Chapter 25 — Probability Optional

Before attempting the harder problems, the rules from the main chapter must be at the fingertips. A random experiment has known possible results but an unpredictable actual result. Its sample space S is the set of all outcomes; an event E is any subset of S. If all outcomes are equally likely, the theoretical probability is P(E) = n(E)/n(S). It always satisfies 0 ≤ P(E) ≤ 1, with P = 0 for an impossible event and P = 1 for a sure event. The complement Ē of E satisfies P(Ē) = 1 − P(E), and the probabilities of all the elementary events add to 1.

Three counts must be remembered: one coin 2, two coins 4, three coins 8; one die 6, two dice 36; a pack of cards 52 with 4 suits of 13, 26 red and 26 black, 12 face cards, 4 of each denomination.

The optional problems differ from the main exercise in three ways. First, most have two stages, so the sample space must be built as a grid or a list of pairs. Second, the event is often described in words that need careful translation: 'at least one six', 'the same day', 'consecutive days', 'a number greater than the other', 'neither a king nor a queen'. Third, some problems give a probability and ask for an unknown count, which requires an equation.

The examiners' scheme for a four-mark question typically gives one mark for the sample space or the total number of outcomes, one for listing the favourable outcomes, one for the fraction and one for the final simplified answer, with the parts of a multi-part question sharing these. Therefore, however sure the answer, write n(S), list n(E), then divide.

Two warnings apply everywhere. Do not assume two outcomes are equally likely merely because there are two of them; 'rain or no rain', 'pass or fail', 'win or lose' are not fair coins. And do not change the total silently: if a card or ball has been removed, the new total must be used for the next draw.

Finally, an answer must be a fraction in lowest terms, a decimal or a percentage; it must lie between 0 and 1; and where an event is described as certain or impossible, the answer 1 or 0 should be stated explicitly with the reason. With these in place, each of the following sections takes up one type of optional problem and solves it completely.

When two dice are thrown, or one die is thrown twice, the outcome is an ordered pair (a, b), a from the first die and b from the second. There are 6 × 6 = 36 outcomes and all are equally likely. The cleanest way to see them is a grid with rows labelled 1 to 6 for the first die and columns 1 to 6 for the second. Each cell is one outcome.

If we write in each cell the sum a + b, the grid becomes the sum table:

Counting each sum in the table: 2 appears once, 3 twice, 4 three times, 5 four times, 6 five times, 7 six times, 8 five times, 9 four times, 10 three times, 11 twice and 12 once. The total is 36. Hence P(sum 2) = 1/36, P(sum 3) = 2/36 = 1/18, P(sum 4) = 3/36 = 1/12, P(sum 5) = 4/36 = 1/9, P(sum 6) = 5/36, P(sum 7) = 6/36 = 1/6, and the values then decrease symmetrically: P(sum 8) = 5/36, P(sum 9) = 1/9, P(sum 10) = 1/12, P(sum 11) = 1/18, P(sum 12) = 1/36.

A well-known problem says: a student argues that there are 11 possible sums, 2 to 12, so each has probability 1/11. Is the argument correct? No. The eleven sums are not equally likely outcomes; the equally likely outcomes are the 36 ordered pairs, and different sums come from different numbers of pairs. Sum 7 comes from six pairs while sum 2 comes from only one, so 7 is six times as likely as 2.

The same grid answers other questions. P(both dice show the same number) = 6/36 = 1/6 (the diagonal). P(sum is even) = 18/36 = 1/2. P(sum is a prime number) = P(sum 2, 3, 5, 7 or 11) = (1 + 2 + 4 + 6 + 2)/36 = 15/36 = 5/12. P(sum greater than 9) = (3 + 2 + 1)/36 = 6/36 = 1/6. P(product is a perfect square) = P({(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,4), (4,1)}) = 8/36 = 2/9. P(first number greater than second) = 15/36 = 5/12, the cells below the diagonal.

