Key Concepts & Formulae at a Glance
- Probability measures the likelihood of an event, on a scale from 0 (impossible) to 1 (certain).
- A random experiment is repeatable, with a known set of possible outcomes but an unpredictable result each time.
- The sample space S is the list of all possible outcomes; the sample size n(S) is the number of elements in it. An event is any subset of the sample space.
- Tree diagrams visualise multi-step experiments and help list every outcome and its probability.
- The Law of Large Numbers: as the number of trials increases, experimental probability tends to get closer to theoretical probability.
- Gambler's Fallacy: past independent outcomes (e.g. previous coin tosses) never change the probability of the next outcome.
TRSuch unpredictability can be useful sometimes! For example, in a cricket match, the fact that a coin is tossed to decide which team will bat first is considered to be a fair method. Can you explain why?
A coin toss is fair because, for an unbiased coin, \(P(\text{Heads})=P(\text{Tails})=\dfrac{1}{2}\) — exactly equal.
Since both teams have exactly the same chance (50%) of winning the toss, neither team is favoured over the other, regardless of team ranking, ground, or any other factor. This makes it an impartial way to decide who bats first.
ActAsk your friend to predict the outcome of a ₹1 coin you toss. Do you see that your friend could guess heads or tails but could not know for certain? That's randomness!
Try this yourself: toss the coin and ask your friend to call heads or tails before it lands. However confident their guess, they cannot know the result in advance — the outcome only becomes certain once the coin has actually landed.
Exercise Set 7.1
1Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.
(i) The next Monday will come after Sunday.
(ii) It will snow in Mumbai in July.
(iii) An elephant will walk through your classroom today.
(iv) You will greet at least one friend at school tomorrow.
(i) Certain (P = 1) — by the structure of the calendar, Monday always comes immediately after Sunday; this is guaranteed, not a matter of chance.
(ii) Impossible (P = 0) — Mumbai has a tropical climate and never experiences snowfall.
(iii) Impossible (P ≈ 0) — elephants are not found in ordinary school classrooms; this practically cannot happen.
(iv) More likely (P close to 1) — on a typical school day, most students interact with at least one friend, though it isn't absolutely guaranteed (e.g. if a student is absent).
TRIf I have rolled a 4 on a die 8 times in succession, is the probability of rolling a 4 again on the next roll still the same?
Yes. Each roll of a fair die is an independent event — the die has no memory of previous rolls.
\(P(\text{rolling a 4}) = \dfrac{1}{6} \approx 0.167\), regardless of how many 4s (or any other number) came up before.
Exercise Set 7.2
1A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets: 10 red, 8 green, 7 yellow, 5 blue.
(i) Calculate the probability that a randomly picked sweet from the sample is green.
(ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.
(i) \(P(\text{green}) = \dfrac{8}{30} = \dfrac{4}{15} \approx 0.267\text{ or }26.7\%\)
(ii) Proportion yellow in sample \(= \dfrac{7}{30}\). Estimated yellow sweets in 600 \(= \dfrac{7}{30}\times 600 = 140\)
2A survey of 40 students about their favourite club: 14 Science Club, 11 Arts Club, 9 Sports Club, 6 Debate Club. Assume there are 800 students in the whole school.
(i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?
(ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.
(i) \(P(\text{Arts}) = \dfrac{11}{40} = 0.275\text{ or }27.5\%\)
(ii) Proportion preferring Sports \(=\dfrac{9}{40}\). Estimated for 800 students \(=\dfrac{9}{40}\times 800 = 180\)
3Toss a coin 20 times and record the result each time (heads or tails).
(i) How many times did you get heads?
(ii) How many times did you get tails?
(iii) Calculate the experimental probability of getting heads.
(iv) If you toss the coin once more, what is the probability of getting tails?
(i)–(iii) This is a hands-on activity — the actual counts of heads and tails depend on your own 20 tosses. Once you have your counts:
Experimental \(P(\text{heads}) = \dfrac{\text{number of heads you got}}{20}\)
(iv) For one more toss, the probability of tails is the theoretical probability, \(\dfrac{1}{2}=0.5\), since each toss of a fair coin is an independent event — it does not depend on your previous 20 results.
4Toss a paper cup into the air 100 times. After each toss record whether the cup lands on its bottom, upside down on its top, or on its side. Assign probabilities to the outcomes using experimental probability.
This is a hands-on activity — record how many of your 100 tosses land in each of the 3 positions (bottom, top, side), then compute:
\(P(\text{bottom}) = \dfrac{\text{number landing on bottom}}{100}\), and similarly for top and side.
