*P*(

*X*)

Watch short & fun videos
**Start Your Free Trial Today**

Discrete Probability Distributions Flashcards

1/23
(missed)
0
0

Start Your Free Trial To Continue Studying

Free
5-day trial
In a game of blackjack, a person is dealt 2 cards: a 4 and a 9. Assuming aces count as 1, find the probability that the next card dealt makes the blackjack total greater than 21.

Current total: 13

13 + ace through 8 ≤ 21 (31 total cards because one 4 was already used: 31/50 chance)

Any other card: 19/50 = 0.38 = 38% chance

Got it

Missed it

In a game of blackjack, a person is dealt 2 cards: a 9 and a 10. Assuming aces count as 1, find the probability that the next card dealt makes the blackjack total greater than 21.

Current total: 9 + 10 = 19

19 + ace or 19 + 2 card ≤ 21 (8 total cards: 8/50 chance)

Any other card: 42/50 = 0.84 = 84% chance

Got it

Missed it

Find the probability of being dealt two jacks in a row from a standard deck of 52 cards.

Number of jacks in the deck: 4

Probability of first jack: 4/52

Probability of second jack: 3/51

(4/52) * (3/51) = 12/2652 = 1/221

Got it

Missed it

Find the probability that 0 ≤ X < 2

0.25 + 0.50 = 0.75 or 75%

Got it

Missed it

Find the probability that X ≤ 2

0.25 + 0.50 + 0.25 = 1 or 100%

Got it

Missed it

Find the probability that X = 2

0.25 or 25%

Got it

Missed it

Find the probability of winning $2 on a roll of two dice if you lose $5 for rolling a 4, 5, or 6, and win $2 for rolling anything else.

Roll a 4: 1-3, 3-1, 2-2 (3); Roll a 5: 1-4, 4-1, 2-3, 3-2 (4); Roll a 6: 1-5, 5-1, 2-4, 4-2, 3-3 (5)

(3 + 4 + 5) / 36 = 12/36 chance of losing $5

36 - 12 = 24/36 = 0.67 = 67% chance of winning $2

Got it

Missed it

Find the expected value (the amount of money you expect to win per roll) when rolling 2 dice, losing $2 for rolling a 4 or 5, winning $1 for rolling a 2 or 3, and winning $3 for any other number.

-$2(7/36) + $1(3/36) + $3(26/36) = $1.86

Got it

Missed it

Find the expected value (the amount of money you are expected to win or lose per roll) when rolling two dice, losing $2 for rolling a 3 or 4, and winning $3 for any other number.

*E(X)* = -$2(5/36) + $3(31/36) = $2.31

Got it

Missed it

Find the expected value of the random variable X:

*E(X)* = sum of *X* * *P(X)* = (0 * 0.4) + (1 * 0.4) + (2 * 0.2) = 0.8

Got it

Missed it

Probability Distribution Function or *P*(*X*)

This is used to assign a probability to all possible values of *X*. It is always a number between 0 and 1.

Got it

Missed it

Go to next set in chapter:
Next set in chapter
Continuous Probability Distributions Flashcards

or choose a specific lesson:
See all lessons in this chapter

23 cards in set

What is the relationship between playing cards, flipping a coin, and rolling dice? The probability of their outcomes can be described by discrete probability distribution. Discrete refers to the fact that when rolling dice, for example, you will be dealing with the probability of rolling exact numbers. The flashcards in this set will help you review the various formulas required to solve problems involving discrete probability distribution.

Front

Back

Probability Distribution Function or *P*(*X*)

*X*. It is always a number between 0 and 1.

Find the expected value of the random variable X:

*E(X)* = sum of *X* * *P(X)* = (0 * 0.4) + (1 * 0.4) + (2 * 0.2) = 0.8

*E(X)* = -$2(5/36) + $3(31/36) = $2.31

-$2(7/36) + $1(3/36) + $3(26/36) = $1.86

Roll a 4: 1-3, 3-1, 2-2 (3); Roll a 5: 1-4, 4-1, 2-3, 3-2 (4); Roll a 6: 1-5, 5-1, 2-4, 4-2, 3-3 (5)

(3 + 4 + 5) / 36 = 12/36 chance of losing $5

36 - 12 = 24/36 = 0.67 = 67% chance of winning $2

Find the probability that X = 2

0.25 or 25%

Find the probability that X ≤ 2

0.25 + 0.50 + 0.25 = 1 or 100%

Find the probability that 0 ≤ X < 2

0.25 + 0.50 = 0.75 or 75%

Find the probability of being dealt two jacks in a row from a standard deck of 52 cards.

