Chapter 13.3 — Exercise 13.2 — Probability Problems

Balls, Marbles, and Coins in Containers

Four separate problems share the identical shape: a container holds a known mix of items, one is drawn or falls out at random, and the question asks for the probability of one category or another.

The "not red" and "not a ₹5 coin" parts are both complementary-event shortcuts from Exercise 13.1 in disguise — rather than counting black balls or non-₹5 coins directly, each answer comes just as fast by subtracting the direct probability from 1. The green-marble question uses the same trick from the other direction: 5 red + 8 white = 13 non-green marbles out of 17, without ever needing to count anything green at all.

A Spinning Arrow

A game's arrow comes to rest pointing at one of eight numbers, 1 through 8, arranged around a dial, each equally likely to be the resting point.

Part (iv) is a certain event dressed up in numbers rather than stated plainly — every single number on an eight-number dial is automatically less than 9, so the event and the entire sample space are actually identical, and the probability has to come out to exactly 1 without any real calculation needed.

A Full Deck of Cards, Six Ways

Parts (ii) and (iii) are worth comparing directly: a face card is 12 cards drawn from all four suits, while a red face card is only the 6 of those 12 that happen to sit in the two red suits — exactly half, since red and black suits split the deck evenly. Whenever a "red" or "black" qualifier gets attached to an existing card category, halving that category's count is usually the fastest route to the new one, though it's always worth confirming the category actually splits evenly before relying on the shortcut.

Drawing Without Replacement

Five cards — the ten, jack, queen, king, and ace of diamonds — sit face down. One is drawn and set aside before a second card is drawn from what remains, which shrinks the sample space for the second draw down to four cards rather than five.

Part (ii)(b) is another impossible event, and it's worth seeing exactly why: once the single queen among these five cards has already been removed, no queen remains anywhere in the four cards left to draw from — the event isn't merely unlikely, it genuinely cannot happen, which is what makes its probability exactly 0 rather than just small.

Good, Defective, and Divisible: Three More

The bulb problem is the second without-replacement scenario in this exercise, and it follows the identical pattern as the card problem above: removing one item first changes both the total count and, depending on what was removed, the favourable count for whatever gets asked next. Here a good bulb is removed, so the defective count stays at 4 while the sample space shrinks from 20 to 19 — a smaller adjustment than the queen-removal problem, where the removed card came directly out of the very category being asked about next.

The discs problem's perfect-square count is worth double-checking by listing rather than estimating: 1, 4, 9, 16, 25, 36, 49, 64, and 81 are the only perfect squares from 1 to 90 (100 would be the next one, past the range), giving exactly 9 — easy to undercount by missing 1 itself, since it's a perfect square too (1 = 1²).

A Die Dropped on a Region

A die is dropped at random onto a rectangular board measuring 3 m by 2 m, which contains a circle of diameter 1 m somewhere inside it. Here the sample space isn't a count of discrete outcomes at all — it's an area, with every point on the board equally likely to be where the die lands.

This is a genuinely different flavour of probability from every other problem in this exercise — n(E)/n(S) still applies in spirit, but both "counts" are areas rather than tallies of discrete outcomes. The same ratio-of-areas idea extends to any shape landing on any region, not just a circle on a rectangle, as long as every point in the larger region is equally likely to be where the landing happens.

Buying a Pen, Losing a Game

The two pen-buying answers are complements of each other by construction — "buys it" and "doesn't buy it" between them cover every possible pen in the lot, so 31/36 + 5/36 checks out to exactly 1 without needing to compute either fraction from scratch a second time. The coin-toss game works the same way in reverse: rather than counting all six ways to lose directly, it's often faster to count the two ways to win (HHH, TTT) and subtract from 1 — 1 − 2/8 = 6/8, matching the direct count.

Two Dice, Every Sum From 2 to 12

Rolling two dice together produces 36 equally likely outcomes — every pairing of a first-die result with a second-die result — even though the sum of the two dice can only take the 11 values 2 through 12. Counting how many of the 36 pairings produce each possible sum gives the complete probability distribution:

A student argues that since sums 2 through 12 give eleven possible values, each should have probability 1/11. The table above shows exactly why that reasoning fails: 2, 3, 4, ... , 11, 12 are the sums the two dice can produce, but they are not themselves the 36 equally likely outcomes the classical probability formula needs — a sum of 7, for instance, can happen six different ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), while a sum of 2 can only happen one way (1+1). Since the eleven sums are not equally likely to begin with, dividing evenly into elevenths is simply the wrong starting assumption, however natural it might look at first glance.

A Die Thrown Twice

Throwing one die twice in a row produces the identical 36-outcome sample space as throwing two dice together in a single toss — the two setups are treated as the same experiment.

These two events are complements of each other — "5 never appears" and "5 appears at least once" between them account for every one of the 36 outcomes with no overlap — so 25/36 + 11/36 = 36/36 = 1 confirms both counts without needing to recount either one independently. Counting the 11 favourable outcomes for part (ii) directly (five pairs with a 5 first, five pairs with a 5 second, plus the single (5,5) pair counted once) matches the complement check exactly.

The Chapter, Complete

Probability opened with a small set of definitions in the chapter introduction, narrowed to the single complementary-event relationship in Exercise 13.1, and closed here with fifteen genuinely different sample spaces, all solved with nothing beyond P(E) = n(E)/n(S) and, wherever it applied, the complement shortcut 1 − P(E). The next chapter, Statistics, turns from single random outcomes to entire collections of data, asking a different kind of question — not how likely one result is, but what an entire spread of results looks like on average.