Sample space and events
When a sound comes from behind we cannot tell if a friend is calling; but when a coin is tossed we already know the result is either head or tail. The list of all results that can occur is the sample space. Building on classes 9–10 : In classes 9–10 we learnt sets and the simple idea of chances; here we use the language of sets to describe chances.
An experiment whose result cannot be fixed beforehand but whose possible results are all known is a random experiment. The set of all possible results is the sample space (S). Any subset of S is an event.
Rule 1: Kinds of events — A simple event has one result. The complement A′ is "A does not happen". Mutually exclusive events cannot happen together (A ∩ B is empty). Exhaustive events have union S. The next formulas work when all results are equally likely.
In the figure : Two dice give 36 results; the event A, "sum is 7", has 6 results (the orange diagonal).
Rule 2: Counting the results — One coin has 2 results; n coins give 2 multiplied n times. One die has 6; two dice 6 × 6 = 36. A pack of 52 cards: 4 suits × 13.
How is the formula obtained? : For every result the whole set of results of the next experiment comes again, so the number multiplies. With two dice, the first die has 6 results and with each of them the second has 6; the total is 6 × 6 = 36.
- Easy : Three coins are tossed: n(S) = 2 × 2 × 2 = 8.
- Medium : Two dice, event "sum is 8": (2,6), (3,5), (4,4), (5,3), (6,2). That is 5 results.
- Exam-type : In 3 coins "at least one head": the complement "all tails" has only 1; so 8 − 1 = 7 results. A = "sum is 7" and B = "both dice equal" are mutually exclusive, because equal numbers always add to an even sum.
The order counts when counting results: (2,5) and (5,2) are two different results; (4,4) is only one.
Taking 12 results (6 + 6) for two dice. The correct count is 6 × 6 = 36.
For an "at least one" event, count the complement "none" and subtract it from S. Memory clue : "at least one = all − none".
In CET if a coin is tossed n times the count is 2ⁿ; do not take 2n or n².
CET-style example : A coin is tossed 4 times. n(S) = ? (A) 16 (B) 8 (C) 4 (D) 32
Answer : (A). 2 × 2 × 2 × 2 = 16. Time-saving method : Double with every toss: 2, 4, 8, 16.
Spot the difference
| Point | Mutually exclusive | Exhaustive |
|---|---|---|
| Meaning | cannot happen together | together they cover S |
| Formula | A ∩ B is empty | A ∪ B = S |
| Example | die shows 1 and die shows 2 | even and odd |
Practice questions
- 1A die is thrown. Write the sample space.
- 24 coins are tossed. Find n(S).
- 3Write the event "an even number" for a die.
- 4Two dice: count the results of the event "sum is 5".
- 5Two dice: count the results of the event "both numbers equal".
- 6Two dice: count the results of the event "sum is 10 or more".
- 73 coins: count the results of the event "exactly two heads".
- 8One card from 52: count the results of the event "king or heart".
Show answers (try them yourself first)
- 1S = {1, 2, 3, 4, 5, 6}.
- 22⁴ = 16.
- 3{2, 4, 6}.
- 4(1,4), (2,3), (3,2), (4,1): 4.
- 5(1,1) to (6,6): 6.
- 6(4,6), (5,5), (6,4), (5,6), (6,5), (6,6): 6.
- 7HHT, HTH, THH: 3.
- 84 kings + 13 hearts − 1 (king of hearts) = 16.
Probability of an event
The weather report says "70% chance of rain today"; that means in 100 such situations it rained 70 times. Chance written as a number is probability. Building on classes 9–10 : In classes 9–10 we learnt fractions and percentages; probability is also just a fraction between 0 and 1.
When all results are equally likely the probability of an event A is the number of results in A ÷ the number of results in S. The probability of an impossible event is 0 and of a sure event is 1.
Rule 1: Basic formula — P(A) = n(A) ÷ n(S); 0 ≤ P(A) ≤ 1; P(A′) = 1 − P(A).
In the figure : Probability always lies between 0 and 1; 0 is impossible, 1 is sure.