The three probabilities should add up to 1 (100%). Since the cup is not a symmetric object like a coin or die, these probabilities are generally not equal — this is exactly why experimental probability (rather than a theoretical formula) is needed here.
5What is the probability of getting an even number when rolling a fair 6-sided die?
Favourable outcomes: \(\{2,4,6\}\), so 3 outcomes. Total outcomes = 6.
\(P(\text{even}) = \dfrac{3}{6}=\dfrac{1}{2}\)
6Suppose you roll a 6-sided die 12 times and get a '3' three times.
(i) What is the experimental probability of rolling a '3'?
(ii) What is the theoretical probability of rolling a '3'?
(iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?
(i) Experimental \(P(3) = \dfrac{3}{12}=\dfrac{1}{4}=0.25\)
(ii) Theoretical \(P(3) = \dfrac{1}{6}\approx 0.167\)
(iii) They differ because 12 rolls is a small sample, so random variation can cause the observed frequency to deviate noticeably from the theoretical value. By the Law of Large Numbers, as the number of rolls increases (60, 600, 6000), the experimental probability should get progressively closer to the theoretical value of 1/6.
Exercise Set 7.3
1When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
\(S=\{1,2,3,4,5,6\}\)
2For the following experiments write down the sample space S.
(i) Rolling a die and tossing a coin together.
(ii) Choosing a random integer between −5 and +5.
(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.
(i) \(S=\{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)\}\) — 12 outcomes.
(ii) Taking "between −5 and +5" to include both endpoints: \(S=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}\) — 11 outcomes.
(iii) Since the outcome is the colour drawn: \(S=\{\text{Green},\text{Red}\}\)
3In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.
(i) List the sample space of all possible snack and drink combinations a person could choose at the fair.
(ii) List the event 'Selecting Samosa as a snack.'
(i) \(S=\{(\text{Samosa},\text{Chai}),(\text{Samosa},\text{Lassi}),(\text{Pakora},\text{Chai}),(\text{Pakora},\text{Lassi}),(\text{Bhaji},\text{Chai}),(\text{Bhaji},\text{Lassi})\}\) — 6 combinations.
(ii) \(E=\{(\text{Samosa},\text{Chai}),(\text{Samosa},\text{Lassi})\}\)
TRCan you calculate the probability of getting one head and one tail (when tossing a fair coin twice)?
Sample space: \(S=\{HH,HT,TH,TT\}\), each equally likely with probability 1/4.
Event "one head and one tail" \(=\{HT,TH\}\) — 2 outcomes.
\(P(\text{one head, one tail}) = \dfrac{2}{4}=\dfrac{1}{2}\)
Exercise Set 7.4
1There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.
(i) Draw a tree diagram showing all possible pairs of fruits.
(ii) List the sample space.
(iii) What is the probability of picking one apple and one banana?
Tree diagram: 3 equally likely picks from Basket A × 2 from Basket B = 6 equally likely paths
(i) Basket A has 3 equally likely items (Apple, Orange₁, Orange₂), so \(P(\text{Apple})=\dfrac{1}{3}\), \(P(\text{Orange})=\dfrac{2}{3}\). Basket B has 2 equally likely items, so \(P(\text{Banana})=P(\text{Mango})=\dfrac{1}{2}\). The tree diagram branches from A (Apple, Orange, Orange) then from each of those to B (Banana, Mango), giving 6 equally likely paths, each with probability \(\frac13\times\frac12=\frac16\).
(ii) Using fruit types: \(S=\{(\text{Apple},\text{Banana}),(\text{Apple},\text{Mango}),(\text{Orange},\text{Banana}),(\text{Orange},\text{Mango})\}\), with respective probabilities \(\dfrac16,\dfrac16,\dfrac13,\dfrac13\) (summing to 1).
(iii) \(P(\text{Apple and Banana}) = P(\text{Apple})\times P(\text{Banana}) = \dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{6}\)
2A box contains 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
Tree diagram: 3 colours × 3 colours (with replacement) = 9 outcomes, each colour weighted by count out of 9
(i) Total pens = 9. \(P(\text{Red})=\dfrac{3}{9}=\dfrac{1}{3}\), \(P(\text{Black})=\dfrac{4}{9}\), \(P(\text{Green})=\dfrac{2}{9}\). Since the pen is replaced, both picks have the same probabilities and are independent. The tree diagram branches from "you" into R, B, G, and from each of those into "friend"'s R, B, G — giving 9 combined outcomes: RR, RB, RG, BR, BB, BG, GR, GB, GG.
