# Conditional Probabilities Examples and Questions

The defintion of conditional probabilities is presented along with examples and their detailed solutions and explanations. More questions are included at the bottom of the page followed by their solutions.

## Conditional Probability DefinitionWe use a simple example to explain conditional probabilities.
Example 1
## More Examples with Detailed Solutions
Example 2
Example 3
If a person is selected at random from the group, what is the probability that he/she a) bought brand B? b) bought a mobile phone from brand B? c) bought a mobile phone given that she/he bought brand B? Solution to Example 3Let event Mp: bought a mobile phone event T: bought a tablet event A: bought brand A event B: bought brand B a) A total of 70 people out of the total of 100 bought brand B; hence \( P(B) = 70/100 = 0.7 \) b) A total of 30 people out of 100 bought a mobile from brand B \( P(Mp \; and \; B) = 30/100 = 0.3 \) c) \( P(Mp|B) = \dfrac{P(Mp \; and \; B)}{P(B)} = 0.3/0.7 = 3/7\) The conditional probability \( P(Mp|B) \) may also be found by restricting the sample space to brand B. The are 30 mobile phones out of a total of 70 brand B; hence \( P(Mp|B) = \dfrac{30}{70} = 3/7 \)
Example 4
Example 5
Solution to Example 5There are 60 Suv's, 40 Sport cars, 50 Vans and 50 Coupe, a total of 200 cards. We answer the questions on finding conditional probabilities using two methods: 1) the definition and 2) restriction of the sample space. a) Using conditional probability definition \( P(black|Van) = \dfrac{P(\text{black and Van)}}{P(Van)} = \dfrac{25/200}{50/200} = 1/2 \) Or restrict sample space to the Vans, there are 50 vans out of which 25 are black. Hence \( P(black|Van) = \dfrac{25}{50} = 1/2 \) b) Using conditional probability definition \( P(Suv|white) = \dfrac{P(\text{Suv and white)}}{P(white)} = \dfrac{10/200}{50/200} = 1/5 \) Restrict sample space to the white cars, there are 50 white out of which 10 are Suv's. Hence \( P(Suv|white) = \dfrac{10}{50} = 1/5 \) c) Using conditional probability definition \( P(blue|Coupe) = \dfrac{P(\text{blue and Coupe)}}{P(Coupe)} = \dfrac{30/200}{50/200} = 3/5 \) Restrict sample space to the Coupe, there are 50 Coupe out of which 30 are blue. Hence \( P(blue|Coupe) = \dfrac{30}{50} = 3/5 \)
Example 6
Example 7
## Questions and their Solutions
## Question 1A die is rolled, find the probability that an even number is obtained knowing the number is greater than 3.
## Question 2A card is drawn at random from a deck of 52 cards. Find the probability of getting a 3 knowing that the card is red.
## Question 3A group 120 people were asked if they own a car or a bicycle. 90 said they owned a car, 40 they owned a bicycle and 10 said they owned neither a car nor a bicycle.If a person is selected at random from this group of people, what is the probability that this person a) owns a car and a bicycle? b) owns a car given that the person selected owns a bicycle? c) owns a bicycle given that the person owns a car or a bicycle but not both?
## Question 4Two companies A and B are offering 70 and 50 products respectively. Company A is offering 40 software products and 30 hardware products. Company B is offering \( x \) hardware products and \( y \) software products to be determined.If a product is selected at random, what is the probability that a) this product is a hardware product given that it is from company B? (in terms of \( y \)) b) this product is a hardware product given that it is from company A? c) For what values of \( y \) will the probability in part a) be greater than the probability in part b)?
