<meta http-equiv="refresh" content="1; url=/nojavascript/"> Probability Distribution ( Read ) | Probability | CK-12 Foundation
Dismiss
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Probability Distribution

%
Best Score
Practice Probability Distribution
Practice
Best Score
%
Practice Now

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:

heads, heads = 25%, heads, tails = 25%, tails, heads = 25%, and tails, tails = 25%.

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)

.04

.12

.16

.16

.12

.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.

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...

Original text