Objective
In this lesson, you will learn about probability distributions and how they describe the probabilities associated with different possible values of a random variable.
Concept
Assume you have an unfair coin that is weighted to land on heads 65% of the time. If you flip that coin 3 times and let \begin{align*}T\end{align*} represent the number of tails you get, what is the probability distribution for \begin{align*}T\end{align*}?
Look to the end of the lesson for the answer.
Watch This
http://youtu.be/s2S1oD3ovps statslectures - Discrete Probability Distributions
Guidance
A probability distribution is a list of each value a random variable can attain, along with the probability of attaining each value. In other words, the probability distribution of an event is sort of a map of how each possible outcome relates to the chance it will happen.
For instance, the probability distribution of flipping a coin twice is:
heads, heads = 25%, heads, tails = 25%, tails, heads = 25%, and tails, tails = 25%.
If we define the random variable \begin{align*}X\end{align*} to be the number of heads you get when you flip a coin twice, we could create the following probability distribution table for \begin{align*}X\end{align*}:
\begin{align*}X\end{align*} |
\begin{align*}0\end{align*} |
\begin{align*}1\end{align*} |
\begin{align*}2\end{align*} |
\begin{align*}P(X)\end{align*} |
\begin{align*}\frac{1}{8}\end{align*} |
\begin{align*}\frac{3}{4}\end{align*} |
\begin{align*}\frac{1}{8}\end{align*} |
There are various ways of visualizing a probability distribution, and we will review that concept in another lesson. For now, we focus on identifying what a probability distribution is, and how to calculate it for a particular event.
Example A
In Chi’s class, 4 students have one parent, 7 have two parents, and 1 student lives with his uncle. Let \begin{align*}P\end{align*} be the number of parents of a randomly selected student from the class. Create a probability distribution for \begin{align*}P\end{align*}.
Solution:
Set random variable \begin{align*}P\end{align*} to be the number of parents:
\begin{align*}P(P)=\% \ probability \ that \ a \ student \ has \ P \ parents\end{align*}
Now find the probability of each \begin{align*}P\end{align*}, noting that there are 12 students total:
\begin{align*}\text{1 student has 0 parents: } & P(0)=\frac{1}{12} \ or \ 8.3 \overline{3} \% \\ \text{4 students have 1 parent: } & P(1)=\frac{4}{12} \ or \ 33.3 \overline{3} \% \\ \text{7 students have 2 parents: } & P(2)=\frac{7}{12} \ or \ 58.3 \overline{3} \% \\\end{align*}
Example B
Roll two fair six-sided dice. Let \begin{align*}D\end{align*} equal the sum of the dice. Create a probability distribution for \begin{align*}D\end{align*}.
Solution: Make a list of the individual probabilities of each of the 36 possible outcomes:
\begin{align*}& \text{1 possibility with a sum of 2: } && P(D= 2) = \frac{1}{36} = 0.0278\\ & \text{2 possibilities with a sum of 3: } && P(D= 3) = \frac{2}{36} = 0.0556\\ & \text{3 possibilities with a sum of 4: } && P(D= 4) = \frac{3}{36} = 0.0833\\ & \text{4 possibilities with a sum of 5: } && P(D= 5) = \frac{4}{36} = 0.1111\\ & \text{5 possibilities with a sum of 6: } && P(D= 6) = \frac{5}{36} = 0.1389\\ & \text{6 possibilities with a sum of 7: } && P(D= 7) = \frac{6}{36} = 0.1667\\ & \text{5 possibilities with a sum of 8: } && P(D= 8) = \frac{5}{36} = 0.1389\\ & \text{4 possibilities with a sum of 9: } && P(D= 9) = \frac{4}{36} = 0.1111\\ & \text{3 possibilities with a sum of 10: } && P(D= 10) = \frac{3}{36} = 0.0833\\ & \text{2 possibilities with a sum of 11: } && P(D= 11) = \frac{2}{36} = 0.0556\\ & \text{1 possibility with a sum of 12: } && P(D= 12) = \frac{1}{36} = 0.0278 \end{align*}
Example C
Janie wants to evaluate the probabilities of pulling various cards from a deck. She sets the discrete random variable \begin{align*}C\end{align*} to be the number of diamonds she gets over the course of three trials, if each trial consists of pulling, recording, and replacing one random card from a standard deck. What is the probability distribution of \begin{align*}C\end{align*}?
Solution: To evaluate the probability distribution of \begin{align*}C\end{align*}, Janie needs to identify the probability of each of the possible values of \begin{align*}C\end{align*}. Note that the chance she will pull a diamond is \begin{align*}\frac{13}{52}\end{align*} or \begin{align*}.25\end{align*}, meaning that the chance she will not pull a diamond is \begin{align*}1-.25=.75\end{align*}:
- For \begin{align*}C=(1)\end{align*}, the total probability is: \begin{align*}.14+.14+.14=.42 \ or \ 42 \%\end{align*} (see the three possible outcomes resulting in \begin{align*}C=1\end{align*} below)
- Diamond, other, other : \begin{align*}.25 \times .75 \times .75=.14\end{align*}
- Other, Diamond, other : \begin{align*}.75 \times .25 \times .75=.14\end{align*}
- Other, other, Diamond : \begin{align*}.75 \times .75 \times .25=.14\end{align*}
- For \begin{align*}C=(2)\end{align*}, the total probability is: \begin{align*}.047+.047+.047=.141 \ or \ 14.1 \%\end{align*}
- Diamond, Diamond, other : \begin{align*}.25 \times .25 \times .75=.047\end{align*}
- Diamond, other, Diamond : \begin{align*}.25 \times .75 \times .25=.047\end{align*}
- Other, Diamond, Diamond : \begin{align*}.75 \times .25 \times .25=.047\end{align*}
- For \begin{align*}C=(3)\end{align*}, the probability is : \begin{align*}.25 \times .25 \times .25=.016 \ or \ 1.6 \%\end{align*}
- Diamond, Diamond, Diamond: \begin{align*}.25 \times .25 \times .25=.016\end{align*}
Concept Problem Revisited
Assume you have an unfair coin that is weighted to land on heads 65% of the time. If you flip that coin 3 times and let \begin{align*}T\end{align*} represent the number of tails you get, what is the probability distribution for \begin{align*}T\end{align*}?
