{1}{2}} , irrespectively of the direction chosen at the previous obstacle. The probability that the ball bounces to the right at the first \latex{ k } obstacles is: \latex{\left(\frac{1}{2} \right)^k}, then the probability that the ball bounces to the left at \latex{ 5 – k } obstacles is: \latex{\left(\frac{1}{2} \right)^{5-k}}.
Out of the \latex{ 5 } bounces the \latex{ k } bounces to the right can be selected in \latex{\binom{5}{k}} different ways, thus the probability that out of the \latex{ 5 } bounces the ball bounces to the right \latex{ k } times, and it bounces to the left \latex{ 5 – k } times, i.e. it lands in slot number \latex{ k } is:
\latex{\binom{5}{k}\times \left(\frac{1}{2}\right)^k\times \left(\frac{1}{2} \right)^{5-k}},
which naturally equals to the result calculated previously.
Example 3
A robot steps two forward or one backward on the number line with equal probability. What is the probability that when starting from the zero point
- it gets to the point \latex{ 4 } in \latex{ 5 } steps?
- it gets to the point \latex{ 32 } in \latex{ 15 } steps?
- it gets to the point \latex{ –32 } in \latex{ 28 } steps?
- it gets to the point \latex{ 4 } in\latex{ 6 } steps?
- it gets to the point \latex{ k } in \latex{ n } steps?
- Let us simulate the steps of the robot by stepping \latex{ 6 } times with a counter. Let us decide to step forward or backward by flipping a coin. Let us write down how many times we get to each of the points while playing \latex{ 20 } games. What is the frequency and the relative frequency of getting to each of the points?
The robot steps two forward or one backward each time with a probability of \latex{\frac{1}{2}}.
Solution (a)
After trial and error we can figure out that the robot can get to the point \latex{ 4 } in \latex{ 5 } steps by stepping forward 3 times and stepping backward twice. The probability that out of the \latex{ 5 } steps \latex{ 3 } are steps forward and \latex{ 2 } are steps backward is:
\latex{\binom{5}{3}\times \left(\frac{1}{2}\right)^3\times \left(\frac{1}{2} \right)^{2}}.
Solution (b)
The robot can only get to the point \latex{ 30 } by \latex{ 15 } forward steps, i.e. it cannot get to the point \latex{ 32 }, so this probability is \latex{ 0 }.
Solution (c)
The robot can only get to the point \latex{ –28 } in \latex{ 28 } steps backward, i.e. it cannot get to the point \latex{ –32 }, so this probability is \latex{ 0 }.
Solution (d)
The trial and error shows that we cannot get to the point \latex{ 4 } in \latex{ 6 } steps. Let us check which points we can get to in 6 steps and what is the probability in each case:
number of steps forward | number of steps backward | point we get to | probability |
\latex{ 0 } | \latex{ 6 } | \latex{ -6 } | \latex{\left(\frac{1}{2}\right)^6} |
\latex{ 1 } | \latex{ 5 } | \latex{ -3 } | \latex{\binom{6}{1}\times \left(\frac{1}{2}\right)^1\times \left(\frac{1}{2} \right)^{5}} |
\latex{ 2 } | \latex{ 4 } | \latex{ 0 } | \latex{\binom{6}{2}\times \left(\frac{1}{2}\right)^2\times \left(\frac{1}{2} \right)^{4}} |
\latex{ 3 } | \latex{ 3 } | \latex{ 3 } | \latex{\binom{6}{3}\times \left(\frac{1}{2}\right)^3\times \left(\frac{1}{2} \right)^{3}} |
\latex{ 4 } | \latex{ 2 } | \latex{ 6 } | \latex{\binom{6}{4}\times \left(\frac{1}{2}\right)^4\times \left(\frac{1}{2} \right)^{2}} |
\latex{ 5 } | \latex{ 1 } | \latex{ 9 } | \latex{\binom{6}{5}\times \left(\frac{1}{2}\right)^5\times \left(\frac{1}{2} \right)^{1}} |
\latex{ 6 } | \latex{ 0 } | \latex{ 12 } | \latex{\left(\frac{1}{2}\right)^6} |
The points we can get to in \latex{ 6 } steps are written in the table. It can be seen that we cannot get to the point \latex{ 4 } in \latex{ 6 } steps; this probability is \latex{ 0 }. It can be assumed that the points divisible by \latex{ 3 } can be reached in \latex{ 6 } steps when starting from the zero point.
