Revised Programme For B.Sc. Part I Exam 2015

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 two
L’s.
Thus, P(L) =
number of favorable outcomes
number of possible outcomes
=
2
12
=
1
6
You can find P(not L) several ways. Since there are 12 possible outcomes and 2 are
L, 12 – 2 = 10 are not L.
Thus, P(not L) =
number of favorable outcomes
number of possible outcomes
=
Also P(not L) = 1 - P(L)
1
= 16
5
=
6
10
12
=
5
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
9. A number divisible by 5
8. An odd number
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
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 the
measure of the central angle for each sector of the
circle. Using the fair spinner, what is the probability of
getting “Ahead 3 spaces?” Express your answer as a
common fraction.
28. What is the probability that a point chosen
inside the largest rectangle is not within a
shaded region? Express your answer as a
common fraction.
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 a
cooler. If a bottle is randomly selected from the cooler, what is the probability
that 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
fundraising activity. Each contestant submitted a photograph of his or her pet
featuring the pet’s nose, along with an entrance fee of $1.00.
Half of the photographs submitted were pictures of cats. A quarter of the
photographs 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 of
birds.
How many photos of pets were entered in the contest?
Extra: If each pet had an equal chance of winning, what’s the probability that a
rabbit’s photograph was the winner?
Probability (Day 1) – Green Solutions
1. 1
4
2.
3. 5
6
4. 1
5. 3
4
6. 0
7.
1
20
9. 1
5
11.
11
20
8.
1
6
1
2
10. 0
12.
2
5
13. 1
4
14. 1
15. 1
5
16.
1
2
17. 2
11
18.
1
11
19. 3
11
20.
10
11
1
18
22.
1
9
24.
5
6
21.
23. 2
9
25. 5
6
26.
1
6
27. Since “Ahead 3 spaces” occupies 100 degrees out of the 360 degrees in the circle,
100 10 5
.
=
=
the probability that the spinner will land there is
360 36 18
28. There are three different-sized regions within the rectangle. Notice that there
1
of each set is shaded,
are five of each size, and one of each size is shaded. Since
5
1
then
of the entire rectangle is shaded, and the probability of choosing a point
5
1
within a shaded region is also . The probability of choosing a point not within a
5
1 4
shaded region, then, is 1 − = .
5 5
29. There are three letters in the word ‘CAT’, only one of which is a vowel. Thus, the
probability 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 randomly
selected bottle is the water bottle is 1/10.
31. There were 416 photographs of pets entered in the contest.
*EXTRA* - The probability that a rabbit’s 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, I
substituted 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 how
many animals were entered in the contest.
*EXTRA* - There are 26 rabbits. I got 1/16 as the probability that a rabbit’s photo
was the winner by doing 26/416 and I simplified the fraction.
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some
cases, sources were not known.
Problems
Bibliography Information
31
The Math Forum @ Drexel
(http://mathforum.org/)
27 - 30
Math Counts (http://mathcounts.org)
1 - 26
Davison, David M. Prentice Hall PreAlgebra Tools for a Changing World.
Needham, Mass: Prentice Hall, 2001.
Print.
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