We will see how to use the multiplication rule by looking at some examples. First, let`s say we roll a six-sided die and then roll a coin. These two events are independent. The probability of throwing a 1 is 1/6. The probability of a head is 1/2. The probability of rolling a 1 and getting a head is 1/6 x 1/2 = 1/12. The common probability of two consecutive heads is 0.25. Since the sample is made with replacement, whether a club was selected in the first selection or not, it is a question of whether a club was selected in the second selection or not. Use the general multiplication rule to calculate common probabilities of independent or dependent events. If you have dependent events, you should use the general multiplication rule because it allows you to consider how the occurrence of event A affects the probability of event B. In fact, they are just as likely. The 10 throws are independent, so we use the multiplication rule for independent events: if the decision is independent of gender, then 45% of the 60% of the class that is female should have a major decided; Thank you! It`s not currently in any of my books.

However, I will write a book of probabilities later! Example 2 A coin is tossed twice. What is the probability of having a tail on the first litter and a tail on the second litter? Solution for Example 2 Two methods for answering the question in Example 2 are presented to demonstrate the benefit of using the product rule specified above. Events A and B are independent and, therefore, the product rule can be used as ( P(E) = P( A ; and ; B) = P(A cap B) = P(A) cdot P(B) = dfrac{1}{2} cdot dfrac{1}{2} = dfrac{1}{4} ) NOTE If you throw a part several times, the sample space contains a large number of elements, and therefore, method 2 is much more convenient to use than method 1, where you have a large number of results. We now present more examples and questions about how to use the product rule of independent events to solve probability questions. Example 3 A coin is rolled and a die is rolled. What is the probability of getting a head and a ( 4 ? Solution to Example 3 We have two independent events to consider: event A “roll a coin and get a head” and event B “roll a die and get a ( 4 ) ” When a coin is rolled, the probability of getting a head is ( P(A) = dfrac{1}{2} ) When the dice are rolled, is the probability of getting a ( 4 ), ( P(B) = dfrac{1}{6} ) ( P ( ) ” gets a head and a ( 4 ) ” ( ) = P( A ; and; B) = P( A cap B) = P(A) cdot P(B) = dfrac{1}{2} cdot dfrac{1}{6} = dfrac{1}{12} ) Example 4 A glass has 3 blue balls, 2 white balls and 5 red balls. A ball is randomly selected and the noted color is then placed back into the glass. A second ball is selected, its color is noted and replaced again in the glass. A third sphere is selected and its color is noted.

What is the probability that a) 3 red balls will be selected b) a blue ball, then a white ball then a blue ball is selected c) a red ball, then a white ball then a blue ball is selected Solution to Example 4 a) Let event A “select a red ball the first time”, Event B “select a red mark” and event C “choose a red ball the third time” The three events A, B and C are independent because the selected ball is reinserted into the glass. The total number of balls is 10 and there are 5 red balls. Now let`s calculate the probability of choosing a red ball.