Probability

(i) \(P(E)+P(\overline{E})=\mathbf{1}\) — this is the complement rule, and it holds for every event E.

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure event (or certain event).

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1, i.e. \(0\le P(E)\le 1\).

(i) NOT equally likely. Whether the car starts depends on its mechanical condition (fuel, battery, engine health) rather than on a symmetric random process — there is no reason at all for ‘starts’ and ‘does not start’ to be equally probable.

(ii) NOT equally likely. Whether the shot goes in depends heavily on the player's skill and the difficulty of the attempt, so ‘shoots’ and ‘misses’ have no reason to be equally probable either.

(iii) Equally likely. If the answer is a pure guess with no information to go on, right and wrong are symmetric — there is nothing to make one more likely than the other.

(iv) Equally likely (the standard assumption used throughout this chapter, ignoring the small real biological variation between the two).

So (iii) and (iv) have equally likely outcomes; (i) and (ii) do not, because in each case one outcome is genuinely more (or less) probable than the other for reasons outside the toss/draw itself.

A coin toss has exactly two possible outcomes, head and tail, and — for a fair, unbiased coin tossed at random — both outcomes are equally likely, each with probability \(\tfrac{1}{2}\). Because neither side is favoured over the other, whichever team calls heads or tails has exactly the same \(\tfrac{1}{2}\) chance of winning the toss as the other team. That symmetry — no team gets an advantage from the method itself — is precisely what makes the toss fair.

Rule to apply. Every probability must satisfy \(0\le P(E)\le 1\). Check each option's value against that rule:

(A), (C) and (D) each lie between 0 and 1, so all three are valid probabilities. Only (B) is negative, which falls outside \([0,1]\), so it cannot be the probability of any event. The answer is (B) −1.5.

By the complement rule,

The probability of ‘not E’ is 0.95.

(i) Orange flavoured candy. The bag contains ONLY lemon flavoured candies, so there are 0 candies of any other flavour — drawing an orange one is impossible.

(ii) Lemon flavoured candy. Every single candy in the bag is lemon flavoured, so drawing a lemon flavoured candy is certain.

These are the two extreme cases from the Key Formulas box: an impossible event (probability 0) and a sure event (probability 1).

Let E be the event ‘the 2 students do not have the same birthday’, so \(P(E)=0.992\). The event ‘the 2 students have the same birthday’ is exactly \(\overline{E}\). By the complement rule,

The probability that the 2 students have the same birthday is 0.008.

Step 1 — the outcomes.

(i) Red. There are 3 red balls.

(ii) Not red. This is the complement of ‘red’: the 5 black balls, or equivalently \(1-\tfrac{3}{8}\).

Step 1 — the outcomes.

(i) Red.

(ii) White.

(iii) Not green. This is the complement of ‘green’: the 5 red plus 8 white marbles, i.e. \(5+8=13\) of the 17, or equivalently \(1-\tfrac{4}{17}\).

Step 1 — the outcomes.

(i) 50p coin.

(ii) Not a ₹5 coin. Complement of ‘₹5 coin’: \(180-10=170\) coins are not a ₹5 coin.

Step 1 — the outcomes. The tank has \(5+8=13\) fish in all, each equally likely to be the one taken out.

Step 2 — favourable outcomes. 5 of the 13 fish are male.

Step 1 — the outcomes.

(i) Points at 8. Exactly 1 outcome (the number 8) is favourable.

(ii) Odd number. The odd numbers are 1, 3, 5, 7 — 4 of them.

(iii) Greater than 2. The numbers 3, 4, 5, 6, 7, 8 are greater than 2 — 6 of them.

(iv) Less than 9. All 8 numbers (1 through 8) are less than 9, so this is a sure event.

Step 1 — the outcomes. 6 equally likely outcomes: 1, 2, 3, 4, 5, 6.