(ii) \(P(\text{same colour}) = P(RR)+P(BB)+P(GG)\)
\(=\left(\dfrac{3}{9}\right)^2+\left(\dfrac{4}{9}\right)^2+\left(\dfrac{2}{9}\right)^2 = \dfrac{9+16+4}{81}=\dfrac{29}{81}\approx 0.358\)
End-of-Chapter Exercises
1Fill in the blanks.
(i) The probability of an impossible event is _______.
(ii) The set of all possible outcomes of a random experiment is called the __________.
(iii) The probability of an event that is certain to happen is _______.
(iv) Tossing a fair coin has a probability of ______ for getting heads.
(i) 0
(ii) sample space
(iii) 1
(iv) 1/2 (or 0.5)
2In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).
15 is the raw count (frequency). The relative frequency is the proportion of the total:
\(\dfrac{15}{50}=0.3\)
3Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) Tossing a fair coin once.
(iii) Rolling a fair 6-sided die.
(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
(v) A baby is born. It is a boy or a girl.
(i) Not equally likely — a car's chance of starting depends on its mechanical condition, not a 50-50 split.
(ii) Equally likely — a fair coin gives heads and tails each probability 1/2.
(iii) Equally likely — a fair die gives each face probability 1/6.
(iv) Not equally likely — there are 3 red and 7 blue marbles, giving different probabilities (3/10 vs 7/10).
(v) Equally likely (by the idealised assumption used in this chapter) — boy and girl are each treated as probability 1/2.
4Write the sample space and calculate the probability based on the given information.
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
(ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(i) \(S=\{HH,HT,TH,TT\}\). At least one head = all except TT: \(P=\dfrac{3}{4}\)
(ii) \(S=\{1,2,...,10\}\). Even numbers: \(\{2,4,6,8,10\}\), \(P=\dfrac{5}{10}=\dfrac{1}{2}\)
(iii) \(S=\{1,...,6\}\). Greater than 4: \(\{5,6\}\), \(P=\dfrac{2}{6}=\dfrac{1}{3}\)
(iv) Total balls = 6. Not red = blue or green = \(2+1=3\); \(P=\dfrac{3}{6}=\dfrac{1}{2}\)
(v) \(S\) has 8 equally likely outcomes: \(\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}\). Exactly two heads: \(\{HHT,HTH,THH\}\) — 3 outcomes. \(P=\dfrac{3}{8}\)
5A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?
\(P(\text{strawberry}) = \dfrac{1}{3}\)
6A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.
| Shirt | Pants |
|---|---|
| Red | Jeans |
| Red | Khakis |
| Red | Shorts |
| Blue | Jeans |
| Blue | Khakis |
| Blue | Shorts |
7A tyre company records distances before replacement in 1000 cases: less than 4000 km — 20; 4001 to 9000 km — 210; 9001 to 14000 km — 325; more than 14000 km — 445. Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.
(i) \(P(<4000) = \dfrac{20}{1000}=0.02\)
(ii) \(P(\text{between }4000\text{ and }14000) = \dfrac{210+325}{1000}=\dfrac{535}{1000}=0.535\)
(iii) \(P(>14000) = \dfrac{445}{1000}=0.445\)
8The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking.
(i) What is the probability that it is a P, E or C?
(ii) What is the probability that it is not an E?
Letters: P, E, A, C, E — 5 cards total, with E appearing twice.
(i) Favourable = P, E, E, C = 4 cards. \(P(P,E,\text{or }C) = \dfrac{4}{5}=0.8\)
(ii) \(P(\text{not }E) = 1-P(E) = 1-\dfrac{2}{5}=\dfrac{3}{5}=0.6\)
9*A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) An odd number?
(iii) A number greater than 2?
(iv) A number less than 9?
(v) A multiple of 3?
Total outcomes = 8.
(i) \(P(8)=\dfrac{1}{8}\)
(ii) Odd numbers \(\{1,3,5,7\}\): \(P=\dfrac{4}{8}=\dfrac{1}{2}\)
(iii) Greater than 2: \(\{3,4,5,6,7,8\}\): \(P=\dfrac{6}{8}=\dfrac{3}{4}\)
(iv) Less than 9: all of \(\{1,...,8\}\) qualify: \(P=\dfrac{8}{8}=1\) (certain)
(v) Multiples of 3: \(\{3,6\}\): \(P=\dfrac{2}{8}=\dfrac{1}{4}\)
10*A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities.
(i) What is the probability of drawing a red ball and then a blue ball?
(ii) What is the probability of drawing 2 blue balls?
