Counting and Probability Practice Questions
ATAR (Australia) · ATAR Mathematics Methods · 138 free MCQs with instant results and detailed explanations.
138
Total
47
Easy
68
Medium
23
Hard
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Sample Questions from Counting and Probability
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Q1
Easy
In how many different ways can the letters in the word 'MATH' be arranged?
A.
24
B.
12
C.
20
D.
16
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Correct Answer: A
The word 'MATH' has 4 distinct letters. The number of arrangements is given by 4! = 4 × 3 × 2 × 1 = 24.
Q2
Easy
A box contains 3 red, 2 blue, and 5 green marbles. If a marble is picked at random, what is the probability that it is green?
A.
1/5
B.
1/3
C.
1/2
D.
5/10
Show Answer & Explanation
Correct Answer: B
The total number of marbles is 3 + 2 + 5 = 10. The number of green marbles is 5. Thus, the probability of picking a green marble is 5/10 = 1/2.
Q3
Easy
If there are 5 shirts and 3 pairs of pants, how many different outfits can be formed if each outfit consists of one shirt and one pair of pants?
A.
8
B.
15
C.
5
D.
18
Show Answer & Explanation
Correct Answer: B
The number of possible outfits is calculated by multiplying the number of shirts by the number of pants: 5 shirts × 3 pants = 15 outfits.
Q4
Medium
A committee of 4 members is to be formed from a group of 10 people. How many different committees can be formed?
A.
210
B.
120
C.
150
D.
200
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Correct Answer: A
The number of ways to choose 4 members from 10 can be calculated using combinations: C(10, 4) = 10! / (4! * (10 - 4)!) = 210.
Q5
Medium
If a coin is tossed three times, what is the probability of getting at least one head?
A.
0.875
B.
0.5
C.
0.25
D.
0.125
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Correct Answer: A
The probability of getting at least one head is 1 minus the probability of getting no heads (only tails). The probability of getting tails in one toss is 0.5. Therefore, the probability of getting tails in three tosses is (0.5)^3 = 0.125. Thus, the probability of at least one head = 1 - 0.125 = 0.875.
Q6
Medium
In a lottery, a player selects 3 numbers from 1 to 20. If the order of selection does not matter, how many different selections can be made?
A.
1140
B.
100
C.
320
D.
440
Show Answer & Explanation
Correct Answer: A
The number of ways to choose 3 numbers from 20 without regard to order is calculated using combinations: C(20, 3) = 20! / (3! * (20 - 3)!) = 1140.
Q7
Medium
A box contains 4 red, 3 blue, and 5 green balls. If two balls are drawn at random, what is the probability that both balls are the same color?
A.
1/18
B.
1/9
C.
2/9
D.
1/6
Show Answer & Explanation
Correct Answer: C
The probability is calculated by considering the combinations of drawing the same colored balls. The probability of drawing two balls of the same color is (6/12)*(5/11) + (3/12)*(2/11) + (10/12)*(9/11) = 2/9.
Q8
Hard
In a box containing 6 red balls, 4 blue balls, and 5 green balls, if 3 balls are drawn at random without replacement, what is the probability that all three balls are of different colors?
A.
0.175
B.
0.206
C.
0.250
D.
0.300
Show Answer & Explanation
Correct Answer: B
To find the probability of drawing 3 balls of different colors, first calculate the total ways to choose any 3 balls from the 15 available (C(15, 3)). Next, calculate the successful combinations for each color configuration (1 red, 1 blue, and 1 green). The number of successful combinations is: C(6, 1) * C(4, 1) * C(5, 1) = 6 * 4 * 5 = 120. Thus, the probability is P = successful outcomes / total outcomes = 120 / 455 = 0.206.
Q9
Hard
A box contains 5 red, 4 blue, and 3 green balls. If 4 balls are drawn at random, what is the probability that exactly 2 balls are red?
A.
0.321
B.
0.267
C.
0.408
D.
0.192
Show Answer & Explanation
Correct Answer: B
To find the probability of drawing exactly 2 red balls, we can use combinations. The total number of ways to choose 4 balls from 12 is C(12,4). The number of ways to choose 2 red balls from 5 is C(5,2) and choose 2 non-red balls from 7 (4 blue + 3 green) is C(7,2). The probability is then (C(5,2) * C(7,2)) / C(12,4), which approximates to 0.267.
Q10
Hard
In a class of 30 students, 18 play soccer, 12 play basketball, and 6 play both sports. If a student is selected at random, what is the probability that the student plays either soccer or basketball?
A.
0.4
B.
0.6
C.
0.8
D.
0.7
Show Answer & Explanation
Correct Answer: C
To find the probability that a student plays either soccer or basketball, we use the principle of inclusion-exclusion: P(S ∪ B) = P(S) + P(B) - P(S ∩ B). Here, P(S) = 18/30, P(B) = 12/30, and P(S ∩ B) = 6/30. Therefore, P(S ∪ B) = 18/30 + 12/30 - 6/30 = 24/30 = 0.8.
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Counting and Probability — ATAR (Australia) ATAR Mathematics Methods Practice Questions Online
This page contains 138 practice MCQs for the chapter Counting and Probability in ATAR (Australia) ATAR Mathematics Methods. The questions are organized by difficulty — 47 easy, 68 medium, 23 hard — so you can choose the right level for your preparation.
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