MDM4U Data Management Practice Exam — Free

The phrase "at least one" is the signal to use the indirect method: count every possible subcommittee, then subtract the ones that break the condition. The only way to break "at least one Grade 11 student" is to have no Grade 11 students at all, so that is the single case to remove.

Order does not matter on a committee (the same four people form the same subcommittee however they are listed), so use combinations, not permutations. From all 12 students (7 from Grade 12 plus 5 from Grade 11),

A subcommittee with no Grade 11 students is chosen entirely from the 7 Grade 12 students:

Subtracting gives the number with at least one Grade 11 student:

Check with the direct method, adding the cases of exactly 1, 2, 3 or 4 Grade 11 students (the rest of each subcommittee comes from Grade 12):

Both methods agree.

The classic trap is 825, which comes from "pick one Grade 11 student to guarantee the condition, then any 3 of the remaining 11": $C(5, 1) \times C(11, 3) = 5 \times 165 = 825$. This double-counts. A subcommittee containing two Grade 11 students, Ava and Ben, is counted once with Ava as the "guaranteed" member and again with Ben, so every subcommittee with more than one Grade 11 student is counted more than once. The answer 495 is the total with the restriction ignored. The answer 11 040 is $P(12, 4) - P(7, 4) = 11\,880 - 840$, which treats the subcommittee as an ordered arrangement; it counts every subcommittee $4! = 24$ times, and indeed $11\,040 \div 24 = 460$.

460 subcommittees (option B).

Let $X$ be the number of free throws she makes. Check the binomial conditions: there is a fixed number of trials ($n = 6$), each shot has only two outcomes (make or miss), the probability of success is the same on every shot ($p = 0.7$, so $q = 1 - p = 0.3$), and the shots are independent. So $X$ is a binomial random variable with

(a) Substitute $n = 6$ and $x = 4$:

Evaluate each factor: $\binom{6}{4} = 15$, $(0.7)^4 = 0.2401$ and $(0.3)^2 = 0.09$. Then

The factor of 15 is there because "exactly 4 makes" can happen in 15 different orders (MMMMXX, MMMXMX, and so on), and each of those sequences has the same probability $(0.7)^4(0.3)^2 \approx 0.0216$.

(b) For a binomial distribution the expected value is

On average she makes 4.2 of her 6 free throws, which is why exactly 4 is the single most likely outcome: its probability, 0.3241, is the largest in the distribution, with exactly 5 close behind at 0.3025.

Two common slips: leaving out the $\binom{6}{4}$ gives 0.0216, the probability of one particular sequence rather than of any sequence with 4 makes; and swapping the exponents, $15(0.7)^2(0.3)^4 = 0.0595$, gives the probability of exactly 2 makes instead of 4. Also read the wording carefully: "exactly 4" is a single term of the distribution, whereas "at least 4" would require adding the probabilities for 4, 5 and 6 makes.

(a) $P(X = 4) \approx 0.3241$; (b) $E(X) = 4.2$ free throws.

Let $X$ be a runner's finishing time in minutes, so $X$ is normal with $\mu = 52$ and $\sigma = 10$. Every part uses the same two moves: convert a time to a z-score with $z = \dfrac{x - \mu}{\sigma}$, then read the area to the left of that z-score from the standard normal table. The z-scores below are rounded to 2 decimal places, and the areas are the 4-decimal table values.

(a) For 40 minutes,

About 11.5% of runners finished in under 40 minutes. For 1200 runners the expected count is $(0.1151)(1200) = 138.1$, so about 138 runners.

(b) Both endpoints must be converted:

The table gives areas to the left of a z-score, so subtract the smaller area from the larger one:

The trap here is to subtract the z-scores first, getting $z = 2.00$, and look up $P(Z < 2.00) = 0.9772$. Areas can be subtracted; z-scores cannot.

(c) This part works backwards, from an area to a time. The fastest 10% are the runners with the lowest times, so we need the z-score with an area of 0.1000 to its left. Searching the body of the table, the closest entry is 0.1003 at $z = -1.28$ (the next entry, 0.0985, belongs to $z = -1.29$), so take $z = -1.28$. Now solve for $x$:

A runner must finish in about 39.2 minutes or less to earn a medal. (Using the more precise $z = -1.2816$ gives 39.18 minutes, the same to one decimal place.) The sign is the thing to watch: using $z = +1.28$ gives 64.8 minutes, which is the cutoff for the slowest 10%, not the fastest. A quick sanity check confirms the answer: part (a) showed that 11.5% of runners finish in under 40 minutes, so the cutoff for the fastest 10% should be slightly below 40 minutes, which it is.

(a) $P(X < 40) = 0.1151$, about 138 runners; (b) $P(45 < X < 65) = 0.6612$; (c) about 39.2 minutes.