(i) Prime number. The prime numbers among 1–6 are 2, 3, 5 — 3 of them (1 is not prime, 4 and 6 are composite).

(ii) Number lying between 2 and 6. ‘Between 2 and 6’ means strictly between the two, i.e. 3, 4, 5 — 3 numbers.

(iii) Odd number. The odd numbers are 1, 3, 5 — 3 of them.

All three happen to come out to \(\tfrac{1}{2}\) here — a coincidence of which particular 3 faces each event picks out, not a general rule.

Step 1 — the deck. 52 cards, 4 suits of 13, so 52 equally likely outcomes.

(i) King of red colour. The red suits are hearts and diamonds, each with exactly one king, so 2 red kings.

(ii) Face card. The face cards are the jacks, queens and kings — \(3\times 4=12\) of them.

(iii) Red face card. Of those 12 face cards, the red ones (hearts and diamonds) number \(3\times 2=6\).

(iv) Jack of hearts. There is exactly 1 such card in the whole deck.

(v) Spade. One whole suit, 13 cards.

(vi) Queen of diamonds. Exactly 1 such card.

(i) First draw. 5 cards, all equally likely, and exactly 1 is the queen.

(ii) Second draw, after the queen is set aside. Only 4 cards remain — ten, jack, king, ace — and every one of them is equally likely to be picked next. This is a NEW, smaller sample space; recount from scratch rather than reusing the first fraction.

(a) An ace. Exactly 1 of the 4 remaining cards is the ace.

(b) A queen. The queen was already removed in the first draw, so 0 of the 4 remaining cards is a queen — this is now an impossible event.

Step 1 — the outcomes. \(12+132=144\) pens in total, each equally likely to be drawn.

Step 2 — favourable outcomes. 132 of the 144 pens are good.

(i) First draw. 20 bulbs, 4 defective, all 20 equally likely to be drawn.

(ii) Second draw, without replacement. The bulb removed in part (i) was NOT defective, so of the original 16 good bulbs, only \(16-1=15\) good bulbs remain, while all 4 defective bulbs are still in the lot. In total, \(20-1=19\) bulbs remain, all equally likely to be drawn next.

Step 1 — the outcomes.

(i) Two-digit number. The single-digit numbers among 1–90 are just 1 to 9 — 9 of them — so the two-digit numbers are the remaining \(90-9=81\).

(ii) Perfect square number. The perfect squares from 1 to 90 are \(1^2,2^2,\dots,9^2\), i.e. 1, 4, 9, 16, 25, 36, 49, 64, 81 (since \(9^2=81\le 90\) but \(10^2=100>90\)) — 9 numbers.

(iii) Divisible by 5. These are 5, 10, 15, ..., 90 — an arithmetic list with \(\frac{90}{5}=18\) terms.

Step 1 — read the faces carefully. The six faces, as printed, read A, B, C, D, E, A — note the letter A appears TWICE, so the six faces are NOT six different outcomes; two of them happen to carry the same letter.

(i) Getting A. A appears on 2 of the 6 faces.

(ii) Getting D. D appears on exactly 1 of the 6 faces.

The five distinct letters A, B, C, D, E do not have equal probability here — A is twice as likely as any of the other four, precisely because it is printed on two faces.

Step 1 — another AREA ratio, as in Example 11. Dropping the die at random on the region means every point of the rectangle is equally likely to be where it lands, so the probability is again a ratio of areas.

Step 2 — the total area. From the figure, the rectangle is 3 m wide and 2 m tall (again, the numbers are given only in the figure):

Step 3 — the favourable area. The circle has diameter 1 m, so its radius is \(\tfrac{1}{2}\) m:

Step 4 — the ratio.

The probability is \(\dfrac{\pi}{24}\) (leave the answer in terms of \(\pi\); numerically, \(\approx 0.131\)).

Step 1 — the outcomes. 144 pens, of which \(144-20=124\) are good.

(i) She will buy it. She buys only a good pen; there are 124 of those.

