Chapter 5 Section 6

1. Chapter 5 Section 6 Further Counting Problems

2. Two types of problems • Two main types of problems in section 6: • Tossing a coin several times. • Draw several colored balls from an urn.

3. Coin Toss Situation: Toss a coin 5 times: Question: How many possible outcomes are there? Answer: 2• 2• 2• 2• 2 = 25 = 32 • Now let us look at some other type of questions dealing with the 5 coin tosses:

4. How many ways can we have… 2 Heads and 3 Tails • Choose ONE of the following: • Look at the number of Heads • Look at the number of Tails • Why? Because if you flip a coin 5 times and 2 are Heads, then we know that 3 are Tails. • This type of question can be answered using combinations

5. Solution to number of ways you can have 2 Heads and 3 Tails • Looking at Heads • We have 5 tosses in which we could get a Head, but we only want 2 Heads, so solution is: • Solution: C ( 5 , 2 ) = P( 5 , 2 ) / 2! = (5 • 4) / (2 • 1) = 10 Note that for this type of problem, n will equal the number of tosses and r will be how many heads (or tails) that you are interested in.

6. Alternate Solution to number of ways you can have 2 Heads and 3 Tails • Looking at Tails • We have 5 tosses in which we could get Tails, but we only want 3 Tails, so solution is: • Solution: C ( 5 , 3 ) = P( 5 , 3 ) / 3! = (5 • 4 • 3) / (3 • 2 • 1) = 10

7. How many different outcomes haveat least 1 heads? • Rephrase “…at least 1 head?” to “1 head or more?” • Recall that the task is to toss the coin 5 times

8. Long solution to “How many different outcomes have at least 1 heads?” • Case I : One Head H T T T T C( 5 , 1 ) = 5 • Case II : Two Heads H H T T T C( 5 , 2 ) = 10 • Case III : Three Heads H H H T T C( 5 , 3 ) = 10

9. Long solution continued • Case IV : Four Heads H H H H T C( 5 , 4 ) = 5 • Case V : Five Heads H H H H H C( 5 , 5 ) = 1 • Final Answer: 5 + 10 + 10 + 5 + 1= 31

10. Short solution to “How many different outcomes have at least 1 heads?” • Note that in the long solution, we had 5 cases, the trick for the short solution is to realize that there is only one other case left: No heads. • C( 5 , 0 ) = 1 is the number of ways to choose 0 Heads, thus the solution is • 25 – C( 5 , 0 ) = 32 – 1 = 31 • Which is the same answer that we got in the long solution

11. Drawing balls from an Urn • An urn contains 18 balls, 10 green and 8 blue. The task is to draw 6 balls. • Notes: • It is assumed that balls are drawn without replacement • Order does not matter! Combinations

12. An urn contains 18 balls, 10 green and 8 blue. The task is to draw 6 balls. • Question: How many different samples are possible? Answer: C( 18 , 6 ) = 18,564

13. Six draws from the urn: 4 green balls and 2 blue balls are drawn. 2. Question: How many samples contain 4 green balls and 2 blue balls? Trick: Break down the 6 draws into two parts. Part 1 looks at the green balls and part 2 looks at the blue balls.

14. 6 draws from the Urn: 4 green balls and 2 blue balls are drawn.(Continued) 4 Green Balls 2 Blue Balls • ___ ___ ___ ___ ___ ___ • C( 8 , 2 ) C ( 10 , 4 ) ( 210 ) • ( 28 ) = 5,880 Answer: There are 5,880 ways to draw 4 green and 2 blue balls!

15. Six draws from the urn and all are green balls. 3. Question: How many samples contain all green balls? Trick: Break down the 6 draws into two parts. Part 1 looks at the green balls and part 2 looks at the blue balls.

16. 6 draws from the urn, how many samples contain all green balls?(Continued) 6 Green Balls 0 Blue Balls • ___ ___ ___ ___ ___ ___ • C( 8 , 0 ) C ( 10 , 6 ) ( 210 ) • (1) (210)(1) = 210 Answer: There are 210 ways to draw all green balls!

17. How many sample contain at most 2 blue balls? • “…at most 2 blue balls?” is the same as “ 2 blue balls or less?” • Like the coin toss, we will break the problems down in cases. • For each case we will break the problem into two parts. The first part concerns the number of blue balls and the second part concerns the number of green balls.

18. Solution to how many sample contain at most 2 blue balls? • Case I: No Blue Balls ( G G G G G G ) C( 8 , 0 ) • C( 10 , 6 ) = 1 • 210 = 210 • Case II: One Blue Ball ( B G G G G G ) C( 8 , 1 ) • C( 10 , 5 ) = 8 • 252 = 2,016

19. Answer to how many sample contain at most 2 blue balls? 3. Case III: Two Blue Balls ( B B G G G G ) C( 8 , 2 ) • C( 10 , 4 ) = 28 • 210 = 5,880 Answer: 210 + 2016 + 5880 = 8106 There are 8,106 different ways to draw at most 2 blue balls.