Tree diagram (without replacement): 9 balls first draw, 8 remaining for the second
Total balls = 9. First draw: \(P(R)=\dfrac{4}{9}\), \(P(B)=\dfrac{5}{9}\).
Since the first ball is not replaced, the second draw's probabilities depend on the first: if first was Red, 8 balls remain (3R, 5B); if first was Blue, 8 remain (4R, 4B).
(i) \(P(\text{Red then Blue}) = \dfrac{4}{9}\times\dfrac{5}{8}=\dfrac{20}{72}=\dfrac{5}{18}\)
(ii) \(P(\text{Blue then Blue}) = \dfrac{5}{9}\times\dfrac{4}{8}=\dfrac{20}{72}=\dfrac{5}{18}\)
11*I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.
Event with probability 0: "the sum of the two dice is 13" — impossible, since the maximum possible sum with two dice is \(6+6=12\).
Outcome with probability 1: "the sum is at least 2" — certain, since the minimum possible sum is \(1+1=2\), so every possible roll satisfies this.
12*Write the sample space and calculate the probability based on the given information.
(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?
(i) Two dice give 36 equally likely outcomes. Primes greater than 5 (within range 2–12): 7 and 11.
Sum = 7: 6 ways; Sum = 11: 2 ways. Total favourable = 8.
\(P = \dfrac{8}{36}=\dfrac{2}{9}\)
(ii) Total balls = 9. Ways to choose 2 balls: \(\binom{9}{2}=36\).
Same-colour ways: \(\binom{4}{2}+\binom{3}{2}+\binom{2}{2}=6+3+1=10\)
\(P(\text{different colours}) = 1-\dfrac{10}{36}=\dfrac{26}{36}=\dfrac{13}{18}\)
(iii) 8 equally likely outcomes. "First = H AND exactly 2 heads total" means first is H and exactly one of the remaining two is H: outcomes HHT, HTH — 2 outcomes.
\(P=\dfrac{2}{8}=\dfrac{1}{4}\)
(iv) Total 4-digit arrangements of 1,2,3,4 (no repetition) = \(4!=24\). For the number to be even, the last digit must be 2 or 4 (2 choices); the other 3 digits arrange in \(3!=6\) ways: favourable \(=2\times6=12\)
\(P(\text{even})=\dfrac{12}{24}=\dfrac{1}{2}\)
(v) This is a binomial situation: \(n=3\), \(p=\dfrac{1}{4}\) (correct), \(q=\dfrac{3}{4}\) (wrong).
\(P(\text{exactly 2 correct}) = \binom{3}{2}p^2q^1 = 3\times\dfrac{1}{16}\times\dfrac{3}{4}=\dfrac{9}{64}\)
13*A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:
(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.
(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.
(iii) What are the sizes of these two sample spaces?
Tree pattern: with replacement, every branch re-offers all 4 numbers; without replacement, each branch excludes the number just drawn
(i) With replacement: each draw has 4 options, independent of the other. \(S=\{(1,1),(1,2),(1,3),(1,4),(2,1),...,(4,4)\}\) — all ordered pairs from \(\{1,2,3,4\}\times\{1,2,3,4\}\).
(ii) Without replacement: first draw has 4 options; second draw has only the remaining 3 options. \(S\) consists of all ordered pairs with distinct entries, e.g. \((1,2),(1,3),(1,4),(2,1),...\) (excluding pairs like (1,1)).
(iii) With replacement: \(4\times4=16\) outcomes. Without replacement: \(4\times3=12\) outcomes.
14*List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.
\(S=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}\)
15*Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?
(i) {1, 2, 3}
(ii) {0, 1, 2}
(iii) {0, 1, 2, 3, 4}
(iv) {0, 1, 2, 3}
With 3 coins, the number of heads can be 0, 1, 2, or 3 — no other value is possible, and each of these values can actually occur.
(i) {1, 2, 3} — invalid: missing 0 (all-tails is a possible outcome, giving 0 heads).
(ii) {0, 1, 2} — invalid: missing 3 (all-heads is possible, giving 3 heads).
(iii) {0, 1, 2, 3, 4} — invalid: includes 4, which is impossible since there are only 3 coins.
(iv) {0, 1, 2, 3} — valid: includes every possible count of heads exactly once, with nothing missing and nothing impossible.
16*Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8 (3 m × 2 m). What is the probability that it will land inside the circle with a diameter of 1 m?
