Green Probability Handout 1 - WordPress.com

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. Since 1 5
of each set is shaded,
then 1 5
of the entire rectangle is shaded, and the probability of choosing a point
within a shaded region is also 1 . The probability of choosing a point not within a
5
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, 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.