Number of jacks in the deck: 4

Probability of first jack: 4/52

Probability of second jack: 3/51

(4/52) * (3/51) = 12/2652 = 1/221

Current total: 9 + 10 = 19

19 + ace or 19 + 2 card ≤ 21 (8 total cards: 8/50 chance)

Any other card: 42/50 = 0.84 = 84% chance

Current total: 13

13 + ace through 8 ≤ 21 (31 total cards because one 4 was already used: 31/50 chance)

Any other card: 19/50 = 0.38 = 38% chance

Find the probability of being dealt a five-card hand containing 4 jacks from a standard 52-card deck.

Favorable outcomes:

4 jacks in deck = 4! / 4!(4 - 4)! = 1

1 card out of 48 remaining cards = 48! / 1!(48 - 1)! = 48

1*48 = 48

Possible outcomes: 2,598,960

Probability: 48/2,598,960 = 1/54,145

Find the probability of being dealt a five-card hand containing 5 red cards from a standard 52-card deck.

5 red cards out of 26 red cards in deck = 26! / 5!(26 - 5)! = 65,780

Possible outcomes: 2,598,960

Probability: 65,780/2,598,960 = 0.0253

Find the probability of being dealt a five-card hand containing at least 2 kings.

Find the probability of being dealt a five-card hand containing at least 3 jacks.

Characteristics of Binomial Experiments

1. The outcomes must be independent, so the probability (*P*) of each trial must be the same.

2. There must be only two possible outcomes.

3. There must be a fixed number of trials (*n*).

Binomial Probability Equation

B(x; n, P)* = *nCx * Px * (1 - P)n - x

*nCx* = combination of n trials with x successes

*x* = number of successes

*n* = number of trials

*P* = probability of success on individual trial

Using the binomial probability formula, find the probability that 1 person voted out of 5 if each person had a 50% chance of voting.

Probability of zero voting:

x=0, n=5, P=0.50

nCx = 5! / 0!(5-0)! = 1

B(x; n, P) = nCx * Px * (1-P)n-x = 1 * (0.50)0 * (1 - 0.50)5-0 = 0.0313

1 - 0.0313 = 0.9687 = 96.9% chance at least one voted

Using the binomial probability formula, find the probability that 2 people voted out of 5 if each person had a 50% chance of voting.

x = 2, n = 5, P = 0.50

nCx = 5! / 2!(5 - 2)! = 10

B(x; n, P) = nCx * Px * (1 - P)n - x = 10 * (0.50)2 * (1 - 0.50)5 - 2 = 0.3125 = 31.2%

Find the probability of winning 4 out of 5 games of craps if there is a 0.493 chance of winning each game.

x = 4, n = 5, P = 0.493

nCx = 5! / 4!(5 - 4)! = 5

B(x; n, P) = nCx * Px * (1 - P)n - x = 5 * (0.493)4 * (1 - 0.493)5 - 4 = 0.1497 = 15%

Using the binomial probability formula, find the probability that x = 3 if P = 0.30 and n = 4.

x = 3, n = 4, P = 0.30

nCx = 4! / 3!(4 - 3)! = 4

B(x; n, P) = nCx * Px * (1 - P)n - x = 4 * (0.30)3 * (1 - 0.30)4 - 3 = 0.0756 = 7.6%

Binomial Experiment

An experiment in which there are only two discrete outcomes: success or failure. The outcomes of individual trials are independent of one another.

To unlock this flashcard set you must be a Study.com Member.

Create
your account

You are viewing lesson
Lesson
5 in chapter 11 of the course:

Back To Course

Statistics 101: Principles of Statistics11 chapters | 141 lessons | 9 flashcard sets

- Go to Probability

- Go to Sampling

- Overview of Statistics Flashcards
- Summarizing Data Flashcards
- Tables and Plots Flashcards
- Probability Flashcards
- Discrete Probability Distributions Flashcards
- Regression & Correlation Flashcards
- Statistical Estimation Flashcards
- Hypothesis Testing in Statistics Flashcards
- Z-Scores & Standard Normal Curve Areas Statistical Table
- Critical Values of the t-Distribution Statistical Table
- Binomial Probabilities Statistical Tables
- Go to Studying for Statistics 101

Browse by subject