ACT Math : Data Analysis
Study concepts, example questions & explanations for ACT Math
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Example Questions
Example Question #24 : Probability
Robert has a lot of loose change in his pocket. He pulled it all out and found that he had 5 nickels, 8 dimes, 3 quarters, and 13 pennies in his pocket. Robert then put all the change back in his pocket except for the quarters. What is the probability of Robert pulling out a dime at random? (assume that the physical size of the coins has no effect on their probability of being picked)
1/13
1/26
8/29
13/26
4/13
4/13
The total number of coins in Robert's pocket is 26. because there are 8 dimes, he has an 8/26 chance of choosing a dime. This reduces down to 4/13 when you divide the numerator and denominator by 2 (GCF)
Example Question #25 : Probability
If there are 490 marbles in a bag, divided evenly into the seven colors of the rainbow, what is the probability of picking out a marble that is one of the three primary colors?
Since the rainbow consists of seven colors, and there are 490 marble, there are 70 of each color (490/7 = 70).
The probability of choosing a marble that is a primary color would be (70 * 3)/490 = 210/490 = 3/7.
Example Question #26 : Probability
A bottle contains 7 yellow jelly beans, 8 blue jelly beans, 9 green jelly beans, and 14 purple jelly beans. If you select one jelly bean at random, what is the probability that it is neither purple nor yellow? Round to the nearest hundredth.
0.45
0.57
0.76
0.43
0.87
0.45
This is a probability question. There are 17 jelly beans that could be selected to satisfy the conditions of selecting a jelly bean that is neither purple or yellow. There are 38 total jelly beans. The proportion should be set up as 17/38 or 0.45, because this represents the probability that a jelly bean that is not purple or yellow will be selected.
Example Question #27 : Probability
What is the probability of rolling a sum of four or less on a pair of regular six-sided dice?
1/2
1/6
1/3
1/4
1/12
1/6
Probability = what you want ÷ total number
There are 6 ways to get a sum of four or less: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)
The total would be 6 * 6 = 36
So the probability of getting four or less (4, 3, 2, 1) is 6/36 = 1/6
The probability of getting less than four (3, 2, 1) is 3/36 = 1/12
Example Question #141 : Data Analysis
If you flip a coin 3 times, what are the odds that the coin will be heads all three times?
If you flip a coin, the chances of you getting heads is 1/2. This is true every time you flip the coin so if you flip it 3 times, the chances of you getting heads every time is 1/2 * 1/2 * 1/2, or 1/8.
Example Question #31 : Probability
What is the probability of getting a sum of four when rolling two six-sided dice?
The sample space for rolling two six-sided dice 
There are three ways to roll a four.
The probability of rolling a four is given by the number of outcomes that produce a four divided by the total number of outcomes.
Example Question #32 : Probability
A bag contains 3 green marbles, 5 red marbles, and 9 blue marbles.
What is the probability of drawing a red marble?
The probability of drawing a red marble is 
The total number of marbles 
Example Question #33 : How To Find The Probability Of An Outcome
In a certain lottery, ten balls are numbered #1 through #10. Two winning balls will be drawn without replacement.
Billy plays the lottery and picks the numbers 9 and 10.
What is the probability Billy Bob will have two winning numbers?
1/90
1/50
2/95
1/45
1/100
1/90
(1/10)(1/9) = 1/90
Example Question #31 : How To Find The Probability Of An Outcome
In a lottery, ten balls are numbered #1 through #10. Two winning balls will be drawn without replacement.
Billy plays the lottery and picks the numbers 9 and 10.
What is the probability that Billy has no winning numbers? (Round to the nearest one percent.)
92%
88%
55%
96%
62%
62%
The correct answer is 62%, obtained from (8/10) (7/9) = 0.62222
Example Question #33 : How To Find The Probability Of An Outcome
A fair coin is tossed 2 times.
When flipping a coin twice, what is the probability of getting one heads and one tails? Without regard to order. I.e., either the heads or the tails could appear first.
0
0.75
1.0
0.25
0.5
0.5
There are 4 possible outcomes for the event {HH, HT, TH, TT}. Two of these four outcomes include exactly one heads and one tails.
Thus the solution is 2/4 = 1/2 = 0.5
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