Solution (e)
The robot surely cannot get to the point \latex{ k } in \latex{ n } steps, if \latex{ 2n\lt k } (because if it always steps forward, then at best it can get to the point \latex{ 2n }), or if \latex{k \lt –n} (because if it always steps backward, then it cannot get further backward than \latex{ –k }), thus in these cases the probability of getting to the point \latex{ k } in \latex{ n } steps is \latex{ 0 }.
If \latex{-n\leq k\leq 2n}, then let \latex{ x } be the number of steps forward, let \latex{ y } be the number of steps backward, then the robot makes a total of \latex{x + y = n} steps, and as it steps \latex{ 2 } forward and 1 backward, it gets to the point \latex{2x – y = k.}
By expressing the number of steps forward from the two equations:
\latex{x=\frac{n+k}{3}}.
We can see that an integer solution is obtained for the number of the steps forward only if \latex{ n + k } is divisible by \latex{ 3 }, then the probability of getting to the point \latex{ k } is:
\latex{\binom{n}{\frac{n+k}{3} }\cdot \left(\frac{1}{2} \right)^{\frac{n+k}{3} }\cdot \left(\frac{1}{2} \right)^{\frac{2n-k}{3} }= \binom{n}{\frac{n+k}{3} }\cdot \frac{1}{2^n} }.
If \latex{ n + k } is not divisible by \latex{ 3 }, then the robot gets to the point \latex{ k } in \latex{ n } steps with a probability of \latex{ 0 }.
Solution (f)
By playing the \latex{ 6 } steps of the robot \latex{ 50 } times based on a possible series of flips:
\latex{ -6 } | \latex{ -3 } | \latex{ 0 } | \latex{ 3 } | \latex{ 6 } | \latex{ 9 } | \latex{ 12 } | |
frequency | \latex{ 0 } | \latex{ 1 } | \latex{ 15 } | \latex{ 19 } | \latex{ 13 } | \latex{ 2 } | \latex{ 0 } |
relative frequency | \latex{ 0 } | \latex{ 0.02 } | \latex{ 0.03 } | \latex{ 0.38 } | \latex{ 0.26 } | \latex{ 0.04 } | \latex{ 0 } |
probability | \latex{ 0.02 } | \latex{ 0.09 } | \latex{ 0.235 } | \latex{ 0.31 } | \latex{ 0.235 } | \latex{ 0.09 } | \latex{ 0.02 } |
Exercises
{{exercise_number}}. Elmer walks his dog every evening on the streets shown in the figure. Starting at the square \latex{ A } he follows the dog, who chooses between the directions South and East at each corner at random. They always walk \latex{ 10 } corners, and then they turn back.
What is the probability that their path leads through
- the square \latex{ B };
- the square \latex{ C };
- the square \latex{ D };
- the square \latex{ E }?
North
\latex{ A }
\latex{ D }
\latex{ E }
\latex{ B }
\latex{ C }
{{exercise_number}}. We flip \latex{ 5 } coins: a \latex{ 1 } penny, a \latex{ 2 } pence, a \latex{ 5 } pence, a \latex{ 10 } pence and a \latex{ 20 } pence coin at once.
- Play the game of flipping the \latex{ 5 } coins \latex{ 50 } times, and before each flip make a guess about the number of heads. Write down the number of heads and also how many times the guess is correct. What is the frequency and the relative frequency of the number of heads?
- What is the probability that out of the \latex{ 5 } coins \latex{ 3 } land on heads?
- What is the probability that there are at least \latex{ 3 } heads?