Fig. 7.8: rectangular region 3 m × 2 m, with a circle of diameter 1 m inside
Area of rectangle \(= 3\times2=6\text{ m}^2\)
Circle diameter = 1 m, so radius = 0.5 m. Area of circle \(=\pi r^2 = \pi(0.5)^2=\dfrac{\pi}{4}\text{ m}^2\approx0.785\text{ m}^2\)
\(P(\text{lands inside circle}) = \dfrac{\text{Area of circle}}{\text{Area of rectangle}} = \dfrac{\pi/4}{6}=\dfrac{\pi}{24}\approx0.131\)
Extra Practice Questions
Seven extra questions in the style of the textbook's own exercises, for independent practice once you've gone through the solved questions above. Attempt each one on paper first, then tap to reveal the full working — exactly like the solved questions above.
1Rank the following events from 0 (Impossible) to 1 (Certain), and give a reason for each:
(i) The sun will rise in the east tomorrow.
(ii) A fair coin lands on its edge and stays there.
(iii) It rains at least once in Mumbai during the monsoon season.
(i) Certain (P = 1) — this is a fixed fact of Earth's rotation, not a matter of chance.
(ii) Impossible (P ≈ 0) — a coin landing and balancing on its thin edge essentially never happens in practice.
(iii) Certain (P close to 1) — Mumbai's monsoon season reliably brings heavy rainfall every year, so at least one rainy day is effectively guaranteed.
2A bag of 50 marbles is sampled, and 40 are drawn out: 18 blue, 14 red, 8 green.
(i) Find the experimental probability of drawing a red marble.
(ii) If the bag actually contains 50 marbles total, estimate how many are green, based on this sample.
(i) Experimental \(P(\text{red}) = \dfrac{14}{40}=\dfrac{7}{20}=0.35\)
(ii) The sample's green proportion is \(\dfrac{8}{40}=\dfrac{1}{5}=0.2\). Applying this proportion to the full bag: \(0.2\times50=10\)
3A spinner is divided into 5 equal sectors, numbered 1 to 5. What is the sample space when it is spun once? What is the theoretical probability of landing on an even number?
Sample space: \(\{1, 2, 3, 4, 5\}\) — 5 equally likely outcomes.
Even numbers in this set: 2 and 4 — that's 2 favourable outcomes out of 5.
\(P(\text{even}) = \dfrac{2}{5}=0.4\)
4A jar contains 5 red pens, 3 blue pens and 2 black pens. One pen is drawn at random. Find the probability that the pen drawn is (i) blue, (ii) not black.
Total pens \(= 5+3+2=10\)
(i) \(P(\text{blue}) = \dfrac{3}{10}=0.3\)
(ii) "Not black" means red or blue: \(5+3=8\) favourable pens, so \(P(\text{not black})=\dfrac{8}{10}=\dfrac{4}{5}=0.8\). Alternatively: \(P(\text{not black})=1-P(\text{black})=1-\dfrac{2}{10}=0.8\).
5A coin is tossed and, at the same time, a die is rolled.
(i) Draw a tree diagram showing all possible outcomes.
(ii) Find the probability of getting Heads and an even number.
Tree diagram: 2 coin outcomes × 6 die outcomes = 12 total combined outcomes
(i) The coin gives 2 outcomes (H, T); the die gives 6 outcomes (1–6). Every branch of the coin splits into all 6 die outcomes, giving \(2\times6=12\) total outcomes in the sample space.
(ii) Favourable outcomes: Heads paired with an even number (2, 4, or 6) — that's 3 outcomes: (H,2), (H,4), (H,6), out of 12 total.
\(P(\text{Heads and even}) = \dfrac{3}{12}=\dfrac{1}{4}=0.25\)
6Two dice are rolled together. Find the probability that the sum of the two numbers shown is 8.
Total outcomes when two dice are rolled \(=6\times6=36\), all equally likely.
Pairs that sum to 8: (2,6), (3,5), (4,4), (5,3), (6,2) — that's 5 favourable outcomes.
\(P(\text{sum}=8) = \dfrac{5}{36}\approx0.139\)
7A box contains 4 red balls and 6 green balls. One ball is drawn and NOT replaced, then a second ball is drawn. Find the probability that both balls drawn are red.
Total balls initially \(= 4+6=10\)
\(P(\text{1st red}) = \dfrac{4}{10}=\dfrac{2}{5}\)
Since the first ball is not replaced, only 9 balls remain, with 3 red left: \(P(\text{2nd red given 1st was red}) = \dfrac{3}{9}=\dfrac{1}{3}\)
\(P(\text{both red}) = \dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{12}{90}=\dfrac{2}{15}\approx0.133\)