SAT Practice 6
- D Put an “x” through any number that is either
even orgreater than 5, or both. This gives us 8, 9,
2, and 6, which is 4 out of the 6 spaces, giving a
probability of^4 ⁄ 6 , or^2 ⁄ 3. - D If the probability of choosing a green marble is
to be^3 ⁄ 4 , then^3 ⁄ 4 of the marbles should be green and
(^1) ⁄ 4 not green. There are 10 blue and 14 red, for a total
of 24 “not green” marbles, and this will not change,
since you are adding only green marbles. If this is^1 ⁄ 4
of the total, then there must be 4(24) =96 marbles
total after you add the extra green marbles. The jar
now contains 10 + 8 + 14 =32 marbles, so you must
add 96 − 32 =64 green marbles.
- B The six sides of a die add up to 1 + 2 + 3 + 4 +
5 + 6 =21. The sum of any five faces can be greater than
18 only if the “down” face is 1 or 2 (so that the sum of
the other faces is either 21 − 1 =20 or 21 − 2 =19. This
is 2 possibilities out of 6 for a probability of^2 ⁄ 6 , or^1 ⁄ 3. - E Sketch the target:
You want to know the proba-
bility of the arrow hitting the
outermost ring, which is the
ratio of the area of the ring to
the entire area of the target.
The area of the whole target is π(3)^2 = 9 π. The area
of the outermost ring is 9π−π(2)^2 (subtract the area
of the middle circle from the area of the big circle)
= 9 π− 4 π= 5 π. So the probability is 5π/9π=5/9.- A Consider a stretch of 100 consecutive nights.
If the probability of a meteor shower is^2 ⁄ 25 , then we
should expect a meteor shower on (^2 ⁄ 25 )(100) =8 of
those nights. If only^1 ⁄ 4 of the nights are cloudless,
though, then (^1 ⁄ 4 )(8) =2 of the nights with a meteor
shower, on average, should be cloudless. This
gives a probability of^2 ⁄ 100 , or^1 ⁄ 50. Mathematically,
we can just multiply the two probabilities (as long
as they are independent) to get the jointprobabil-
ity: (^2 ⁄ 25 )(^1 ⁄ 4 ) =^1 ⁄ 50.
6.^3 ⁄ 28 Call the probability of choosing a yellow ball x.
If there are three times as many red balls as yellow
balls, the probability of choosing a red ball must
be 3x.The probability of choosing a green ball is
(^4) ⁄ 7. These probabilities must have a sum of 1:
x+ 3 x+^4 ⁄ 7 = 1
Simplify: 4 x+^4 ⁄ 7 = 1
Subtract^4 ⁄ 7 :4x=^3 ⁄ 7
Divide by 4: x=^3 ⁄ 28
- D Most people would say that this probability is
quite high, because the test is so reliable. But intu-
ition is often wrong. Imagine that you test 101 peo-
ple. Of these, on average, one will have the disease,
and 100 will not. Since the test is 100% accurate
for those who have the disease, that person will test
positive. Of the 100 who do not have the disease,
99 will test negative, but one will test positive, be-
cause of the 1% “false positive” rate. So of those
two who test positive, only one will have the dis-
ease; thus, the probability is^1 ⁄ 2.
CHAPTER 9 / SPECIAL MATH PROBLEMS 357
Concept Review 6
- 1
- red marble:^3 ⁄ 12 , or^1 ⁄ 4
white marble:^4 ⁄ 12 , or^1 ⁄ 3
blue marble:^5 ⁄ 12 - What is the probability of drawing a red marble?
(^5) ⁄ 15 , or (^1) ⁄ 3
If 3 more red marbles are added, what is the prob-
ability of drawing a red marble?^8 ⁄ 18 , or^4 ⁄ 9
- If the jar contains 24 red and blue marbles and the
probability of selecting a red marble at random is^3 ⁄ 8 ,
there must be (^3 ⁄ 8 )(24) =9 red marbles, and 24 − 9 = 15
blue marbles. If the probability of drawing a red
marble is to be^1 ⁄ 2 , there must be as many red as blue
marbles, so you must add 15 − 9 =6 marbles.
6. Let’s say that the probability of drawing a black
marble is x.Since there are 4 times as many red
marbles as black marbles, the probability of
drawing a red marble must be 4x.The probabil-
ity of choosing a white marble is^2 ⁄ 3. Since we are
certain to pick one of these colors, the sum of all
of these probabilities must be 1: x+ 4 x+^2 / 3 = 1
Simplify: 5 x+^2 ⁄ 3 = 1
Subtract^2 ⁄ 3 :5x=^1 ⁄ 3
Divide by 5: x=^1 ⁄ 15
Therefore,^1 ⁄ 15 of the marbles are black,^4 ⁄ 15 of the
marbles are red, and^2 ⁄ 3 of the marbles are white.
The least common denominator of these fractions
is 15, which means that 15 is the least possible
number of marbles. In that case, there are 1 black
marble, 4 red marbles, and 10 white marbles.
Answer Key 6: Probability Problems
123