- How do the answers given to the previous question change, if we flip one coin \latex{ 5 } times in succession in one experiment?
{{exercise_number}}. At the ice hockey matches the goaltender of the Finnish team stops a shot of the puck with a probability of \latex{ 0.85 }.
- What is the probability that if \latex{ 25 } shots are on goal in the first period, then no goals will be scored in the Finnish net?
- At least how many times shall the opponent shoot on goal so that the probability of scoring a goal will be at least \latex{ 75 }%?
{{exercise_number}}. We draw \latex{ 8 } cards from the Hungarian pack of \latex{ 32 } cards in succession so that we replace the cards drawn after each draw. What is the probability that
- we draw Hearts at least \latex{ 7 } times?
- we do not draw Hearts?
- we draw Aces at most \latex{ 3 } times?
{{exercise_number}}. Three dice are rolled at once. What is the probability that we roll at most \latex{ 4 } with at least two dice?
{{exercise_number}}. There are \latex{ 3 } letters (of the English alphabet) and \latex{ 3 } digits in the Hungarian car number plates. We see a car in the street and we note its number plate. What is the probability that there are two \latex{ 0 } digits on its number plate, if we assume that all the digits are equally probable (we disregard the last incomplete thousand, i.e. \latex{ 000 })?
{{exercise_number}}. The final of a football championship is played by the teams Timbertoes and Shortlegs. The score is still tied after extra time; therefore there will be \latex{ 5 } penalty shots for each team to decide who the champion will be. If the Timbertoes score a penalty shot with a chance of \latex{ 80 }%, and the Shortlegs score a penalty shot with a chance of \latex{ 85 }%, then what is the probability that the Timbertoes win \latex{ 5 : 3 }? (The penalty shoot-out is continued even if it is already obvious that one of the teams is the winner.)
{{exercise_number}}. There are \latex{ 35 } numbers on the wheel of roulette: \latex{ 17 } numbers are red, \latex{ 17 } are black, and the \latex{ 0 } is green. If we spin \latex{ 10 } times in succession, then what is the probability that
- we spin red numbers and black numbers in turns?
- the first five spins are red and the second five spins are black?
- five spins are red and five spins are black?
{{exercise_number}}. \latex{ 15 } passengers get in \latex{ 4 } railway carriages at random; every passenger gets in a railway carriage with a probability of \latex{ 0.25 }. What is the probability that
- all of them get in the first carriage?
- all of them get in the same carriage?
- exactly two of them get in the first carriage?
- at most two of them get in the first carriage?
{{exercise_number}}. A flea is jumping around on the number line. It jumps forward \latex{ 6 } units long or backward \latex{ 4 } units long with equal probability. To which points can it get with \latex{ 5 } jumps starting from the \latex{ 0 } point, and what is the probability in each case?
{{exercise_number}}. There are two fleas, Alistair and Ben at the zero minute at the points \latex{ 0 } and \latex{ 3 } on the number line. Both fleas jump one unit forward or backward in every integer minute with equal probability. What is the probability that
- they swap places in the second minute?
- they jump to the same point in the fifth minute?
Puzzle
Three closed doors are shown to the player in a game. There are goats behind two of the doors, and there is a prize car behind one of the doors. The player chooses a door. This door is not opened, but one of the other two doors is opened. There is a goat behind it, and it is shown to the player. After this the player can decide whether to open the originally chosen door or the other one.
According to Andrew as there is a goat behind one of the remaining two doors and there is a car behind the other one, the door with the car behind can be chosen with a probability of \latex{\frac{1}{2}}. According to Bruce we chose the door with the car behind it with a probability of \latex{\frac{1}{3}} at first, therefore the car is behind the other two doors with a probability of \latex{\frac{2}{3}}. As one of these (the one there is no car behind) is opened, the car is behind the remaining door with a probability of \latex{\frac{2}{3}}, so it is worth choosing that one.
Who do you agree with? Experiment: which tactic do we win with a greater chance: if we stick to our first choice or if we change our mind?