3.7 Joint Probabilities

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

1. What is the probability that a randomly selected person is a cell phone user and had no speeding violations in the last year?
2. What is the probability that a randomly selected person had a speeding violation in the last year and does not use a cell phone?

Solution:

1. [latex]\displaystyle{\mbox{Probability}=\frac{\mbox{number of  cell phone users and no violations}}{\mbox{total number in study}}=\frac{280}{755}}[/latex]
2. [latex]\displaystyle{\mbox{Probability}=\frac{\mbox{number of violations and not cell phone users}}{\mbox{total number in study}}=\frac{45}{755}}[/latex]

Which of the following are repeated trial experiments?  For the repeated trial experiments, identify the trial and the number of repetitions.

1. Finding the probability of rolling an odd number in the roll of die.
2. Finding the probability of drawing five spades from a deck of cards.
3. Finding the probability a randomly selected person has blue eyes and blond hair.
4. Finding the probability that three women from a pool of candidates are selected for a committee.
5. Finding the probability that a student answers ten multiple choice questions correctly.

Solution:

1. Single trial experiment.  The die is rolled one time.
2. Repeated trial experiment.  The trial is selecting a card from the deck and this trial is repeated five times.
3. Single trial experiment.  A single person is selected.
4. Repeated trial experiment.  The trial is selecting a women from the candidate pool and this trial is repeated three times.
5. Repeated trial experiment.  The trial is answering an individual question and this trial is repeated ten times.

For each of the following determine if the experiment is done with or without replacement.

1. Flipping a coin three times.
2. Selecting three women for a committee from a pool of candidates.
3. Drawing five cards from deck of cards.
4. Rolling a die six times.
5. Selecting the members of the student executive committee from the student council.

Solution:

1. With replacement.  The probability of heads or tails stays the same with each flip.
2. Without replacement.  In this case, we want three different women on the committee, so we must select them without replacement.  (Selecting with replacement would mean a possibility of the same women being selected three times and then the committee would consist of just a single person).
3. Depending on the context, this could be with or without replacement.  If each card is replaced after it is selected, this would be with replacement.  If each card is not replaced after it is selected, this would be without replacement.  In this situation, the question would probably include a statement about whether the cards are drawn with or without replacement.
4. With replacement.  The probability of rolling any of the numbers stays the same with each roll of the die.
5. Without replacement.  In this case, we want the members of the executive committee to be all different, so we must select them without replacement.

A small local high school has 25 students in its graduating class.  18 of the students are going to college next year and the remaining 7 are not going to college next year.  Suppose two students are selected at random from the graduating class.

1. What is the probability that both students are going to college next year?
2. What is the probability that exactly one of the students is going to college next year?

Solution:

This is a repeated trial experiment.  A trial is selecting a student and there are two trials.  The assumption here is that the experiment is done without replacement because we do not want to get the same student twice.

1. We want to get college-bound students on both trials.  In other words, college-bound student on trial one AND college-bound student on trial two.  On the first trial, the probability of getting a college-bound student is [latex]\displaystyle{\frac{18}{25}}[/latex].  We are selecting without replacement, so after the first trial we assume that we have removed one of the college-bound students.  This means that on the second trial, there are only 24 students to pick from (one student was removed on the first trial) and there are only 17 college-bound students left (on the first trial we removed one of the 18 college-bound students).  So on the second trial, the probability of getting a college-bound student is [latex]\displaystyle{\frac{17}{24}}[/latex].  The probability of getting two college-bound students is

   [latex]\begin{eqnarray*}  \\ \mbox{Probability} & = & \frac{18}{25} \times \frac{17}{24}=0.51 \\ \\ \end{eqnarray*}[/latex]

2. We want one college-bound student (denoted [latex]C[/latex]) and one non college-bound student (denoted [latex]N[/latex]).  In this case, we have to think about the order of the selections—there is a difference between college-bound on trial one, non-college bound on trial two([latex]CN[/latex]) and non-college bound on trial one, college bound on trial two ([latex]NC[/latex]).  All possible orders must be accounted for when we calculate the probability.  One of the two possible orders must occur:  [latex]CN[/latex] OR [latex]NC[/latex].  For each of the individual orders, we multiply the probabilities as we move from trial to trial.  The “or” means that we add the probabilities of the different orders.  In other words:

   [latex]\begin{eqnarray*} \\ \mbox{Probability} & = & \mbox{Probability of }CN+\mbox{Probability of }NC \\ \end{eqnarray*}[/latex]

   For the [latex]CN[/latex] order (college-bound on trial one, non college-bound on trial two), we want a college-bound student on trial one and the probability of getting a college-bound student is [latex]\displaystyle{\frac{18}{25}}[/latex].  We are selecting without replacement, so after the first trial we assume that we have removed one of the college-bound students.  This means that on the second trial, there are only 24 students to pick from (a college-bound student was removed on the first trial) and there are 7 non college-bound students (none of the non college-bound students were removed after the first trial).  So on the second trial, the probability of getting a non college-bound student is [latex]\displaystyle{\frac{7}{24}}[/latex]. So the probability of getting the [latex]CN[/latex] order is [latex]\displaystyle{\frac{18}{25} \times \frac{7}{24}}[/latex].

   Similarly for the [latex]NC[/latex] order (non college-bound on trial one, college-bound on trial two), we want a non college-bound student on trial one and the probability of getting a non college-bound student is [latex]\displaystyle{\frac{7}{25}}[/latex].  We are selecting without replacement, so after the first trial we assume that we have removed one of the non college-bound students.  This means that on the second trial, there are only 24 students to pick from (a non college-bound student was removed on the first trial) and there are 18 college-bound students (none of the college-bound students were removed after the first trial).  So on the second trial, the probability of getting a college-bound student is [latex]\displaystyle{\frac{18}{24}}[/latex]. So the probability of getting the [latex]NC[/latex] order is [latex]\displaystyle{\frac{7}{25} \times \frac{18}{24}}[/latex].

   The probability of getting exactly one college bound student is

   [latex]\begin{eqnarray*}  \mbox{Probability} & = & \mbox{Probability of }CN+\mbox{Probability of }NC \\ & = & \left(\frac{18}{25} \times \frac{7}{24}\right)+\left(\frac{7}{24} \times \frac{18}{25} \right) \\ & = & 0.42 \end{eqnarray*}[/latex]

Suppose a fair die is rolled two times.

1. What is the probability of getting two 5’s?
2. What is the probability of getting exactly one 2 and one 6 in the two rolls?

Solution:

This is a repeated trial experiment.  A trial is rolling a die and there are two trials.  The trials are independent (what happens on the first roll does not affect what happens on the second roll).

1. We want to get a 5 on both rolls.  In other words, a 5 on roll one AND a 5 on roll two.  On the first roll, the probability of getting a 5 is [latex]\displaystyle{\frac{1}{6}}[/latex].  On the second roll, the probability of getting a 5 is [latex]\displaystyle{\frac{1}{6}}[/latex].  Because the rolls are independent, the probability of getting a 5 on the second roll is not affected by what happens on the first roll.  The probability of getting two 5’s is

   [latex]\begin{eqnarray*}  \\ \mbox{Probability} & = & \frac{1}{6} \times \frac{1}{6}=\frac{1}{36} \\ \\ \end{eqnarray*}[/latex]

2. We want one 2 and one 6. In this case, we have to think about the order of the rolls—there is a difference between 2 on roll one, 6 on roll two and 6 on roll one, 2 on roll two.  All possible orders must be accounted for when we calculate the probability. One of the two possible orders must occur:  [latex]26[/latex] OR [latex]62[/latex].  For of the individual orders, we multiply the probabilities as we move from trial to trial.  The “or” means that we add the probabilities of the different orders.  In other words,

   [latex]\begin{eqnarray*} \\ \mbox{Probability} & = & \mbox{Probability of }26+\mbox{Probability of }62\end{eqnarray*}[/latex]

   For the [latex]26[/latex] order (2 on roll one, 6 on roll two), the probability of getting a 2 on roll one is [latex]\displaystyle{\frac{1}{6}}[/latex] and the probability of getting a 6 on roll two is [latex]\displaystyle{\frac{1}{6}}[/latex]. So the probability of getting the [latex]26[/latex] order is [latex]\displaystyle{\frac{1}{6} \times \frac{1}{6}}[/latex].

