In a multi-step random experiment, the order of the sequence of possible outcomes can matter or not matter.
|2| marbles are drawn from a bag containing |3| blue marbles and |1| red marble.
The order of the marbles matters
The probability of getting a red marble |(R)| followed by a blue marble |(B)| is ||\dfrac{3}{16}.|| We obtain this probability as follows:
||\begin{alignat}{2}P(R,B)&=\dfrac{1}{4}&&\times\dfrac{3}{4}\\&=&&\!\dfrac{3}{16}\end{alignat}||
The event "getting a red marble followed by a blue marble" |(R,B)| is not the same as the event "getting a blue marble followed by a red marble" |(B,R).|
The order of the marbles does not matter
The probability of obtaining a red marble |(R)| and a blue marble |(B)| is ||\dfrac{6}{16}.|| We obtain this probability as follows:
||\begin{alignat}{2}P(B\text{ and }R)&=P(B,R)&&+P(R,B)\\&=\dfrac{1}{4}\times\dfrac{3}{4}&&+\dfrac{3}{4}\times\dfrac{1}{4}\\&=\!\phantom{\dfrac{1}{1}}\dfrac{3}{16}&&+\!\phantom{\dfrac{1}{1}}\dfrac{3}{16}\\&=&&\!\dfrac{6}{16}\end{alignat}||
The event "getting a red marble and a blue marble" |(R,B)| is the same as the event "getting a blue marble and a red marble" |(B,R).| This is why they must be added together to calculate |P(B\ \text{et}\ R).|
It is not always clear whether order matters or not in a random experiment. The context must always be analyzed to deduce this.
When analyzing a multi-step random experiment, where order matters or does not matter, it is also important to check whether the experiment is with or without replacement.
A multi-step random experiment where order matters is an experiment where the sequence of the outcomes is important.
Pascal draws |2| marbles from a bag that contains |7| coloured marbles. The 1st marble he draws is purple and the 2nd is green.
If the order in which Pascal draws the marbles matters, then the event "draw a purple marble followed by a green marble" is not the same as the event "draw a green marble followed by a purple marble."
In this case, the sample space is represented by the following set, which contains more possible outcomes than if the order had not mattered.
||\Omega=\left\lbrace\begin{matrix}\begin{aligned}&(P,P)&&(P,B)&&(P,G)&&(P,O)\\&(B,P)&&(B,B)&&(B,G)&&(B,O)\\&(G,P)&&(G,B)&&(G,G)&&(G,O)\\&(O,P)&&(O,B)&&(O,G)&&(O,O)\end{aligned}\end{matrix}\right\rbrace||
To find the probability of an event in a multi-step random experiment where the order matters, the multiplication rule can be used.
Isabelle participates in a contest for a chance to win a computer. Isabelle must draw |3| coloured cards from an opaque jar containing |3| green and |5| red cards. To win the contest, Isabelle must draw a green card, then a red card, then a green card in that order.
What is the probability that Isabelle wins the computer?
This random experiment is done with replacement, since Isabelle has to put the card back in the jar after each draw. Therefore, the total number of cards does not change from one step to the next. Furthermore, the order of the cards drawn matters in order to win the computer.
First, the probabilities of all possible outcomes of the first step are calculated.
||\begin{align}P(\text{Green})&=\dfrac{\text{Number of Green Cards}}{\text{Total Number of Cards}}\\&=\dfrac{3}{8}\\\\P(\text{Red})&=\dfrac{\text{Number of Red Cards}}{\text{Total Number of Cards}}\\&=\dfrac{5}{8}\end{align}||
Since the experiment is performed with replacement, the probabilities for the 2nd and 3rd steps stay the same.
Then, the probability associated with each possible outcome is calculated by multiplying the probability of each step.
||\begin{align}P(\text{Green, Green, Green})&=\dfrac{3}{8}\times\dfrac{3}{8}\times\dfrac{3}{8}=\dfrac{27}{512}\\P(\text{Green, Green, Red})&=\dfrac{3}{8}\times\dfrac{3}{8}\times\dfrac{5}{8}=\dfrac{45}{512}\\P(\text{Green, Red, Red})&=\dfrac{3}{8}\times\dfrac{5}{8}\times\dfrac{5}{8}=\dfrac{75}{512}\\P(\text{Red, Red, Red})&=\dfrac{5}{8}\times\dfrac{5}{8}\times\dfrac{5}{8}=\dfrac{125}{512}\end{align}||
To help answer the question, a tree diagram can be constructed.
