1. DO NOW 4/27/2016 Find the theoretical probability of each outcome. 1. rolling a 6 on a number cube. 2. rolling an odd number on a number cube. 3. flipping two coins and both landing head up Compare to Informal Probability, this is the calculated probability.

2. Suppose an experiment is flipping two fair coins. All the possible outcomes are shown by the tree diagram. Determine the theoretical probability of both coins landing heads up.

3. To determine the probability of two independent events, multiply the probabilities of the two events. Now look at the separate probabilities of each coin landing heads up. The probability in each case is. The product of these two probabilities is

4. Example 1: Randomly selecting a marble from a bag, replacing it, and then selecting another marble. The bag contains 3 red marbles and 12 green marbles. What is the probability of selecting a red marble and then a green marble? Because the first marble is replaced after it is selected, the sample space for each selection is the same. The events are independent.

5. P(red, green) = P(red)  P(green) The probability of selecting red is, and the probability of selecting green is.

6. Example 2: A coin is flipped 4 times. What is the probability of flipping 4 heads in a row. Because each flip of the coin has an equal probability of landing heads up, or a tails, the sample space for each flip is the same. The events are independent. P(h, h, h, h) = P(h) P(h) P(h) P(h) The probability of landing heads up is with each event.

7. Try It Now An experiment consists of spinning the spinner twice. What is the probability of spinning two odd numbers? The result of one spin does not affect any following spins. The events are independent. With 6 numbers on the spinner, 3 of which are odd, the probability of landing on two odd numbers is P(odd, odd) = P(odd) P(odd).

8. Suppose an experiment involves drawing marbles from a bag. Determine the theoretical probability of drawing a red marble and then drawing a second red marble without replacing the first one. Probability of drawing a red marble on the first draw

9. Probability of drawing a red marble on the second draw Suppose an experiment involves drawing marbles from a bag. Determine the theoretical probability of drawing a red marble and then drawing a second red marble without replacing the first one.

10. To determine the probability of two dependent events, multiply the probability of the first event times the probability of the second event after the first event has occurred.

11. Example 3: A snack cart has 6 bags of pretzels and 10 bags of chips. Grant selects a bag at random, and then Iris selects a bag at random. What is the probability that Grant will select a bag of pretzels and Iris will select a bag of chips?

12. P(pretzel and chip) = P(pretzel) P(chip after pretzel) Grant selects one of 6 bags of pretzels from 16 total bags. Then Iris selects one of 10 bags of chips from 15 total bags. The probability that Grant selects a bag of pretzels and Iris selects a bag of chips is.

13. Try It Now A bag has 10 red marbles, 12 white marbles, and 8 blue marbles. Two marbles are randomly drawn from the bag. What is the probability of drawing a blue marble and then a red marble?

14. P(blue and red) = P(blue) P(red after blue) One of 8 blue marbles is selected from a total of 30 marbles. Then one of 10 red marbles is selected from the 29 remaining marbles. The probability that first a blue marble is selected and then a red marble is selected is.

15. Three separate boxes each have one blue marble and one green marble. One marble is chosen from each box. What is the probability of choosing a blue marble from each box? Additional Examples The outcome of each choice does not affect the outcome of the other choices, so the choices are independent. P(blue, blue, blue) = In each box, P(blue) =. 1212 1212 · 1212 · 1212 = 1818 = 0.125 Multiply. Course 3

16. What is the probability of choosing a blue marble, then a green marble, and then a blue marble? P(blue, green, blue) = 1212 · 1212 · 1212 = 1818 = 0.125 Multiply. In each box, P(blue) =. 1212 In each box, P(green) =. 1212

17. What is the probability of choosing at least one blue marble? 1 – 0.125 = 0.875 Subtract from 1 to find the probability of choosing at least one blue marble. Think: P(at least one blue) + P(not blue, not blue, not blue) = 1. In each box, P(not blue) =. 1212 P(not blue, not blue, not blue) = 1212 · 1212 · 1212 = 1818 = 0.125Multiply.

18. Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box. What is the probability of choosing a blue marble from each box? The outcome of each choice does not affect the outcome of the other choices, so the choices are independent. In each box, P(blue) =. 1414 P(blue, blue) = 1414 · 1414 = 1 16 = 0.0625 Multiply.

19. Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box. What is the probability of choosing a blue marble and then a red marble? In each box, P(blue) =. 1414 P(blue, red) = 1414 · 1414 = 1 16 = 0.0625 Multiply. In each box, P(red) =. 1414 Course 3

20. Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box. What is the probability of choosing at least one blue marble? In each box, P(blue) =. 1414 P(not blue, not blue) = 3434 · 3434 = 9 16 = 0.5625Multiply. Think: P(at least one blue) + P(not blue, not blue) = 1. 1 – 0.5625 = 0.4375 Subtract from 1 to find the probability of choosing at least one blue marble. Course 3

21. The letters in the word dependent are placed in a box. If two letters are chosen at random, what is the probability that they will both be consonants? P(first consonant) = 2323 6969 = Because the first letter is not replaced, the sample space is different for the second letter, so the events are dependent. Find the probability that the first letter chosen is a consonant.

22. If the first letter chosen was a consonant, now there would be 5 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant. P(second consonant) = 5858 5 12 5858 2323 ·= The probability of choosing two letters that are both consonants is. 5 12 Multiply.

23. If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels? There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Example 3A. Now find the probability of getting 2 vowels. Find the probability that the first letter chosen is a vowel. If the first letter chosen was a vowel, there are now only 2 vowels and 8 total letters left in the box. P(first vowel) = 1313 3939 =

24. Find the probability that the second letter chosen is a vowel. The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities. P(second vowel) = 1414 2828 = 1 12 1414 1313 ·= Multiply. 1212 5 12 1 12 + = 6 12 = The probability of getting two letters that are either both consonants or both vowels is. 1212 P(consonant) + P(vowel)

25. The letters in the phrase I Love Math are placed in a box. If two letters are chosen at random, what is the probability that they will both be consonants? P(first consonant) = 5959 Because the first letter is not replaced, the sample space is different for the second letter, so the events are dependant. Find the probability that the first letter chosen is a consonant.

26. Course 3 Independent and Dependent Events P(second consonant) = 5 18 1212 5959 ·= The probability of choosing two letters that are both consonants is. 5 18 Multiply. If the first letter chosen was a consonant, now there would be 4 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant. 1212 4848 =

27. If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels? There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Try This 3A. Now find the probability of getting 2 vowels. Find the probability that the first letter chosen is a vowel. If the first letter chosen was a vowel, there are now only 3 vowels and 8 total letters left in the box. Course 3 Independent and Dependent Events P(first vowel) = 4949

28. Find the probability that the second letter chosen is a vowel. The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities. Course 3 Independent and Dependent Events P(second vowel) = 3838 12 72 3838 4949 ·= Multiply. 1616 = 4949 5 18 1616 + = 8 18 = P(consonant) + P(vowel) The probability of getting two letters that are either both consonants or both vowels is. 4949

29. Determine if each event is dependent or independent. 1. drawing a red ball from a bucket and then drawing a green ball without replacing the first 2. spinning a 7 on a spinner three times in a row 3. A bucket contains 5 yellow and 7 red balls. If 2 balls are selected randomly without replacement, what is the probability that they will both be yellow? independent dependent Insert Lesson Title Here 5 33