1. Probability Question 6 Solutions

2. Part I: Shinobi has 16 ninja stars. Four (4) of them are red, five (5) of them are blue and seven (7) of them are black. If Shinobi had thrown 3 stars without replacing them, what would be the probability of the following? a) What is the probability of at least 2 stars being black? b) What is the probability of the third star being red?

3. Part I: a) What is the probability of at least 2 stars beeing black? The colour of a succeeding star depends on the colour of the star previous to it. We call this dependent probability where the outcome of the second step depends on the outcome of the first step . This is oppose to independent probability where the outcome of the second step does not depend on the outcome of the first . The most efficient way to look at a problem like this visually is to draw a tree diagram.On the tree diagram, you would draw out all the possiblities in each step and then insert the probability for each outcome as shown in the next slide.

4. R Bl B R Bl B R Bl B R Bl B B 5 16 R Bl B R Bl B R Bl B R Bl B R 4 16 3 15 7 15 R Bl B R Bl B R Bl B R Bl B Bl 5 15 2 14 7 14 5 14 3 14 6 14 7 14 3 14 15 4 7 15 4 15 3 14 7 14 6 14 7 14 4 15 4 14 4 14 4 14 4 14 4 14 3 14 6 15 5 15 3 14 7 16 5 14 5 14 4 14 5 14 4 14 6 14 6 14 4 14 R = Red (4) B = Blue (5) Bl = Black (7)

5. Please note that where the Red, Blue, and Black start branching off individually they themselves are suppose to have branched off from the same initial spot like this: However, we did not have room on the slide to illustrate this at its best size. For this we apologize, but we hope it does not confuse you too much. R B Bl

6. R Bl B R Bl B R Bl B R Bl B B Look for the sequences where there are two (2) black stars. 5 16 R Bl B R Bl B R Bl B R Bl B R 4 16 3 15 7 15 R Bl B R Bl B R Bl B R Bl B Bl 5 15 2 14 7 14 5 14 3 14 6 14 7 14 3 14 15 4 7 15 4 15 3 14 7 14 6 14 7 14 4 15 4 14 4 14 4 14 4 14 4 14 3 14 6 15 5 15 3 14 7 16 5 14 5 14 4 14 5 14 4 14 6 14 6 14 4 14 5 14

7. For the probability of 2 black stars in a sequence look for the sequence of branches with at least 2 blacks. -Since it starts of being out of 16 stars, the denominator of the first branch is 16. As the stars aren't replaced, the denominator goes down one unit because of the star we took. -As you move down each branch you can calculate the probability of each sequence. -Moving down the branch, you take the probability of each outcome in a sequence and multiply it by each other. -You the take that total and add it to the total of the other sequences that have 2 black

8. = = = = = = P( R , Bl, Bl) P( B , Bl, Bl) P(Bl, Bl, Bl) P(Bl, Bl, B ) P(Bl, Bl, R ) P(Bl, B , Bl) P(Bl, R , Bl) Calculating the probability of at least two stars being black... = = = = = = = = 4 7 6 16 15 14 ( ) ( ) ( ) 168 3360 ( ) 5 7 6 16 15 14 ( ) ( ) ( ) 7 6 5 16 15 14 ( ) ( ) ( ) 7 6 5 16 15 14 ( ) ( ) ( ) 7 6 4 16 15 14 ( ) ( ) ( ) 7 5 6 16 15 14 ( ) ( ) ( ) 7 4 6 16 15 14 ( ) ( ) ( ) ( ) 210 3360 168 3360 ( ) 168 3360 ( ) ( ) 210 3360 ( ) 210 3360 ( ) 210 3360

9. = = 0.4% = 40% There is a 40% probability that when three stars are thrown, at least two stars are black. 2 5 = + ( ) 210 3360 168 3360 ( ) + ( ) 210 3360 + 168 3360 ( ) ( ) 210 3360 + ( ) 210 3360 + + 168 3360 ( ) 1344 3360

10. Shinobi has 16 ninja stars. Four (4) of them are red, five (5) of them are blue and seven (7) of them are black. If Shinobi had thrown 3 stars without replacing them, what would be the probability of the following? b) What is the probability of the third star being red?

11. This is again, dependent probability because the outcome of a star depends on the star preceeding it. We will be using another tree diagram to help you visualize the problem. We are looking for the probability of the third star being red. In total, the number of stars we have is 16. That's 4 red, 5 blue, and 7 black. Remember that as each star is taken, it is not replace.