The possible outcome sought is the outcome |(\text{Green, Red, Green}).| The outcomes |(\text{Green, Green, Red})| and |(\text{Red, Green, Green})| are not considered, since they contain the correct colours, but do not correspond to the precise order that allows Isabelle to win the contest.
Answer: The probability that Isabelle wins the computer is |\dfrac{45}{512},| or about |9\ \%.|
When we are interested in the number of possible outcomes of a random experiment where order matters, the arrangement formulas can also be used.
A multi-step random experiment where the order does not matter is an experiment where the sequence of outcomes is not important.
Pascal draws |2| marbles from a bag containing |7| coloured marbles. The 1st marble he draws is purple and the 2nd is green.
If the order in which Pascal draws the marbles does not matter, then the event "draw a purple marble followed by a green marble" is the same as the event "draw a green marble followed by a purple marble."
In this case, the sample space is represented by the following set, which contains fewer possible outcomes than if the order had mattered.
||\begin{align}\Omega&=\left\lbrace\begin{matrix}\begin{aligned}&(P,P)&&(P,B)&&(P,G)&&(P,O)\\&\cancel{(B,P)}&&(B,B)&&(B,G)&&(B,O)\\&\cancel{(G,P)}&&\cancel{(G,B)}&&(G,G)&&(G,O)\\&\cancel{(O,P)}&&\cancel{(O,B)}&&\cancel{(O,G)}&&(O,O)\end{aligned}\end{matrix}\right\rbrace\\\\&=\left\lbrace\begin{matrix}\begin{aligned}&(P,P)&&(P,B)&&(P,G)&&(P,O)\\&&&(B,B)&&(B,G)&&(B,O)\\&&&&&(G,G)&&(G,O)\\&&&&&&&(O,O)\end{aligned}\end{matrix}\right\rbrace
\end{align}||
To find the probability of an event in a multi-step random experiment where the order does not matter, the multiplication rule can be used.
Isabelle takes part in a contest for a chance to win a computer. During the contest, Isabelle must draw |3| coloured cards from an opaque jar containing |3| green and |5| red cards. In order to win the contest, Isabelle must draw |2| green cards and |1| red card.
What is the probability that Isabelle wins the computer?
This random experiment is carried out with replacement, since Isabelle must put the card back into the jar after each draw. Therefore, the total number of cards does not change from one step to the next. Furthermore, since the sequence of the cards picked is not important, the order does not influence the winning of the computer.
First, the probability of each possible choice at the 1st step is calculated.
||\begin{align}P(\text{Green})&=\dfrac{\text{Number of Green Cards}}{\text{Total Number of Cards}}\\&=\dfrac{3}{8}\\\\P(\text{Red})&=\dfrac{\text{Number of Red Cards}}{\text{Total Number of Cards}}\\&=\dfrac{5}{8}\end{align}||
Since the experiment is with replacement, the probabilities for the 2nd and 3rd steps stay the same.
Next, the probability associated with each possible outcome is calculated by multiplying the probability at each step.
||\begin{align}P(\text{Green, Green, Green})&=\dfrac{3}{8}\times\dfrac{3}{8}\times\dfrac{3}{8}=\dfrac{27}{512}\\P(\text{Green, Green, Red})&=\dfrac{3}{8}\times\dfrac{3}{8}\times\dfrac{5}{8}=\dfrac{45}{512}\\P(\text{Green, Red, Red})&=\dfrac{3}{8}\times\dfrac{5}{8}\times\dfrac{5}{8}=\dfrac{75}{512}\\P(\text{Red, Red, Red})&=\dfrac{5}{8}\times\dfrac{5}{8}\times\dfrac{5}{8}=\dfrac{125}{512}\end{align}||
To help answer the question, a tree diagram can be constructed.