A die is thrown twice. What is the probability that 5 will not come up either time, and that 5 will come up at least once? Direct counting for 'at least once' means listing the pairs with a 5 in the first place (6 pairs), the pairs with 5 in the second place (6 pairs), and then remembering that (5,5) was counted twice: 6 + 6 − 1 = 11. The complement is easier. '5 does not come up either time' means the first die shows one of the five numbers other than 5 and so does the second; 5 × 5 = 25 outcomes. So P(no 5) = 25/36 and P(at least one 5) = 1 − 25/36 = 11/36.

The general pattern for two independent stages is: P(a particular number on neither die) = (5/6)², P(on at least one die) = 1 − (5/6)² = 11/36, P(on exactly one die) = 10/36 = 5/18, P(on both) = 1/36. The student can verify that 25 + 10 + 1 = 36.

'Neither…nor' problems arise with cards too. P(neither a king nor a queen) means the card is not one of the 8 kings and queens, so 44/52 = 11/13. P(neither red nor a face card): the red cards are 26 and the black face cards are 6, so the excluded cards number 32 and the probability is 20/52 = 5/13.

For three coins, P(at least one head) = 1 − P(no head) = 1 − 1/8 = 7/8, and P(at least two heads) = P(exactly two) + P(three) = 3/8 + 1/8 = 1/2. Note that 'at least two heads' and 'at most one head' are complements, each with probability 1/2.

A related question: two dice are thrown; what is the probability that the two numbers are different? The complement 'both the same' has 6 outcomes, so P(different) = 1 − 6/36 = 5/6. And the probability that at least one of the dice shows an even number is 1 − P(both odd) = 1 − 9/36 = 27/36 = 3/4, since each die has 3 odd numbers and 3 × 3 = 9 both-odd outcomes.

The rule to remember: whenever the event contains the words 'at least one' or 'not', first look at the opposite event; it is usually a single, easily counted case. Always state the complement in words, count it, and subtract from 1. Never subtract from 36 by mistake — the subtraction is of probabilities from 1, or of counts from n(S), but not a mixture of the two.

Two customers, Shyam and Ekta, visit a particular shop in the same week, Tuesday to Saturday. Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day, (ii) consecutive days, (iii) different days?

Each customer has 5 choices, so the sample space consists of 5 × 5 = 25 ordered pairs (day of Shyam, day of Ekta), all equally likely. Draw a 5 × 5 grid with Tue, Wed, Thu, Fri, Sat along both sides.

(i) Same day: the pairs (Tue, Tue), (Wed, Wed), (Thu, Thu), (Fri, Fri), (Sat, Sat): 5 outcomes on the diagonal. P = 5/25 = 1/5.

(ii) Consecutive days: the pairs where the two days are next to each other in either order: (Tue, Wed), (Wed, Tue), (Wed, Thu), (Thu, Wed), (Thu, Fri), (Fri, Thu), (Fri, Sat), (Sat, Fri): 8 outcomes. P = 8/25.

(iii) Different days: this is the complement of 'same day', so P = 1 − 1/5 = 4/5, or directly 20/25.

The structure of this problem is exactly that of two dice, with 5 faces instead of 6. Any problem in which two people independently choose one of n things — a day, a seat, a number from 1 to n, a colour — has a sample space of n² ordered pairs, of which n lie on the diagonal (same choice). So P(same choice) = n/n² = 1/n and P(different choices) = 1 − 1/n. For instance, if two friends each pick a number from 1 to 10, P(same number) = 1/10; if two students in a class of the same section are each assigned one of 7 days for a duty, P(same day) = 1/7.

A variation: two friends were both born in the year 1990 (not a leap year, 365 days). What is the probability that they have the same birthday? By the same reasoning, P = 365/365² = 1/365, and P(different birthdays) = 364/365. A student who says 'either they have the same birthday or not, so 1/2' has again assumed equally likely outcomes where there are none.

Another variation from the same family: three friends each choose one of the five weekdays. The sample space now has 5³ = 125 outcomes; P(all three on the same day) = 5/125 = 1/25. The grid method is replaced by the multiplication rule, but the idea is identical.

In some problems the number of objects of one colour is not given; instead a probability, or a relation between two probabilities, is given, and the count must be found. The method is to name the unknown x, write each probability in terms of x, form the equation and solve.