12. R Bl B R Bl B R Bl B R Bl B B 5 16 R Bl B R Bl B R Bl B R Bl B R 4 16 3 15 7 15 R Bl B R Bl B R Bl B R Bl B Bl 2 14 7 14 5 14 3 14 6 14 7 14 3 14 15 4 7 15 4 15 3 14 7 14 6 14 7 14 4 15 4 14 4 14 4 14 4 14 4 14 3 14 6 15 5 15 3 14 7 16 5 14 5 14 4 14 5 14 4 14 6 14 6 14 4 14 5 14 Note: Look for the the sequences where the red is at the last branch. 5 15

13. When you are looking for the probability of the third star being red, you look for the sequences where the last branch has a red. -Since it starts of being out of 16 stars, the denominator of the first branch is 16. As the stars aren't replaced, the denominator goes down one unit because of the star we took. -As you move down each branch you can calculate the probability of each sequence. -Moving down the branch, you take the probability of each outcome in a sequence and multiply it by each other. -You the take that total and add it to the total of the other sequences that have red for the last star.

14. Calculating the probability of the third star being red... P ( R , R , R ) P ( R , B , R ) P ( R , Bl, R ) P ( B , R , R ) P ( B , B , R ) P ( B , Bl, R ) P (Bl, Bl, R ) P (Bl, B , R ) P (Bl, R , R ) = = = = = = = = = = = = = = = = = = 24 3360 ( ) 16 15 14 4 3 2 ( ) ( ) ( ) 4 5 3 16 15 14 ( ) ( ) ( ) 4 7 3 16 15 14 ( ) ( ) ( ) 5 4 3 16 15 14 ( ) ( ) ( ) 5 4 4 16 15 14 ( ) ( ) ( ) 5 7 4 16 15 14 ( ) ( ) ( ) 7 6 4 16 15 14 ( ) ( ) ( ) 7 5 4 16 15 14 ( ) ( ) ( ) 7 4 3 16 15 14 ( ) ( ) ( ) 60 3360 ( ) 84 3360 ( ) 60 3360 ( ) 80 3360 ( ) 140 3360 ( ) 168 3360 ( ) 140 3360 ( ) 84 3360 ( )

15. = = 0.25 = 25% 1 4 = 3360 840 24 3360 ( ) 60 3360 ( ) 84 3360 ( ) 60 3360 ( ) 80 3360 ( ) 140 3360 ( ) 168 3360 ( ) 140 3360 ( ) 84 3360 ( ) + + + + + + + + There is a 25% probability that the third star is red.

16. Part II: Shinobi has 16 ninja stars. Four (4) of them are red, five (5) of them are blue and seven (7) of them are black. 2 stars are thrown. If a blue star thrown, it is discarded and replaced with a red star. If a red star is thrown, it is discarded and replaced with a black star. If a black star is thrown, it is discarded and replaced with a blue star. What is the probability of the second star being thrown is blue?

17. What we know from the problem... This is a dependent probability because the outcome of the second star depends on the outcome of the second star. We also know that when a star is thrown: Blue is replaced with Red . Red is replaced with Black . Black is replaced with Blue . We will start off by drawing another tree diagram and insert the probability of each outcome. Remember, for this question we are looking for the probability of the second star being blue .

18. R Bl B R R Bl B Bl R Bl B B 7 16 4 16 5 16 16 3 8 16 5 16 4 16 6 16 6 16 5 16 7 16 4 16 Note: look for the sequences where the second star is blue.

19. -Since we are looking for the probabilty of the second star being blue, we look at the tree diagram for the sequence where the second branch is blue. -There are 16 stars. The denominator of the probability fractions always stays 16 because the stars are being replaced. -Moving down the branch, you take the probability of each outcome in a sequence and multiply it by each other. -You the take that total and add it to the total of the other sequences that have the second star as blue.

20. Calculating the probability of the second star being blue... P ( R , B ) P (Bl, B ) P ( B , B ) = = = = = = 4 ( ) 16 5 16 ( ) 256 20 ( ) 7 16 ( ) 6 16 ( ) 42 256 ( ) 16 5 ( ) 4 16 ( ) 256 20 ( )

21. = 82 256 = 0.3203 = 32.0% There is a 32% probability that the second star will be blue. = 256 20 ( ) 42 256 ( ) 256 20 ( ) + + 41 128