If each throw has a 65% chance of heads, then it has a 35% chance of tails:
- For \begin{align*}T=1\end{align*}, we could have THH, HTH, or HHT. Each of those has a \begin{align*}.35 \times .65 \times .65=.15\end{align*} chance of occurring, so \begin{align*}P(T=1)=.15 \times 3=.45 \ or \ 45 \%\end{align*}
- For \begin{align*}T=2\end{align*}, we could have TTH, THT, or HTH. Each has a \begin{align*}.35 \times .35 \times .65=.08\end{align*} chance, so \begin{align*}P(T=2)=.08 \times 3=.24 \ or \ 24 \%\end{align*}
- For \begin{align*}T=3\end{align*}, we could have only TTT, with a chance of \begin{align*}.35 \times .35 \times .35=.043 \ or \ 4.3 \%\end{align*}
Vocabulary
A probability distribution is a list of each value a random variable can attain, along with the probability of attaining each value.
Guided Practice
- Create a probability distribution for number of heads when you flip a coin 3 times.
- Let \begin{align*} C\end{align*} be the number of chocolate chip cookies you get if you randomly pull and replace two cookies from a jar containing 6 chocolate chip, 4 peanut butter, 8 snickerdoodle, and 12 sugar cookies. Create a probability distribution for \begin{align*}C\end{align*}.
- Let \begin{align*}S\end{align*} be the score of a single student chosen at random from Mr. Spence’s class. Create a probability distribution for \begin{align*}S\end{align*}, given the following:
Number of Students |
Test |
11 |
87 |
7 |
89 |
13 |
92 |
9 |
94 |
6 |
96 |
Solutions:
1. Write out all the possibilities:
\begin{align*}& \text{TTT has 0 heads.} && \text{TTH has 1 heads.} && \text{THT has 1 heads.}\\ & \text{THH has 2 heads.} && \text{HTT has 1 heads.} && \text{HTH has 2 heads.}\\ & \text{HHT has 2 heads.} && \text{HHH has 3 heads.} \end{align*}
\begin{align*}&\text{So we have 1 possibility with 0 heads:} && P(0) = \frac{1}{8} = 0.125\\ &\text{3 possibilities with 1 heads:} && P(1) = \frac{3}{8} = 0.375\\ &\text{3 possibilities with 2 heads:} && P(2) = \frac{3}{8} = 0.375\\ &\text{1 possibility with 3 heads:} && P(3) = \frac{1}{8} = 0.125\end{align*}
2. There are a total of 30 cookies, the probability of pulling a chocolate chip cookie is \begin{align*}\frac{6}{30}=.20\end{align*}, so the probability of not pulling a chocolate chip is \begin{align*}\frac{24}{30}=.80\end{align*}
- For \begin{align*}C=0\end{align*} we have to pull a non-chocolate chip both times: \begin{align*}.8 \times .8=.64 \ or \ 64 \%\end{align*}
- For \begin{align*}C=1\end{align*} we could either pull the chocolate chip cookie first or second, so we get \begin{align*}(.2 \times .8)+(.8 \times .2)=.32 \ or \ 32 \% \end{align*}
- For \begin{align*}C=2\end{align*} we have to pull chocolate chip both times, so we have \begin{align*}.2 \times .2=.04 \ or \ 4 \%\end{align*}
3. There are a total of 46 students in Mr. Spence’s class, so there are 46 scores. The probability of a random student having score \begin{align*}S\end{align*} is the same as that score’s portion of the total number of scores:
- \begin{align*}P(S=87)=\frac{11}{46}\end{align*}
- \begin{align*}P(S=89)=\frac{7}{46}\end{align*}
- \begin{align*}P(S=94)=\frac{9}{46}\end{align*}
- \begin{align*}P(S=96)=\frac{6}{46}\end{align*}
Practice
1. What is a probability distribution?
2. What is a random variable?
3. What is the difference between a discrete and a continuous random variable?
For problems 4-7, refer to the following table:
\begin{align*}S\end{align*} |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
\begin{align*}P(S)\end{align*} |
.04 |
.12 |
.16 |
.16 |
.12 |
.04 |
4. Assuming the table is a probability distribution for discrete random variable \begin{align*}S\end{align*}, which is the sum of two dice rolled once, how many sides does each die have?
5. What is \begin{align*}P(3)\end{align*}?
6. What is \begin{align*}P(6)\end{align*}?
7. What is \begin{align*}P(9)\end{align*}?
8. Roll two seven-sided dice once. Let \begin{align*}S\end{align*} be the sum of the two dice. Create a probability distribution for \begin{align*}S\end{align*}.
9. Flip a fair coin 3 times, let \begin{align*}H\end{align*} be the number of heads. Create a probability distribution for \begin{align*}H\end{align*}.
10. Let \begin{align*}S\end{align*} be the sum of two standard fair dice. Create a probability distribution for \begin{align*}S\end{align*}, if the experiment consists of a single roll of both dice.
Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 7.3.