Chapter 14.2 — Exercise 14.1 — Real Life Applications
Applications of probability in real life. This is Lesson 2 of 2 in Chapter 14: Probability.
The Same Formula, Genuinely Different Situations
Every problem here still reduces to favourable outcomes over total outcomes. What actually varies from one problem to the next is how much work counting those two numbers correctly takes — from a straightforward die roll to an area-based dart-board calculation needing geometry before probability even begins. Nine genuinely different real-world sources run through this exercise — dice, coins, spinners, marbles, letters, flour bags, insurance records, and a dart board — and every single one still resolves to the exact same fraction once its outcomes are correctly identified.
A Straightforward Count
A die is rolled once; find the probability of a composite number turning up. Possible outcomes: 1, 2, 3, 4, 5, 6 — all equally likely, since each face has an identical chance of landing up. Composite numbers among them: 4 and 6 (composite means having a factor besides 1 and itself; 1 has no such factor and the primes 2, 3, 5 don't either). P(composite) = 2/6 = 1/3.
Checking "equally likely" explicitly before applying the formula is worth doing as a genuine habit, not a formality — an ordinary fair die guarantees it here, since no face is weighted or shaped differently from any other, but this exact assumption is exactly what fails later in this same exercise, once the marbles and letters no longer split into equal-sized groups.
Real Recorded Outcomes Instead of Theoretical Ones
A coin is tossed 100 times: 45 heads, 55 tails recorded. P(head) = 45/100 = 9/20. P(tail) = 55/100 = 11/20. Sum: 9/20 + 11/20 = 20/20 = 1.
This is genuinely experimental probability, not theoretical — nothing here assumes the coin is perfectly fair, and the 45/55 split, noticeably off from a clean 50/50, is exactly the kind of real-world imperfection experimental probability is built to handle. The sum still checks out to exactly 1 regardless, confirming every recorded toss landed in one of the two counted categories with nothing left over or double-counted. This is really the "sum of all probabilities equals 1" fact from the introduction, checked directly against genuine, imperfect experimental data rather than against a clean theoretical example — and it holds here regardless of how far the coin's actual behaviour strays from a perfectly fair 50/50 split.
Reading Likelihood Straight Off a Spinner
A four-colour spinner is spun once. Without any numeric calculation at all, five direct observations follow purely from comparing each colour's share of the spinner: the pointer is most likely to stop on whichever colour occupies the largest region (red), least likely on the smallest (yellow), equally likely between any two regions of matching size (blue and green here), has no chance at all of landing on a colour not present on the spinner (white), and has no colour it's certain to land on, since every region is smaller than the whole.
Every one of these five answers is really a comparison of areas, not a formula — larger region, higher probability; equal regions, equal probability; zero region, zero probability — the same favourable-over-total idea, just read visually instead of computed with a fraction. This is a genuinely useful reminder that the formula itself is never the actual point of any probability problem — comparing sizes, whether regions on a spinner or counts in a data table, is the real underlying task, and the fraction is simply the final, formal way of writing that same comparison down.
A Bag of Marbles, Four Unequal Colours
| Colour | Count | Probability |
|---|---|---|
| Green | 5 | 5/12 |
| Blue | 3 | 3/12 |
| Red | 2 | 2/12 |
| Yellow | 2 | 2/12 |
| Total | 12 | 12/12 = 1 |
The four colour outcomes are explicitly not equally likely here, since their counts genuinely differ — 5 green marbles give green a real, measurable advantage over the 2 red or 2 yellow marbles on every single draw. Unlike the coin and die examples earlier, "equally likely" always needs checking rather than assumed, and this table is exactly what that check looks like once the assumption fails: each colour gets its own share proportional to how many marbles of that colour are actually present.
Letters Instead of Numbers
A letter is chosen at random from the 26-letter English alphabet. P(vowel) = 5/26 (A, E, I, O, U). P(letter after P) = 10/26 (Q through Z). P(vowel or consonant) = 26/26 = 1, since every single letter is necessarily one or the other. P(not a vowel) = 21/26, the remaining 21 consonants that aren't vowels.
The third result, probability exactly 1, is really the sure-event fact from the introduction showing up in a new disguise: "vowel or consonant" covers the entire 26-letter alphabet with nothing left over, so the event is certain by definition, exactly the same way "a number less than or equal to 6" was already shown to be certain for a single ordinary die roll.
When the Sample Space Comes From Real Records
Eleven flour bags, each marked 5 kg, actually weigh: 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00 kg. Find the probability a randomly chosen bag weighs more than 5 kg. Counting directly through the list: 7 of the 11 bags exceed 5 kg. P(more than 5 kg) = 7/11.
A related insurance-records problem works identically at a larger scale: among 2000 drivers, probabilities are read straight off a data table rather than a clean formula — P(age 18–29 with exactly 3 accidents) = 61/2000, P(age 30–50 with 1+ accidents) = 225/2000 = 9/80 (adding four sub-categories from the table first), and P(no accidents at all) = 1305/2000 = 261/400 (adding three further sub-categories). Every one of these is still favourable over total; the only added difficulty is that "favourable" now means correctly summing several rows of a real data table before dividing, not simply reading a single number straight off the table directly.
Probability Built From Area, Not From Counting
A dart is thrown randomly at a square board with an inscribed circle of radius 2 cm; find the exact probability it lands somewhere in the shaded region between the circle and the square.
Area of circle = πr² = (22/7)(2²) = 88/7 sq.cm
Side of square = 2r = 4 cm → Area of square = 16 sq.cm
Shaded area = 16 − 88/7 = 24/7 sq.cm
P(shaded region) = (24/7) / 16 = 3/14 ≈ 21.4%With no discrete outcomes to count at all, area itself stands in for "number of outcomes" — the entire square board is the full sample space here, and the shaded region is precisely the favourable event being asked about, with the identical favourable-over-total logic carrying through completely unchanged, even though nothing here is being counted one discrete item at a time the way a die roll or a marble draw would be.
Where Class 9 Mathematics Ends
This problem closes Class 9 Mathematics on a genuinely fitting note — a single probability calculation reaching back into Chapter 11's area formulas to finish itself, tying two entirely separate chapters neatly together in one final, closing worked example. Class 10 Mathematics, Chapter 1 begins with Euclid's Division Lemma, returning to number theory with the same careful, deliberate, step-by-step reasoning built up patiently across every single chapter covered here.