Problem A. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls. Let the blue balls be x. Total = 5 + x. P(red) = 5/(5 + x), P(blue) = x/(5 + x). Since P(blue) = 2 P(red), x/(5 + x) = 10/(5 + x), so x = 10.

Problem B. A box contains 12 balls of which x are black. If 6 more black balls are put in, the probability of drawing a black ball is double what it was. Before: P = x/12. After: (x + 6)/18. Equation: (x + 6)/18 = 2x/12 = x/6. Then 6x + 36 = 18x, 12x = 36, x = 3.

Problem C. A jar contains 24 marbles, some green and some blue. If a marble is drawn at random, the probability that it is green is 2/3. Find the number of blue marbles. Let green = g. Then g/24 = 2/3, so g = 16, and blue = 24 − 16 = 8.

Problem D. A bag contains 18 balls out of which x are red. If one ball is drawn, P(red) = x/18. If 2 more red balls are added, the probability of red becomes 9/8 of what it was: (x + 2)/20 = (9/8)(x/18) = x/16. So 16x + 32 = 20x, 4x = 32, x = 8.

Problem E. In a bag of 30 balls, the probability of a white ball is 2/5 and of a black ball is 1/3; the rest are red. White = 2/5 × 30 = 12; black = 1/3 × 30 = 10; red = 30 − 22 = 8, and P(red) = 8/30 = 4/15.

Points to check: x must come out as a whole number, positive, and not larger than the total; if it does not, the equation was set up wrongly. Also, when balls are added the total changes, so the new denominator is the old total plus the number added. The equation is usually linear; cross-multiply and collect the x-terms on one side. Verify the answer by substituting back and comparing the two probabilities, as was done at the end of Problem B in the main chapter: 3/12 = 1/4 and 9/18 = 1/2, which is double.

When an object is drawn and not put back, the second draw is from a smaller collection, and the probability of the second draw depends on what was taken out. The rule is simple: recount both the favourable objects and the total before every draw.

Bulbs. A lot of 20 bulbs has 4 defective. P(first bulb defective) = 4/20 = 1/5. Suppose the first bulb drawn is good and is kept aside. Now 19 bulbs remain, 4 defective and 15 good; P(second bulb not defective) = 15/19. If instead the first bulb had been defective, the second draw would be from 19 bulbs with 3 defective, and P(defective) = 3/19.

Cards. Five cards, the ten, jack, queen, king and ace of diamonds, are shuffled face down. P(the card picked is a queen) = 1/5. If the queen is drawn and put aside, from the remaining four, P(an ace) = 1/4 and P(a queen) = 0/4 = 0. With a full pack: the ten, jack, queen, king and ace of diamonds are removed and one card is drawn from the remaining 47. P(a heart) = 13/47; P(a diamond) = 8/47; P(a king) = 3/47; P(a face card) = (12 − 2)/47 = 10/47.

Pens. A lot of 144 pens has 20 defective. Nuri buys a pen only if it is good. P(she buys) = 124/144 = 31/36; P(she does not buy) = 5/36. If a good pen is sold and the shopkeeper draws another for the next customer, P(defective) = 20/143.

Names in a box. Slips with the names of 6 girls and 4 boys are in a box. One is drawn: P(girl) = 6/10 = 3/5. If that slip is a girl's and is not returned, P(the next slip is a boy's) = 4/9.

Class 10 does not treat the probability of two events happening one after another as a product; the examination asks only for the probability of the second draw given what happened in the first. The student should therefore read the problem to see whether the first object was replaced. 'With replacement' means the collection is restored and the second probability equals the first; 'without replacement' or 'is not replaced' or 'is kept aside' means the collection shrinks.

An aside that often appears in the optional exercise: a die thrown twice is the same as two dice, and the second throw is not affected by the first; a die has no memory. So 'a die is thrown twice; the first shows 4; what is the probability that the second shows 4?' has the answer 1/6, not 1/36. Drawing without replacement changes the odds; throwing again does not.

Many card and number problems ask for the probability that the object has one property or another. The favourable outcomes are those that have at least one of the two properties, and objects having both must be counted only once.