## Solution to Question 5A financial advisor believes that the probability that the stock market will go down is 0.8 given that the economy deteriorates. The advisor also believes that the probability that the economy will deteriorates is 0.5.Taking the above information into account, what is the probability that the economy will deteriorate and the stock market will go down? ## Solutions to the Above Questions
## Solution to Question 1The sample space in throwing a die: \( S = \{1,2,3,4,5,6\} \)event E: even number , \( E = \{2,4,6\} \) event G number greater than 3: \( G = \{4,5,6\} \) \( E \cap G = \{4,6\} \) The question is to find the probability that an even number is obtained knowing the number is greater than 3 written as a conditional probability \( P(E|G) \). \( P(E|G) = \dfrac{P( E \cap G)}{P(G)} = \dfrac{2/6}{3/6} = 2/3 \) This question might be answered as follows There are 2 elements in \( E \cap G \) and 3 elements in \( G \) We restrict the sample space to event \( G \); hence \( P(E|G) = \dfrac{n(E \cap G)}{n(G)} = 2/3 \)
## Solution to Question 2Let event T: getting a 3 , there are 4 3's in a deck of cards, hence \( P(T) = 4/52 = 1/13\).Let event R: getting a red card , there are 26 red cards in a deck of cards, hence \( P(R) = 26/52 = 1/2 \) There are 2 3's cards that are red, hence \( P(T \cap R) = 2/52 = 1/26 \) We are asked to find the conditional probability of getting a 3 knowing that the card is red written as \( P(T|R) \) \( P(T|R) = \dfrac{P(T \cap R)}{P(R)} = \dfrac{1/26}{1/2} = 1/13 \) This question might be answered as follows There are 2 3's cards that are red We restrict the sample space to red cards which are 26, hence \( P(T|R) = 2/26 = 1/13 \)
## Solution to Question 3a)Let C the event: owns a car , Let B the event: owns a bicycle We are asked to find \( P(C \cap B) \) Let \( x \) be the number of people who owns a car and a bicycle. Using the Venn diagram, below we have \( 90 - x\) own a car only, \( 40 - x\) own a bicycle only and \( 10 \) own neither and the total is \( 120\); hence \( (90 - x) + (40 - x) + x + 10 = 120 \) Solve for \( x \) to obtain \( x = 20 \) \( P(C \cap B) = 20/120 = 1/6\) b) owns a car given that the person selected owns a bicycle? We are asked to find the conditional probability \( P(C|B) \). \( P(C|B) = \dfrac{P(C \cap B)}{P(B)} = \dfrac{1/6}{40/120} = 20/40 = 1/2 \) c) owns a bicycle given that the person owns a car or a bicycle but not both? Let event COB : own a car or a bicycle but not both The people who own a car or a bicycle but not both are included in the union without the intersection and their number is \( N = 70 + 20 = 90 \) We are asked to find the probability \( P(B|COB) \). \( P(B|COB) = \dfrac{P(B \cap COB)}{P(COB)} = \dfrac{20/120}{90/120} = 2/9 \)
## Solution to Question 4Let us arrange all the given information on a table as follows
a) The total number of products is: \( 70 + 50 = 120 \) We also have \( 70+x+y = 120\) which gives \( x + y = 50 \) Let event S: product selected is a software product , let event H: product selected is a hardware product Let event A: product selected is from company A , let event B: product selected is from company B We are asked to find the conditional probability \( P(H|B) \). \( P(H|B) = \dfrac{P(H \cap B)}{P(B)} = \dfrac{\dfrac{y}{120}}{\dfrac{x+y}{120}} = \dfrac{y}{x+y} = y / 50\) b) We are asked to find the conditional probability \( P(H|A) \). \( P(H|A) = \dfrac{P(H \cap A)}{P(A)} = \dfrac{\dfrac{30}{120}}{\dfrac{70}{120}} = \dfrac{3}{7}\) c) We need to solve the inequality \( y / 50 \gt 3 / 7 \) Multiply all terms by 50 and simplify \( y \gt 150 / 7 \) Simplify \( y \gt 21.42 \) \( y \) is a positive integer, hence \( y \ge 22 \)
## Solution to Question 5Let event E: the economy will deteriorate , let event S: the stock market will go down.We are given the conditional probability \( P(S|E) = 0.8 \) and the probability \( P(E) = 0.5\) We are asked to find \( P(S \cap E) \). The definition of the conditional probability gives \( P(S|E) = \dfrac{P(S \cap E)}{P(E)} = 0.8 \) hence \( P(S \cap E) = 0.8 \times P(E) = 0.8 \times 0.5 = 0.4 \) ## More References and linksprobability questionsclassical formula for probability Binomial Probabilities Examples and Questions mutually exclusive events Introduction to Probabilities sample space event elementary statistics and probabilities. |