Discrete Random Variables Practice Questions
ATAR (Australia) · ATAR Mathematics Methods · 147 free MCQs with instant results and detailed explanations.
147
Total
51
Easy
68
Medium
28
Hard
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Sample Questions from Discrete Random Variables
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Q1
Easy
What is the expected value of a discrete random variable X if it takes the values 1, 2, and 3 with probabilities 0.2, 0.5, and 0.3 respectively?
A.
2.1
B.
1.8
C.
2.5
D.
1.5
Show Answer & Explanation
Correct Answer: A
The expected value (E(X)) is calculated as E(X) = ฮฃ [xi * P(xi)], where xi are the values and P(xi) are their probabilities. Therefore, E(X) = (1 * 0.2) + (2 * 0.5) + (3 * 0.3) = 0.2 + 1 + 0.9 = 2.1.
Q2
Easy
A company produces light bulbs with a 90% success rate of passing quality control. If 5 bulbs are selected at random, what is the probability that exactly 4 will pass quality control?
A.
0.2059
B.
0.3281
C.
0.4101
D.
0.5000
Show Answer & Explanation
Correct Answer: B
Using the binomial probability formula P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where n=5, k=4, and p=0.9. Thus, P(X=4) = C(5, 4) * (0.9)^4 * (0.1)^1 = 5 * 0.6561 * 0.1 = 0.3281.
Q3
Easy
What is the expected value of a discrete random variable X, which takes values 2, 4, and 6 with probabilities 0.2, 0.5, and 0.3 respectively?
A.
4.2
B.
4.0
C.
5.0
D.
3.5
Show Answer & Explanation
Correct Answer: A
The expected value E(X) is calculated as E(X) = ฮฃ (x * P(x)). Here, E(X) = (2 * 0.2) + (4 * 0.5) + (6 * 0.3) = 0.4 + 2.0 + 1.8 = 4.2.
Q4
Medium
A bag contains 4 red, 3 blue, and 5 green marbles. If one marble is drawn at random, what is the expected value of the number of marbles drawn if red marbles are worth 3 points, blue marbles are worth 2 points, and green marbles are worth 1 point?
A.
2.5
B.
3.0
C.
1.8
D.
2.0
Show Answer & Explanation
Correct Answer: A
The expected value is calculated as follows: E(X) = (4/12)*3 + (3/12)*2 + (5/12)*1 = (1/3)*3 + (1/4)*2 + (5/12)*1 = 1 + 0.5 + 0.4167 = 2.5.
Q5
Medium
A six-sided die is rolled twice. What is the probability of getting a sum of 7?
A.
1/6
B.
1/12
C.
1/36
D.
5/36
Show Answer & Explanation
Correct Answer: D
There are 6 ways to achieve a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) out of 36 possible outcomes from rolling two dice, which simplifies to 5/36.
Q6
Medium
A jar contains 3 red and 5 green marbles. If two marbles are drawn without replacement, what is the probability that both marbles are red?
A.
3/8
B.
1/6
C.
1/12
D.
1/4
Show Answer & Explanation
Correct Answer: C
The probability of drawing the first red marble is 3/8, and for the second red marble (after one red is removed) it is 2/7. Therefore, the overall probability is (3/8)*(2/7) = 6/56 = 1/12.
Q7
Medium
A factory produces light bulbs with a 90% chance of being functional. If 10 bulbs are randomly selected, what is the expected number of functional bulbs?
A.
7
B.
9
C.
8
D.
10
Show Answer & Explanation
Correct Answer: B
The expected value for functional bulbs is calculated by multiplying the total number of bulbs (10) by the probability of being functional (0.9), which gives an expected value of 9.
Q8
Hard
A discrete random variable X represents the number of heads obtained when flipping a fair coin 5 times. What is the probability of obtaining exactly 3 heads?
A.
0.3125
B.
0.5
C.
0.2
D.
0.7
Show Answer & Explanation
Correct Answer: A
The probability of getting exactly k heads in n flips of a fair coin is given by the binomial formula P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Here, n = 5, k = 3, and p = 0.5. Thus, P(X=3) = (5 choose 3) * (0.5)^3 * (0.5)^(5-3) = 10 * 0.125 * 0.25 = 0.3125.
Q9
Hard
A factory produces light bulbs, and the probability that a randomly selected bulb is defective is 0.1. If 12 bulbs are selected at random, what is the expected number of defective bulbs?
A.
1.2
B.
0.8
C.
3.0
D.
1.0
Show Answer & Explanation
Correct Answer: A
The expected value E(X) of a discrete random variable is calculated as E(X) = n * p, where n is the number of trials and p is the probability of success. Here, n = 12 and p = 0.1. Thus, E(X) = 12 * 0.1 = 1.2 defective bulbs.
Q10
Hard
A die is rolled three times. Let X be the random variable representing the number of times a 6 appears. What is the expected value of X?
A.
1
B.
1.5
C.
2
D.
3
Show Answer & Explanation
Correct Answer: A
The expected value E(X) for a binomial distribution is given by E(X) = n * p, where n is the number of trials and p is the probability of success. Here, n = 3 and p = 1/6 (the probability of rolling a 6). Therefore, E(X) = 3 * (1/6) = 1.
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Discrete Random Variables โ ATAR (Australia) ATAR Mathematics Methods Practice Questions Online
This page contains 147 practice MCQs for the chapter Discrete Random Variables in ATAR (Australia) ATAR Mathematics Methods. The questions are organized by difficulty โ 51 easy, 68 medium, 28 hard โ so you can choose the right level for your preparation.
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