P(a card is a spade or an ace): spades 13, aces 4, but the ace of spades is in both, so n(E) = 13 + 4 − 1 = 16 and P = 16/52 = 4/13. P(a card is a 10 or a spade) = (4 + 13 − 1)/52 = 16/52 = 4/13. P(a red card or a king) = (26 + 4 − 2)/52 = 28/52 = 7/13, since two kings are red. When the two properties cannot both hold — a spade and a heart, an ace and a king — nothing is subtracted: P(an ace or a king) = 8/52 = 2/13; P(a red card or a black face card) = (26 + 6)/52 = 32/52 = 8/13.

'And' asks for objects with both properties. P(a red face card) = 6/52 = 3/26; P(a black queen) = 2/52 = 1/26; P(a number on a die that is even and prime) = P({2}) = 1/6.

Number problems use the same idea. Cards numbered 1 to 20: P(a multiple of 3 or 5) = P({3, 5, 6, 9, 10, 12, 15, 18, 20}) = 9/20, where 15 is counted once. P(a multiple of 3 and 5) = P({15}) = 1/20. Tickets numbered 1 to 100: P(a number divisible by 7 or by 10) = (14 + 10 − 1)/100 = 23/100, because 70 is divisible by both. P(a number divisible by 2 and 3) = P(multiple of 6) = 16/100 = 4/25. P(a two-digit perfect square) = P({16, 25, 36, 49, 64, 81}) = 6/100 = 3/50.

The Class 10 course does not name the addition rule of probability, but the counting principle it uses — count the first set, add the second, subtract the overlap — is the same as the rule for the union of two sets that the student learned in the Sets chapter: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Recognising this connects two chapters and prevents double counting.

A frequent examination item: 'a card is drawn; find the probability that it is neither a red card nor a queen'. Red or queen: 26 + 4 − 2 = 28. Neither: 52 − 28 = 24, so P = 24/52 = 6/13. Compute the 'or' first, then take the complement.

Experiments may combine different objects. A coin is tossed and a die is thrown. The outcomes are pairs: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6, so n(S) = 12. P(a head and an even number) = P({H2, H4, H6}) = 3/12 = 1/4. P(a tail or a 6) = (6 + 2 − 1)/12 = 7/12. P(a head and a number greater than 4) = 2/12 = 1/6.

Three coins tossed together, or one coin tossed three times, have 8 outcomes; the counts of heads 0, 1, 2, 3 have frequencies 1, 3, 3, 1. Thus P(exactly two heads) = 3/8, P(at least two heads) = 4/8 = 1/2, P(at most one head) = 4/8 = 1/2, P(exactly one tail) = 3/8, P(the same face on all three) = 2/8 = 1/4, P(heads and tails both appear) = 6/8 = 3/4. The last is a complement of 'all the same'.

Three dice give 6³ = 216 outcomes; the SSC course does not go beyond noting the count, but a simple question can be asked: P(all three dice show 6) = 1/216, and P(all three show the same number) = 6/216 = 1/36.

A family with three children: assuming a boy or a girl is equally likely at each birth, the 8 outcomes are like three coins. P(all girls) = 1/8; P(exactly two boys) = 3/8; P(at least one girl) = 7/8; P(no girl) = 1/8; P(the eldest is a boy) = 4/8 = 1/2.

Spinners and wheels: a wheel divided into 8 equal sectors numbered 1 to 8 behaves like a die with 8 faces; P(a number greater than 2) = 6/8 = 3/4; P(a prime number) = P({2, 3, 5, 7}) = 4/8 = 1/2. A spinner with 4 equal sectors coloured red, blue, green, yellow, spun twice, gives 16 outcomes; P(the same colour both times) = 4/16 = 1/4; P(red at least once) = 1 − 9/16 = 7/16.

The general principle behind all these counts is the multiplication rule: if the first stage has m outcomes and the second has n, the two-stage experiment has m × n outcomes, and so on for more stages. It is the reason two coins give 4, two dice 36, a coin and a die 12, three coins 8. Whatever the objects, list the outcomes in a fixed order — vary the last object fastest — so that none is missed and none is repeated, then mark the favourable ones.