   Similarly for the [latex]62[/latex] order (6 on roll one, 2 on roll two), the probability of getting a 6 on roll one is [latex]\displaystyle{\frac{1}{6}}[/latex] and the probability of getting a 2 on roll two is [latex]\displaystyle{\frac{1}{6}}[/latex].  So the probability of getting the [latex]62[/latex] order is [latex]\displaystyle{\frac{1}{6} \times \frac{1}{6}}[/latex].

   The probability of getting exactly one 2 and one 6 is

   [latex]\begin{eqnarray*}  \mbox{Probability} & = & \mbox{Probability of }26+\mbox{Probability of }62 \\ & = & \left(\frac{1}{6} \times \frac{1}{6}\right)+\left(\frac{1}{6} \times \frac{1}{6} \right) \\ & = & \frac{1}{18} \end{eqnarray*}[/latex]

A box contains 5 red cards and 12 white cards.  Suppose three cards are drawn at random from the box without replacement.

1. What is the probability that all three cards are white?
2. What is the probability that exactly one of the cards is red?
3. What is the probability that at least one card is red?
4. What is the probability that at most one card is white?

Solution:

This is a repeated trial experiment.  A trial is selecting a card and there are three trials.  There are 17 cards in the box.

1. We want to get a white card on all three draws.  In other words, white on draw one AND white on draw two AND white on draw three.  On the first draw, the probability of getting a white card is [latex]\displaystyle{\frac{12}{17}}[/latex].  We are selecting without replacement, so after the first draw we assume that we removed a white card from the box.  This means that on the second draw, there are only 16 cards left in the box (one card was removed on the first draw) and there are only 11 white cards left (on the first draw we removed one of the 12 white cards).  So on the second draw, the probability of getting a white card is [latex]\displaystyle{\frac{11}{16}}[/latex].  After the second draw we assume that we removed white cards from the box on draws one and two.  This means that on the third draw, there are only 15 cards left in the box (two cards were removed on the first two draws) and there are only 10 white cards left (white cards were removed on draws one and two).  So on the third draw, the probability of getting a white card is [latex]\displaystyle{\frac{10}{15}}[/latex]. The probability of getting three white cards is

   [latex]\begin{eqnarray*}  \\ \mbox{Probability} & = & \frac{12}{17} \times \frac{11}{16} \times \frac{10}{15}=0.3235 \\ \\ \end{eqnarray*}[/latex]

2. We want one red card ([latex]R[/latex]), so the other two cards must be white ([latex]W[/latex]).  In this case, we have to think about the order of the selection.  All possible orders must be accounted for when we calculate the probability.  One of three possible orders must occur:  [latex]RWW[/latex] OR [latex]WRW[/latex] OR [latex]WWR[/latex].  For each of the individual orders, we multiply the probabilities as we move from draw to draw.  The “or” means that we add the probabilities of the different orders.  In other words,

   [latex]\begin{eqnarray*} \\ \mbox{Probability} & = & P(RWW)+P(WRW)+P(WWR)\end{eqnarray*}[/latex]

   For the [latex]RWW[/latex] order, the probability of red on draw one is [latex]\displaystyle{\frac{5}{17}}[/latex].  We are selecting without replacement, so after the first trial we assume that we have removed one of the red cards. This means that on the second draw, there are only 16 cards left in the box (one card was removed on the first draw) and all 12 white cards are left (on the first draw we removed a red card).  So on the second draw, the probability of getting a white card is [latex]\displaystyle{\frac{12}{16}}[/latex].  After the second draw we assume that we removed a red card on draw one and a white card on draw two.  This means that on the third draw, there are only 15 cards left in the box (two cards were removed on the first two draws) and there are only 11 white cards left (one white card was removed on draw two).  So on the third draw, the probability of getting a white card is [latex]\displaystyle{\frac{11}{15}}[/latex].  So the probability of getting the [latex]RWW[/latex] order is [latex]\displaystyle{\frac{5}{17} \times \frac{12}{16} \times \frac{11}{15}}[/latex].

   Using similar logic, the probability of getting the [latex]WRW[/latex] order is [latex]\displaystyle{\frac{12}{17} \times \frac{5}{16}\times\frac{11}{15}}[/latex] and the probability of getting the [latex]WWR[/latex] order is [latex]\displaystyle{\frac{12}{17} \times \frac{11}{16} \times \frac{5}{15}}[/latex].

