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Probability (Day 1) Green Problems

Suppose you select a letter at random from the words MIDDLE SCHOOL.

Find P(L) and P(not L). First determine the number of possible outcomes. There are

12 letters in the two words, so there are 12 possible outcomes when you select a

letter at random. Next determine of favorable outcomes for P(L). There are twoLs.

You can find P(not L) several ways. Since there are 12 possible outcomes and 2 are

L, 12 2 = 10 are not L.number of favorable outcomes 10 5

Thus, P(not L) =number of possible outcomes

=12

=6

Also P(not L) = 1 - P(L)

= 11-6

=5

6

number of favorable outcomes 2 1Thus, P(L) =

number of possible outcomes=

12=

6

A drawer contains 6 red socks, 4 blue socks, and 14 white socks. A sock is pulled

from the drawer at random. Find the probability for each case.

1. Red 2. Blue

3. Red or white 4. Red, white or blue

5. Not red 6. Green

A spinner numbered from 1 to 20 is spun randomly. Find the probability of where

the spinner lands for each case.

7. 17 8. An odd number

9. A number divisible by 5 10. 26

11. A number with a 1 in it 12. A prime number

13. A number less than 6 14. A number

15. A number that is not less than 17 16. A number divisible by 3 or 4

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Find each probability for choosing a letter at random from the word PROBABILITY.

17. P(B) 18. P(P)

19. P(A or I) 20. P(not P)

A box contains 7 red, 14 yellow, 21 green, and 84 purple marbles. A marble is

drawn at random from the box. Find each probability.

21. P(red) 22. P(yellow)

23. P(green or red) 24. P(purple, yellow or red)

25. P(not green) 26. P(not purple, yellow or red)

27. The numerical values around the spinner indicate themeasure of the central angle for each sector of thecircle. Using the fair spinner, what is the probability ofgetting Ahead 3 spaces? Express your answer as acommon fraction.

28. What is the probability that a point choseninside the largest rectangle is not within ashaded region? Express your answer as acommon fraction.

2 2 2 2 2

3

1

2

2 2 2 2 2

3

1

2

29. What is the probability of Jonah picking a vowel if he randomly chooses a letter

from the word CAT? Express your answer as a common fraction.

30. There are six bottles of soda, three bottles of juice and one bottle of water in acooler. If a bottle is randomly selected from the cooler, what is the probabilitythat it is the bottle of water? Express your answer as a common fraction.

31. Top Notch Nose Contest

Schnoz Elementary School decided to hold a Top Notch Nose Contest as a

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fundraising activity. Each contestant submitted a photograph of his or her petfeaturing the pets nose, along with an entrance fee of $1.00.

Half of the photographs submitted were pictures of cats. A quarter of thephotographs received were pictures of dogs, 1/8 were pictures of horses, 1/16 were

pictures of rabbits, and 13 were gerbils. Only 1/32 of the photos were picture ofbirds.

How many photos of pets were entered in the contest?

Extra: If each pet had an equal chance of winning, whats the probability that arabbits photograph was the winner?

Probability (Day 1) Green Solutions

1. 1

4

2.1

6

3. 56

4. 1

5. 34

6. 0

7. 120

8.12

9. 15

10. 0

11. 11

20 12.

2

5

13. 14

14. 1

15. 1

5 16.

1

2

17. 211

18.111

19. 311

20.1011

21. 118

22.19

23. 2

9 24.

5

6

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25. 5

6 26.

1

6

27. Since Ahead 3 spaces occupies 100 degrees out of the 360 degrees in the circle,

the probability that the spinner will land there is 100 10 5 .360 36 18= =

28. There are three different-sized regions within the rectangle. Notice that there

are five of each size, and one of each size is shaded. Since15

of each set is shaded,

then15

of the entire rectangle is shaded, and the probability of choosing a point

within a shaded region is also1

5. The probability of choosing a point not within a

shaded region, then, is

1 4

1 .5 5 =

29. There are three letters in the word CAT, only one of which is a vowel. Thus, theprobability of Jonah picking a vowel at random is one out of three or 1/3.

30. The bottle of water is one of the 10 bottles, so the probability that a randomlyselected bottle is the water bottle is 1/10.

31. There were 416 photographs of pets entered in the contest.*EXTRA* - The probability that a rabbits photograph was the winner is 26/416 =

1/16.What I did was do the problem in language first. X = Total number of pets. So, X =Cats (x) + Dogs (x) + Horses (x) + Rabbits (x) + Birds (x) + Gerbils. Then, Isubstituted them for numbers. Then I added 1/2x + 1/4x + 1/8x + 1/16x + 1/32x +13. When I added all of the fractions, the sum was 31/32. The equation was now X =31/32x from 31/32x and from X. Now, the equation was 1/32x = 13. To get X alone,I divided 1/32 from both X and 13. For X, the quotient I got was 416. That was howmany animals were entered in the contest.

*EXTRA* - There are 26 rabbits. I got 1/16 as the probability that a rabbits photo

was the winner by doing 26/416 and I simplified the fraction.

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Bibliography Information

Teachers attempted to cite the sources for the problems included in this problem set. In somecases, sources were not known.

Problems Bibliography Information

31The Math Forum @ Drexel(http://mathforum.org/)

27 - 30 Math Counts (http://mathcounts.org)

1 - 26

Davison, David M. Prentice Hall Pre-Algebra Tools for a Changing World.Needham, Mass: Prentice Hall, 2001.

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