Problems in which an object bearing a number is drawn at random are answered by counting numbers with a property. The total is the number of tickets; the favourable count is found by listing or by a divisibility argument.

Discs 1 to 90. P(a two-digit number) = 81/90 = 9/10, because the one-digit numbers are 1 to 9. P(a perfect square) = P({1, 4, 9, 16, 25, 36, 49, 64, 81}) = 9/90 = 1/10. P(divisible by 5) = 18/90 = 1/5, because 90 ÷ 5 = 18. P(a prime number less than 20) = P({2, 3, 5, 7, 11, 13, 17, 19}) = 8/90 = 4/45.

Cards 2 to 101. n(S) = 100. P(an even number) = 50/100 = 1/2 (2, 4, …, 100). P(a perfect square) = P({4, 9, 16, 25, 36, 49, 64, 81, 100}) = 9/100. P(a number that is a multiple of 7) = 14/100 = 7/50 (7, 14, …, 98).

Tickets 1 to 20. P(a multiple of 3 or 7) = P({3, 6, 7, 9, 12, 14, 15, 18}) = 8/20 = 2/5. P(a number with the digit 1) = P({1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}) = 11/20.

Counting multiples. The number of multiples of k from 1 to N is the integer part of N/k. From 1 to 100 there are 100 ÷ 4 = 25 multiples of 4, 100 ÷ 6 = 16 (remainder 4) multiples of 6, and 100 ÷ 12 = 8 multiples of 12, so P(multiple of 4 or 6) = (25 + 16 − 8)/100 = 33/100. Between two limits, subtract: multiples of 3 from 10 to 50 are 12, 15, …, 48; count = (48 − 12)/3 + 1 = 13.

Lottery. 1000 tickets are sold and 5 prizes are given; P(a ticket wins a prize) = 5/1000 = 1/200; P(no prize) = 199/200. A raffle with 2 prizes in 200 tickets; a person bought 10 tickets; P(at least one of them wins) is not required in Class 10, but P(a particular one of his tickets wins the first prize) = 1/200.

Piggy bank. 100 fifty-paise coins, 50 one-rupee coins, 20 two-rupee coins and 10 five-rupee coins; one falls out. Total 180. P(50-paise) = 100/180 = 5/9; P(not five-rupee) = 170/180 = 17/18; P(a coin of at least one rupee) = 80/180 = 4/9.

These problems exercise number sense more than probability. Write the list for small ranges; use the division rule for large ones; and check the boundaries (does the list start at 1 or 2? is the last number included?).

When outcomes are points of a region, P(E) = (favourable area)/(total area), or a ratio of lengths on a line. The optional problems use regions requiring the area formulas of Class 10.

Circle in a rectangle. A die is dropped at random on a rectangular region 3 m by 2 m that contains a circle of diameter 1 m. P(inside the circle) = π × (0.5)² / (3 × 2) = π/24 ≈ 0.131.

Concentric circles. A dartboard has an outer circle of radius 10 cm and an inner circle of radius 4 cm. P(a random hit lands in the inner circle) = π × 16/(π × 100) = 4/25; P(in the ring between) = 21/25. The π cancels, so the answer is exact.

Square in a square. A square target of side 8 cm has a central square of side 2 cm. P(hit the centre square) = 4/64 = 1/16.

Triangle in a rectangle. A rectangular field 20 m by 12 m has a triangular flower bed with base 10 m and height 6 m. P(a stone thrown at random lands in the bed) = (½ × 10 × 6)/(20 × 12) = 30/240 = 1/8.

Semicircle on a rectangle. A window is a rectangle 2 m by 1.5 m with a semicircle of diameter 2 m on top. Total area = 3 + π × 1/2 = 3 + 1.571 = 4.571 sq m. P(a raindrop hits the semicircular part) = 1.571/4.571 ≈ 0.344.