   The probability of getting exactly one red card is

   [latex]\begin{eqnarray*} \mbox{Probability} & = & P(RWW)+P(WRW) +P(WWR) \\ & = & \left(\frac{5}{17} \times \frac{12}{16} \times \frac{11}{15}\right)+\left(\frac{12}{17} \times \frac{5}{16} \times \frac{11}{15} \right)+\left(\frac{12}{17} \times \frac{11}{16} \times \frac{5}{15} \right) \\ & = & 0.4853 \\ \\ \end{eqnarray*}[/latex]

3. We want at least one red card in the three draws.  This means we can have exactly one red card or exactly two red cards or exactly three red cards.  As before, we have to think about the order of the selection.  All possible orders must be accounted for when we calculate the probability.  Here, there are seven possible ways of getting at least one red card:  [latex]RWW[/latex] OR [latex]WRW[/latex] OR [latex]WWR[/latex] OR [latex]RRW[/latex] OR [latex]RWR[/latex] OR [latex]RRW[/latex] OR [latex]RRR[/latex].  Of course, we could work out the probabilities of each of these orders and add them all up.  But there is a faster way to find this probability—use the complement.  The complement of “at least one red card” is “exactly zero red cards.”  When we look at the seven possible orders that make up the “at least one red card” event, the complement consists of all of the missing orders.  In this case there is only one missing order, [latex]WWW[/latex], which is the event “exactly zero red cards.”  Using the complement, the probability of at least one red card is

   [latex]\begin{eqnarray*} \\  P(\mbox{at least one red card}) & = & 1-P(\mbox{exactly zero red card}) \\ & = & 1-P(WWW)\end{eqnarray*}[/latex]

   In part 1 of this question, we found the probability of [latex]WWW[/latex]:  [latex]\displaystyle{\frac{12}{17} \times \frac{11}{16} \times \frac{10}{15}}[/latex].  So the probability of at least one red card is

   [latex]\begin{eqnarray*}   P(\mbox{at least one red card}) & = & 1-P(WWW) \\ & = & 1-\left(\frac{12}{17} \times \frac{11}{16} \times \frac{10}{15}\right) \\ & = & 0.6765 \\ \\ \end{eqnarray*}[/latex]

4. We want at most one white card in the three draws.  This means we can have exactly zero white cards or exactly one white card.  As before, we have to think about the order of the selection.  All possible orders must be accounted for when we calculate the probability.  Here, there are four possible ways of getting at most one white card:  [latex]RRR[/latex] OR [latex]RRW[/latex] OR [latex]RWR[/latex] OR [latex]WRR[/latex].  Using similar logic to above, the probability of getting the [latex]RRR[/latex] order is [latex]\displaystyle{\frac{5}{17} \times \frac{4}{16}\times \frac{3}{15}}[/latex],  the probability of getting the [latex]RRW[/latex] order is [latex]\displaystyle{\frac{5}{17} \times \frac{4}{16}\times\frac{12}{15}}[/latex],the probability of getting the [latex]RWR[/latex] order is [latex]\displaystyle{\frac{5}{17} \times \frac{12}{16}\times\frac{5}{15}}[/latex], and the probability of getting the [latex]WRR[/latex] order is [latex]\displaystyle{\frac{12}{17} \times \frac{5}{16}\times\frac{4}{15}}[/latex].  The probability of getting at most one white card is

   [latex]\begin{eqnarray*}\\ \mbox{Probability} & = & P(RRR)+P(RRW)+P(RWR)+P(WRR) \\ & = & \left(\frac{5}{17} \times \frac{4}{16} \times \frac{3}{15}\right)+\left(\frac{5}{17} \times \frac{4}{16} \times \frac{12}{15} \right)+\left(\frac{5}{17} \times \frac{12}{16} \times \frac{4}{15} \right) \\ & & +  \left(\frac{12}{17} \times \frac{5}{16} \times \frac{4}{15} \right)\\ & = & 0.1912 \end{eqnarray*}[/latex]

A company produces a popular brand of sports drink.  The company is currently running a contest where winning symbols are placed under the bottle caps.  7% of all the bottle caps contain winning symbols.  You buy three bottles of the sports drink.

