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# Probability Distribution

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Practice Probability Distribution
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Probability Distribution

#### 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  $T$ represent the number of tails you get, what is the probability distribution for $T$ ?

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:

If we define the random variable  $X$ to be the number of heads you get when you flip a coin twice, we could create the following probability distribution table for $X$

 $X$ $0$ $1$ $2$ $P(X)$ $\frac{1}{8}$ $\frac{3}{4}$ $\frac{1}{8}$

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  $P$ be the number of parents of a randomly selected student from the class. Create a probability distribution for $P$ .

Solution:

Set random variable  $P$ to be the number of parents:

$P(P)=\% \ probability \ that \ a \ student \ has \ P \ parents$

Now find the probability of each   $P$ , noting that there are 12 students total:

$\text{1 student has 0 parents: } & P(0)=\frac{1}{12} \ or \ 8.3 \overline{3} \% \\\text{4 students have 1 parent: } & P(1)=\frac{4}{12} \ or \ 33.3 \overline{3} \% \\\text{7 students have 2 parents: } & P(2)=\frac{7}{12} \ or \ 58.3 \overline{3} \% \\$

Example B

Roll two fair six-sided dice. Let  $D$ equal the sum of the dice. Create a probability distribution for $D$ .

Solution: Make a list of the individual probabilities of each of the 36 possible outcomes:

$& \text{1 possibility with a sum of 2: } && P(D= 2) = \frac{1}{36} = 0.0278\\& \text{2 possibilities with a sum of 3: } && P(D= 3) = \frac{2}{36} = 0.0556\\& \text{3 possibilities with a sum of 4: } && P(D= 4) = \frac{3}{36} = 0.0833\\& \text{4 possibilities with a sum of 5: } && P(D= 5) = \frac{4}{36} = 0.1111\\& \text{5 possibilities with a sum of 6: } && P(D= 6) = \frac{5}{36} = 0.1389\\& \text{6 possibilities with a sum of 7: } && P(D= 7) = \frac{6}{36} = 0.1667\\& \text{5 possibilities with a sum of 8: } && P(D= 8) = \frac{5}{36} = 0.1389\\& \text{4 possibilities with a sum of 9: } && P(D= 9) = \frac{4}{36} = 0.1111\\& \text{3 possibilities with a sum of 10: } && P(D= 10) = \frac{3}{36} = 0.0833\\& \text{2 possibilities with a sum of 11: } && P(D= 11) = \frac{2}{36} = 0.0556\\& \text{1 possibility with a sum of 12: } && P(D= 12) = \frac{1}{36} = 0.0278$

Example C

Janie wants to evaluate the probabilities of pulling various cards from a deck. She sets the discrete random variable $C$ 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 $C$ ?

Solution: To evaluate the probability distribution of $C$ , Janie needs to identify the probability of each of the possible values of $C$ . Note that the chance she will pull a diamond is  $\frac{13}{52}$ or $.25$ , meaning that the chance she will not pull a diamond is $1-.25=.75$ :

• For   $C=(1)$ , the total probability is:   $.14+.14+.14=.42 \ or \ 42 \%$ (see the three possible outcomes resulting in  $C=1$ below)
• Diamond, other, other :   $.25 \times .75 \times .75=.14$
• Other, Diamond, other :   $.75 \times .25 \times .75=.14$
• Other, other, Diamond :   $.75 \times .75 \times .25=.14$
• For   $C=(2)$ , the total probability is:   $.047+.047+.047=.141 \ or \ 14.1 \%$
• Diamond, Diamond, other :   $.25 \times .25 \times .75=.047$
• Diamond, other, Diamond :  $.25 \times .75 \times .25=.047$
• Other, Diamond, Diamond :   $.75 \times .25 \times .25=.047$
• For   $C=(3)$ , the probability is :   $.25 \times .25 \times .25=.016 \ or \ 1.6 \%$
• Diamond, Diamond, Diamond:   $.25 \times .25 \times .25=.016$
##### 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  $T$ represent the number of tails you get, what is the probability distribution for $T$ ?

If each throw has a 65% chance of heads, then it has a 35% chance of tails:

• For   $T=1$ , we could have THH, HTH, or HHT. Each of those has a   $.35 \times .65 \times .65=.15$  chance of occurring, so $P(T=1)=.15 \times 3=.45 \ or \ 45 \%$
• For   $T=2$ , we could have TTH, THT, or HTH. Each has a   $.35 \times .35 \times .65=.08$  chance, so $P(T=2)=.08 \times 3=.24 \ or \ 24 \%$
• For   $T=3$ , we could have only TTT, with a chance of   $.35 \times .35 \times .35=.043 \ or \ 4.3 \%$

#### 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

1. Create a probability distribution for number of heads when you flip a coin 3 times.
2. Let $C$ 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 $C$ .
3. Let  $S$ be the score of a single student chosen at random from Mr. Spence’s class. Create a probability distribution for $S$ , given the following:
 Number of Students Test 11 87 7 89 13 92 9 94 6 96

Solutions:

1. Write out all the possibilities:

$& \text{TTT has 0 heads.} && \text{TTH has 1 heads.} && \text{THT has 1 heads.}\\& \text{THH has 2 heads.} && \text{HTT has 1 heads.} && \text{HTH has 2 heads.}\\& \text{HHT has 2 heads.} && \text{HHH has 3 heads.}$

$&\text{So we have 1 possibility with 0 heads:} && P(0) = \frac{1}{8} = 0.125\\&\text{3 possibilities with 1 heads:} && P(1) = \frac{3}{8} = 0.375\\&\text{3 possibilities with 2 heads:} && P(2) = \frac{3}{8} = 0.375\\&\text{1 possibility with 3 heads:} && P(3) = \frac{1}{8} = 0.125$

2. There are a total of 30 cookies, the probability of pulling a chocolate chip cookie is   $\frac{6}{30}=.20$ , so the probability of not pulling a chocolate chip is   $\frac{24}{30}=.80$

• For  $C=0$ we have to pull a non-chocolate chip both times:  $.8 \times .8=.64 \ or \ 64 \%$
• For  $C=1$ we could either pull the chocolate chip cookie first or second, so we get $(.2 \times .8)+(.8 \times .2)=.32 \ or \ 32 \%$
• For  $C=2$ we have to pull chocolate chip both times, so we have  $.2 \times .2=.04 \ or \ 4 \%$

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  $S$ is the same as that score’s portion of the total number of scores:

• $P(S=87)=\frac{11}{46}$
• $P(S=89)=\frac{7}{46}$
• $P(S=94)=\frac{9}{46}$
• $P(S=96)=\frac{6}{46}$

#### 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:

 $S$ 2 3 4 5 6 7 8 9 10 $P(S)$ 0.04 0.12 0.16 0.16 0.12 0.04

4. Assuming the table is a probability distribution for discrete random variable   $S$ , which is the sum of two dice rolled once, how many sides does each die have?

5. What is   $P(3)$ ?

6. What is   $P(6)$ ?

7. What is   $P(9)$ ?

8. Roll two seven-sided dice once. Let  $S$ be the sum of the two dice. Create a probability distribution for $S$ .

9. Flip a fair coin 3 times, let  $H$ be the number of heads. Create a probability distribution for $H$ .

10. Let  $S$ be the sum of two standard fair dice. Create a probability distribution for $S$ , if the experiment consists of a single roll of both dice.