Time intervals. A bus arrives at a stop every 15 minutes and a passenger arrives at a random moment. P(he waits less than 5 minutes) = 5/15 = 1/3; P(he waits more than 10 minutes) = 5/15 = 1/3. The music in a game stops at random within 2 minutes; P(within the first half-minute) = 30/120 = 1/4; P(between 1 and 1.5 minutes) = 30/120 = 1/4.

A point on a segment. A point is chosen on a 10 cm line AB; P(it is nearer A than B) = 5/10 = 1/2; P(it is within 2 cm of the midpoint) = 4/10 = 2/5.

Points to note: the favourable region must lie completely inside the total region; areas must be in the same units; π usually cancels when both regions are circles, and otherwise 22/7 or 3.14 is used as the problem directs. Geometric probability obeys all the usual rules: 0 ≤ P ≤ 1, P(outside) = 1 − P(inside), and the whole region has probability 1.

A probability of 1/6 does not say what a single throw will show; it says what proportion of a large number of throws will show a particular number. If an experiment is repeated N times, the expected number of times an event E occurs is N × P(E). The actual number will be close to this but usually not exactly equal.

A die is thrown 600 times: expected number of sixes = 600 × 1/6 = 100. Two dice are thrown 360 times: expected number of times the sum is 7 = 360 × 6/36 = 60; the expected number of doublets = 360 × 1/6 = 60; the expected number of times the sum is 12 = 360 × 1/36 = 10.

A coin is tossed 1000 times: expected heads = 500. If it actually shows 455 heads, the experimental probability is 0.455 and the coin may still be fair; small departures are normal. If it shows 700 heads, one would suspect a biased coin.

Applications. A factory finds that the probability of a bulb being defective is 0.02; in a lot of 5000 bulbs about 5000 × 0.02 = 100 defective bulbs are expected. An insurance company knows from records that the probability of a claim in a year on a certain policy is 0.005; on 20000 policies it expects about 100 claims, and sets the premium accordingly. A seed company states a germination probability of 0.9; a farmer sowing 2000 seeds expects about 1800 plants.

The reverse computation is also asked: if a die is thrown 300 times and shows 4 on 55 occasions, the experimental probability of 4 is 55/300 = 11/60, close to 1/6. If a batsman has hit a boundary 6 times in 30 balls, the probability that he does not hit a boundary on the next ball is estimated as 24/30 = 4/5.

Surveys give experimental probabilities too. 1500 families were surveyed: 475 have one girl, 814 have two girls, 211 have none. P(a family has exactly one girl) = 475/1500 = 19/60; P(two girls) = 814/1500 = 407/750; P(no girl) = 211/1500. These three add up to 1, as they must, because every family falls in exactly one class. In a survey of 2400 students, 1200 like cricket, 720 like football and the rest like other games; P(a student picked at random likes neither) = 480/2400 = 1/5.

This section closes the circle of the chapter. Theoretical probability, obtained by reasoning about equally likely outcomes, predicts experimental frequencies; experimental frequencies, obtained by counting, estimate probabilities where no theory is available. Both give a number between 0 and 1, and both obey the same rules.

The optional exercise is designed around mistakes that students habitually make. Knowing them in advance is the best preparation.

Error 1: assuming equal likelihood. 'A match is won or lost, so P(win) = 1/2.' 'Either a baby is a boy or a girl.' The second is acceptable as an assumption; the first is not, because teams differ in strength. Examiners ask: 'Apoorv throws two dice and squares the number he gets; Peehu throws one die and squares the number; who has the better chance of getting 36?' Apoorv needs (6,6), P = 1/36; Peehu needs 6, P = 1/6; Peehu has the better chance.

Error 2: miscounting two-coin outcomes. A student argues that two coins give three outcomes, two heads, two tails, one of each, so each has probability 1/3. Wrong: HT and TH are separate outcomes; the probabilities are 1/4, 1/4, 1/2.

Error 3: treating sums of dice as equally likely. Eleven sums, so 1/11 each. Wrong: use the 36-cell table.

Error 4: forgetting the changed total. After removing cards, dividing by 52 instead of the new total. After a ball is taken out, forgetting that both counts drop.

Error 5: double counting in 'or'. P(spade or ace) = 17/52 instead of 16/52.

