# Multiplication Rule for Probabilities of Independent Events

Examples on using the multiplication rule to find the probability of two or more independent events occurring are presented along with detailed solutions.

## Independent Events
In probabilities, two events are independent if the occurence of one does not affect the probability of occurence of the other.
- A = "roll a die and get a \( 1 \)" , B = "flip a coin and get a tail".
- A = "draw a card from a deck and get a King", replace it back into the deck, B = "draw another card and get a Queen"
- A = "roll a die and get a \( 4 \)" , B = "roll the same die (or another one) and get a "6"
- A = "flip a coin and get a head" , B = roll the same coin (or another one) and get a tail"
A jar has 3 blue balls, 2 white balls and 5 red balls - A = Pick a ball at random from the jar and get a red ball, replace it back into the jar, B = Pick a ball at random from the jar and get a white ball
The events C and D are NOT independent. - C = "draw a card from a deck and get a King", D = "draw a second card from the same deck and get a Queen".
A jar contains 3 blue balls, 2 white balls and 3 red balls - C = "Pick a ball at random from the jar and get a red ball", D = "Pick a second ball at random from the same jar and get a white ball".
## Examples with Detailed SolutionsExample 2
A coin is tossed twice. What is the probability of getting a tail in the first toss and a tail in the second toss?
Example 6 A poll found that 25% of the people in a certain country have heart problems. If three people are selected at random, find the probability that all three have heart problems. Solution to Example 6Event A " the first person have heart problems" , event B " the second person have heart problems " and C " the third person have heart problems " \( P(A) = 0.25\) , \( P(B) = 0.25 \) and \( P(C) = 0.25 \). These are independent events, hence \( P (A and B and C ) = P(A) \cdot P(B) \cdot P(C) = 0.25 \cdot 0.25 \cdot 0.25 = 0.015625 \)
## More Questions with Solutions1) A die is rolled twice. Find the probability of getting an even number in the first throw and a number greater than 4 in the second throw.2) A card is drawn from a deck of 52 cards and then replaced and a second card is drawn. Find the probability of getting a "King" and then a "Queen of hearts". 3) In a country, 45% of the people smoke. If 4 people are selected at random, what is the population that they are all smokers? ## Solutions to above exercises1)Let event A: get an even number , event B: get a number greater than 4 A = \( \{2,4,6\} \) , B = \( \{5,6\} \) \( P(A) = 3/6 = 1/2 \) , \( P(B) = 2/6 = 1/3 \) Events A and B are independent; hence \( P(A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6} \) 2) \( P(King) = P(A) = 4/52 = 1/13 \) , \( P(Queen \; of \; hearts) = P(B) = 1/52 \) Events A and B are independent; hence \( P(A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{13} \cdot \dfrac{1}{52} = \dfrac{1}{676} \) 3) \( P(smoker) = 0.45 \) All 4 events are independent; hence \( P(\text{all 4 smokers}) = P(smoker \; and \; smoker \; and \; smoker \; and \; smoker) \\\\ = P(smoker) \cdot P(smoker) \cdot P(smoker) \cdot P(smoker) \\\\ = (0.45)^4 = 0.04100625 \) ## More References and linksAddition Rule for Probabilities.Binomial Probabilities Examples and Questions probability questions classical formula for probability mutually exclusive events Introduction to Probabilities sample space eventelementary statistics and probabilities. Home Page |