The favourable outcomes are |(\text{Green, Green, Red}),| |(\text{Green, Red, Green})| and |(\text{Red, Green, Green})| since they all correspond to the outcomes that allow Isabelle to win the contest. All that is left is to add their probabilities.
||\begin{align}P(\text{2 Green and 1 Red})&=\dfrac{45}{512}+\dfrac{45}{512}+\dfrac{45}{512}\\&=\dfrac{135}{512}\end{align}||
Answer: The probability that Isabelle wins the computer is |\dfrac{135}{512},| or about |26\ \%.|
When we are interested in the number of possible outcomes of a random experiment where order does not matter, we can also use the combination formulas.
Here is an example of a random experiment without replacement where the order does not matter.
On his birthday, Forrest is given a box of |10| chocolates by his mother. She knows that Forrest would eat all the chocolates at once, so his mother suggests that he close his eyes and eat |3| at random. In the box, there are |4| dark chocolates |(D),| |4| milk chocolates |(M)| and |2| white chocolates |(W).|
What is the probability that Forrest will eat one chocolate of each kind?
This random experiment is carried out without replacement, since Forrest does not return the chocolates to the box after eating them. Therefore, the total number of chocolates changes from one step to the next. As well, the order of the chocolates does not matter, since we are only interested in the type of chocolates Forrest eats.
We start by calculating the probability of each possible choice of the 1st step.
||\begin{align}P(N)&=\dfrac{\text{Number of Dark Chocolates}}{\text{Total Number of Chocolates}}\\&=\dfrac{4}{10}\\\\P(L)&=\dfrac{\text{Number of Milk Chocolates}}{\text{Total Number of Chocolates}}\\&=\dfrac{4}{10}\\\\P(B)&=\dfrac{\text{Number of White Chocolates}}{\text{Total Number of Chocolates}}\\&=\dfrac{2}{10}\end{align}||
Since this experiment is carried out without replacement, the probabilities change at the 2nd step. This is because there is |1| less chocolate in the box, which implies that the denominator of the fractions changes from |10| to |9.| As well, the numerator decreases by |1| depending on the result obtained in the first step. The same is true for step 3.
To help answer the question, a tree diagram can be constructed.
Next, the possible outcomes that correspond to the event "eat one of each type of chocolate" are identified in the tree diagram. These possible outcomes are as follows:
||\begin{aligned}&(D,M,W)&&(D,W,M)\\&(M,D,W)&&(M,W,D)\\&(W,D,M)&&(W,M,D)\end{aligned}||
The probabilities of each of these outcomes are calculated first using the multiplication rule.
||\begin{align}P(D,M,W)&=\dfrac{4}{10}\times\dfrac{4}{9}\times\dfrac{2}{8}=\dfrac{32}{720}\\P(D,W,M)&=\dfrac{4}{10}\times\dfrac{2}{9}\times\dfrac{4}{8}=\dfrac{32}{720}\\P(M,D,W)&=\dfrac{4}{10}\times\dfrac{4}{9}\times\dfrac{2}{8}=\dfrac{32}{720}\\P(M,W,D)&=\dfrac{4}{10}\times\dfrac{2}{9}\times\dfrac{4}{8}=\dfrac{32}{720}\\P(W,D,M)&=\dfrac{2}{10}\times\dfrac{4}{9}\times\dfrac{4}{8}=\dfrac{32}{720}\\P(W,M,D)&=\dfrac{2}{10}\times\dfrac{4}{9}\times\dfrac{4}{8}=\dfrac{32}{720}\\\end{align}||
All the fractions are added together.
||\begin{align}P\left(\begin{gathered}\text{Eating one chocolate}\\\text{of each kind}\end{gathered}\right)&=\dfrac{32}{720}+\dfrac{32}{720}+\dfrac{32}{720}+\dfrac{32}{720}+\dfrac{32}{720}+\dfrac{32}{720}\\&=6\times\dfrac{32}{720}\\&=\dfrac{192}{720}\end{align}||
Finally, reduce the fraction.
||\begin{align}P\left(\begin{gathered}\text{Eating one chocolate}\\\text{of each kind}\end{gathered}\right)&=\dfrac{192\boldsymbol{\color{#ec0000}{\div48}}}{720\boldsymbol{\color{#ec0000}{\div48}}}\\&=\dfrac{4}{15}\end{align}||
Answer: The probability that Forrest eats one chocolate of each kind is |\dfrac{4}{15},| or |26.\overline{6}\ \%.|