Error 6: misreading 'at least' and 'at most'. 'At least two heads' among three coins means 2 or 3 heads (4 outcomes); 'at most two heads' means 0, 1 or 2 heads (7 outcomes); 'more than two' means only 3 (1 outcome); 'exactly two' means 3 outcomes.

Error 7: the gambler's fallacy. 'Five heads in a row, so a tail is due.' Each toss is independent; P(head) stays 1/2.

Error 8: an impossible answer. A probability of 7/6 or −1/4 signals an arithmetic slip; recheck.

Error 9: leaving the answer unsimplified or as a mixture. 18/36 should be written 1/2; 0.5 and 50 per cent are acceptable but 18/36 should be reduced.

Error 10: not stating the sample space. Even a correct answer loses the method marks.

Examiners' favourite statements for one-mark 'true or false' or 'fill in the blank' items: the probability of a sure event is 1; of an impossible event is 0; P(E) + P(not E) = 1; the probability of an event lies between 0 and 1 inclusive; when a die is thrown, the probability of a number greater than 6 is 0; when two coins are tossed, the probability of two heads is 1/4; a card drawn from a pack is a face card with probability 3/13; the sum of the probabilities of all elementary events is 1.

Below are complete solutions written the way a four-mark answer should appear.

Problem 1. Two dice are thrown simultaneously. Find the probability of getting (i) a doublet, (ii) a sum of 10, (iii) a sum greater than 10, (iv) numbers whose product is 6.
Solution: n(S) = 36. (i) Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6); P = 6/36 = 1/6. (ii) Sum 10: (4,6), (5,5), (6,4); P = 3/36 = 1/12. (iii) Sum 11 or 12: (5,6), (6,5), (6,6); P = 3/36 = 1/12. (iv) Product 6: (1,6), (2,3), (3,2), (6,1); P = 4/36 = 1/9.

Problem 2. A bag contains 4 red, 5 white and some blue balls. If the probability of drawing a white ball is 1/3, find the number of blue balls and the probability of drawing a blue ball.
Solution: Let blue = x. Total = 9 + x. P(white) = 5/(9 + x) = 1/3 gives 9 + x = 15, x = 6. P(blue) = 6/15 = 2/5.

Problem 3. A card is drawn from a well-shuffled pack. Find the probability that it is (i) a red king, (ii) a face card of spades, (iii) neither a jack nor a heart, (iv) a card bearing a number between 2 and 10 (both excluded) of any suit.
Solution: n(S) = 52. (i) 2/52 = 1/26. (ii) 3/52. (iii) Jack or heart: 4 + 13 − 1 = 16; neither: 36; P = 36/52 = 9/13. (iv) Numbers 3 to 9 are 7 per suit, 28 in all; P = 28/52 = 7/13.

Problem 4. Three coins are tossed. Find the probability of getting (i) exactly two heads, (ii) at least one tail, (iii) heads and tails alternately.
Solution: n(S) = 8. (i) HHT, HTH, THH: 3/8. (ii) 1 − P(HHH) = 7/8. (iii) HTH, THT: 2/8 = 1/4.

Problem 5. A game of chance: a wheel with numbers 1 to 8 in equal sectors is spun. Find the probability that the pointer stops at (i) 8, (ii) an odd number, (iii) a number greater than 2, (iv) a number less than 9.
Solution: n(S) = 8. (i) 1/8. (ii) 1, 3, 5, 7: 4/8 = 1/2. (iii) 3 to 8: 6/8 = 3/4. (iv) All: 8/8 = 1.

Problem 6. A rectangular garden 8 m by 6 m has a circular pond of radius 1.4 m. A ball thrown into the garden lands at random. Find the probability that it falls in the pond (take π = 22/7).
Solution: Area of pond = 22/7 × 1.4 × 1.4 = 6.16 sq m. Area of garden = 48 sq m. P = 6.16/48 = 0.1283, about 77/600.

Each solution states n(S), lists the favourable outcomes, divides and simplifies. Follow the same order in the examination, and write the answer to each part on a separate line with the part number, so that the examiner can award the marks part by part.