1. What is the probability that all bottles have winning symbols?
2. What is the probability that exactly one of the bottles has a winning symbol?
3. What is the probability that at least one bottle has a winning symbol?

Solution:

This is a repeated trial experiment.  A trial is selecting a bottle and there are three trials.  This is an experiment without replacement (you do not want to select the same bottle three times).  However, because the population of bottles is very, very large, we can treat the experiment as if the selections are made with replace.  This means that we can treat the selection of the bottles as independent and so the probability of getting a winning bottle will be 7% on every draw.

1. We want to get a winning symbol on all three bottles.  In other words, win on bottle one AND win on bottle two AND win on bottle three.  The probability of winning on the first bottle is 7%, the probability of winning on the second bottle is 7%, and the probability of winning on the third bottle is 7%.  Because we can treat the selections as independent, the probability of winning does not change from draw to draw.  The probability of getting three winning bottles is

   [latex]\begin{eqnarray*}  \\ \mbox{Probability} & = & 0.07 \times 0.07 \times 0.07=0.0003 \\ \\ \end{eqnarray*}[/latex]

2. We want one winning bottle ([latex]W[/latex]), so the other two bottles must be non-winners ([latex]N[/latex]).  The probability of winning on any bottle is 7%, so the probability of losing on any bottle is 93%.  In this case, we have to think about the order of the selection.  All possible orders must be accounted for when we calculate the probability.  One of the three possible orders must occur:  [latex]WNN[/latex] OR [latex]NWN[/latex] OR [latex]NNW[/latex].  For each of the individual orders, we multiply the probabilities as we move from draw to draw.  The “or” means that we add the probabilities of the different orders.  In other words,

   [latex]\begin{eqnarray*} \\ \mbox{Probability} & = & P(WNN)+P(NWN)+P(NNW)\\ \\ \end{eqnarray*}[/latex]

   For the [latex]WNN[/latex] order (win on bottle one, non-wins on bottles two and three), the probability of getting a win on bottle one is [latex]0.07[/latex] and the probability of getting a non-win on bottle two or bottle three is [latex]0.93[/latex]. So the probability of getting the [latex]WNN[/latex] order is [latex]\displaystyle{0.07 \times 0.93 \times 0.93}[/latex].Using similar logic, the probability of getting the [latex]NWN[/latex] order is [latex]\displaystyle{0.93 \times 0.07\times 0.93}[/latex] and the probability of getting the [latex]NNW[/latex] order is [latex]\displaystyle{0.93 \times 0.93 \times 0.07}[/latex].

   The probability of getting exactly one winning bottle is

   [latex]\begin{eqnarray*} \mbox{Probability} & = & P(WNN)+P(NWN) +P(NNW) \\ & = & \left(0.07 \times 0.93 \times 0.93 \right)+\left(0.93 \times 0.07 \times 0.93 \right)+\left(0.93 \times 0.93 \times 0.07 \right) \\ & = & 0.1816\\ \\ \end{eqnarray*}[/latex]

3. We want at least one winning bottle.  This means we can have exactly one winning bottle or exactly two winning bottles or exactly three winning bottles.  Of course, we could work out the probabilities of each of these orders and add them all up.  But the a faster way to find this probability is to use the complement.  The complement of “at least one winning bottle” is “exactly zero winning bottles”  The “exactly zero winning bottle” is the case [latex]NNN[/latex] (all three bottles are non-winners).  Using the complement, the probability of at least one winning bottle is

   [latex]\begin{eqnarray*} \\ P(\mbox{at least one winner}) & = & 1-P(\mbox{exactly zero winners}) \\ & = & 1-P(NNN)\\ \\ \end{eqnarray*}[/latex]

   The probability of zero winning bottles ([latex]NNN[/latex]) is :  [latex]\displaystyle{0.93 \times 0.93 \times 0.93}[/latex].  So the probability of at least one winning bottle is

   [latex]\begin{eqnarray*}  P(\mbox{at least one winner}) & = & 1-P(NNN) \\ & = & 1-\left(0.93 \times 0.93 \times 0.93\right) \\ & = & 0.1956 \\ \\ \end{eqnarray*}[/latex]