(ii) She will not buy it. This is the complement of ‘will buy’ — the 20 defective pens.

(i) Filling the table. Using the 36-outcome table from Example 13, count how many of the 36 ordered pairs give each sum:

and by the same symmetric pattern, sum = 9 has 4 pairs, sum = 10 has 3 pairs, sum = 11 has 2 pairs, sum = 12 has 1 pair (sum = 8 is already given as 5 pairs). Dividing every count by 36 gives the completed table:

Check: adding all 11 probabilities gives \(\tfrac{1+2+3+4+5+6+5+4+3+2+1}{36}=\tfrac{36}{36}=1\), as it must, since the sum of the probabilities of all the elementary events of an experiment is always 1.

(ii) Is the argument correct? No, the argument is NOT correct. The GENUINELY equally likely outcomes of this experiment are the 36 ordered pairs listed in Example 13's table — not the 11 possible SUMS. The 11 sums are a coarser grouping of those 36 outcomes, and from the table above they clearly do NOT occur equally often: a sum of 7 comes from 6 different pairs, while a sum of 2 comes from only 1 pair. Treating ‘11 possible outcomes, hence each \(\tfrac{1}{11}\)’ assumes those 11 sums are equally likely, which the completed table shows to be false — for instance \(P(\text{sum}=7)=\tfrac{6}{36}=\tfrac{1}{6}\), which is nowhere near \(\tfrac{1}{11}\). The lesson: before applying \(P(E)=\tfrac{\text{favourable}}{\text{total}}\), always check that the outcomes you are counting are truly equally likely — here that means the 36 dice pairs, not the 11 sums.

Step 1 — list all the outcomes. Each toss gives H or T, and the three tosses are independent, so the outcomes are ordered triples. All \(2\times2\times2=8\) outcomes are equally likely:

Step 2 — Hanif WINS. He wins only if all three tosses match: HHH or TTT — 2 outcomes.

Step 3 — Hanif LOSES. Losing is the complement of winning.

The probability that Hanif loses the game is 3/4.

Step 1 — the outcomes. Per the hint, treat ‘a die thrown twice’ exactly like Example 13's two dice: 36 equally likely ordered pairs \((a,b)\), where \(a\) is the result of the first throw and \(b\) of the second.

(i) 5 does not come up either time. Neither \(a\) nor \(b\) may be 5, so each of them has 5 allowed values (1, 2, 3, 4, 6) instead of 6:

(ii) 5 comes up at least once. This is the complement of ‘5 does not come up either time’:

(i) Two coins, ‘three outcomes each \(\tfrac{1}{3}\)’ — NOT CORRECT. The genuinely equally likely outcomes are the 4 ordered pairs \(\{HH,HT,TH,TT\}\) (as in Example 9), not the 3 DESCRIPTIONS ‘two heads’, ‘two tails’, ‘one of each’. Counting how many of the 4 equally likely outcomes fall under each description:

So the true probabilities are

— NOT \(\tfrac{1}{3}\) each. The argument's error is treating three UNEQUALLY-sized groups of outcomes as if they were themselves equally likely elementary events.

(ii) Die, ‘odd or even, each \(\tfrac{1}{2}\)’ — CORRECT. Here the same check works OUT in the argument's favour: the genuinely equally likely outcomes are the 6 faces \(\{1,2,3,4,5,6\}\), and the two groups ‘odd’ = \(\{1,3,5\}\) and ‘even’ = \(\{2,4,6\}\) each contain exactly 3 of those 6 outcomes — they ARE equally sized. So

and the stated conclusion happens to be correct here — but only because odd and even really do split the die's 6 faces into two equal-sized halves, not merely because there were ‘two named options’. Comparing (i) and (ii) side by side is the whole point of this question: counting how many DESCRIPTIONS an experiment has tells you nothing about their probabilities on its own — you always have to go back to the genuinely equally likely outcomes and count those.