These eight original questions mix the skills from choosing the right probability model. Use a normal table for Q(z) values. Write the model you choose before you calculate.
Plan your time with the timed original practice session builder if you want to work under a limit.
Questions
Question 1
A quality inspector checks 10 items from a long production line. Each item is defective with probability 0.03, independently of the others. Let X be the number of defective items.
(a) Name the model and state why it fits. (b) Find P(X = 0) to 4 decimal places.
Answer
(a) X ~ B(10, 0.03). There is a fixed number of trials (10), each item is defective or not, the probability is the same for each, and the items are independent.
(b) P(X = 0) = 0.97¹⁰ = 0.7374.
Question 2
A bag holds 8 tokens, of which 3 are gold. Three tokens are drawn one at a time without replacement. Explain why the number of gold tokens is not binomial, then find the probability that all three are gold.
Answer
The tokens are not replaced, so the probability of gold changes from draw to draw. The constant-probability condition fails, and the draws are not independent.
P(all gold) = (3/8) × (2/7) × (1/6) = 6/336 = 1/56 ≈ 0.0179.
Question 3
X ~ B(5, 0.25). Find P(X ≥ 1).
Answer
The complement of X ≥ 1 is X = 0.
P(X ≥ 1) = 1 − 0.75⁵ = 1 − 0.2373 = 0.7627.
Question 4
The mass of a fictional mango is normally distributed with mean 400 g and standard deviation 30 g. Find the probability that a mango chosen at random has mass above 445 g.
Answer
z = (445 − 400) ÷ 30 = 1.5. The shaded region is the right tail, so P = Q(1.5) = 0.0668.
The tail is thin, so a small answer is expected.
Question 5
Each attempt at a game is won with probability 0.3, independently. Find the smallest number of attempts n so that the probability of at least one win exceeds 0.95.
Answer
Write 1 − 0.7ⁿ > 0.95, so 0.7ⁿ < 0.05.
Taking logarithms and reversing the inequality, n > log 0.05 ÷ log 0.7 = 8.40.
Check: 0.7⁸ = 0.0576 is not below 0.05, and 0.7⁹ = 0.0404 is. So n = 9.
Question 6
Scores in a fictional test are normally distributed with mean 72 and standard deviation 12. The top 10% of scores are above k. Find k.
Answer
The shaded region is the right tail with area 0.10, so z > 0. Q(z) = 0.10 gives z = 1.28.
k = 72 + 1.28 × 12 = 72 + 15.36 = 87.36.
This describes a position in the fictional group and does not predict any real SPM grade.
Question 7
X ~ N(25, 2²). A diagram shades the region between X = 24 and X = 27.5. A student writes the answer as Q(1.25) − Q(0.5) = −0.2029. Explain what went wrong, then find the correct probability.
Answer
A probability cannot be negative, so the subtraction is wrong. The student treated both ends as right tails.
Standardise: for 24, z = (24 − 25) ÷ 2 = −0.5. For 27.5, z = (27.5 − 25) ÷ 2 = 1.25.
The strip crosses the mean, so expect a probability above 0.5.
P = 1 − Q(0.5) − Q(1.25) = 1 − 0.3085 − 0.1056 = 0.5859.
Question 8
On a fictional farm, each egg is cracked with probability 0.1, independently. A tray holds 20 eggs. Let X be the number of cracked eggs.
(a) Write the model. (b) Find P(X ≥ 2) to 4 decimal places.
Answer
(a) X ~ B(20, 0.1).
(b) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1).
P(X = 0) = 0.9²⁰ = 0.121 58. P(X = 1) = 20 × 0.1 × 0.9¹⁹ = 0.270 17.
P(X ≥ 2) = 1 − 0.121 58 − 0.270 17 = 1 − 0.391 75 = 0.6083.
A common slip is to write 1 − P(X = 1) and forget the P(X = 0) term.
If you got these wrong
| Question | Skill to revisit |
|---|---|
| 1 and 2 | Why repeated events do not always form a binomial model |
| 3, 5 and 8 | Using the complement for an at-least probability |
| 6 | Recovering a raw score from a normal percentile |
| 4 and 7 | Reconciling a calculation with the shaded diagram |
Record each wrong step in the mistake log and paper-error review. If the same slip keeps returning, see online one-to-one Additional Mathematics tuition.