Rule 2: Complementary event — The probabilities of A happening and not happening always add up to 1. So for "at least one" use P = 1 − P(none).
How is the formula obtained? : The results of A and those of A′ together fill the whole of S, so n(A) + n(A′) = n(S). Divide both sides by n(S): P(A) + P(A′) = 1, that is P(A′) = 1 − P(A).
- Easy : A die is thrown; a number greater than 4: {5, 6} so P = 2/6 = 1/3.
- Medium : Two dice; sum 7: P = 6/36 = 1/6. Three coins; at least one head: P = 1 − 1/8 = 7/8.
- Exam-type : One card from 52; a face card (king, queen, jack): 12 cards. P = 12/52 = 3/13.
Probability is never negative or greater than 1. If it comes so, the calculation is wrong.
Dividing favourable by unfavourable results. If 1 out of 4 is favourable, P = 1/4 (not 1/3); the denominator is the total number of results.
Use the same rule for the numerator and denominator (order counted or not). If it is hard, go through the complement. Memory clue : "favourable ÷ total; hard? then 1 − complement".
When CET says "at least one 6" do not count directly; use 1 − (5/6)².
CET-style example : Two dice are thrown. The probability of at least one 6? (A) 11/36 (B) 1/6 (C) 1/3 (D) 25/36
Answer : (A). There are 5 × 5 = 25 results with no 6; P = 1 − 25/36 = 11/36. Time-saving method : 36 − 25 = 11 directly.
Spot the difference
| Point | Impossible event | Sure event |
|---|---|---|
| P | 0 | 1 |
| Results | none | all of S |
| Example | 7 on a die | one of 1 to 6 on a die |
Practice questions
- 1A die is thrown. Find the probability of a prime number.
- 2One card from 52; the probability that it is an ace?
- 3A bag has 3 red and 7 blue balls; one is drawn. Probability that it is red?
- 4Two coins; probability of both heads?
- 5Two dice; probability that the sum is 9?
- 6Three coins; probability of exactly two heads?
- 7Two dice; probability that the product of the numbers is 12?
- 8One card from 52; probability that it is neither a king nor a heart?
Show answers (try them yourself first)
- 1{2, 3, 5}: 3/6 = 1/2.
- 24/52 = 1/13.
- 33/10.
- 41/4.
- 5(3,6), (4,5), (5,4), (6,3): 4/36 = 1/9.
- 6HHT, HTH, THH: 3/8.
- 7(2,6), (3,4), (4,3), (6,2): 4/36 = 1/9.
- 8King or heart: 16 cards; the rest 36. P = 36/52 = 9/13.
Addition theorem
In a class 20 students like cricket and 15 like football; if those who like both are counted twice the total is wrong. "Or" means the common part must be counted only once. Building on classes 9–10 : In classes 9–10 we learnt n(A ∪ B) = n(A) + n(B) − n(A ∩ B) for sets; the same formula applies to probability.
A or B happening is A ∪ B; A and B both happening is A ∩ B. The addition theorem gives the probability of A ∪ B.
Rule 1: The theorem — P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A, B are mutually exclusive then P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B). "Neither": P(A′ ∩ B′) = 1 − P(A ∪ B).
In the figure : In 52 cards A = king, B = heart: only one is common (the king of hearts), so it is subtracted once.
How is the formula obtained? : When counting the results of A ∪ B, taking n(A) of A and n(B) of B counts the n(A ∩ B) common results twice. So n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Dividing by n(S) gives the probability formula.
- Easy : Mutually exclusive: P(A) = 0.4, P(B) = 0.3. P(A ∪ B) = 0.7.
- Medium : 52 cards: king or heart. 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
- Exam-type : P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.7: P(A ∩ B) = 0.5 + 0.4 − 0.7 = 0.2.
"Or" means addition (minus the common part), "and" means intersection. After counting, check whether the common part was subtracted.
Taking just P(A) + P(B) when the events are not mutually exclusive; the answer may even exceed 1.
If three numbers are given and one is missing, find it from the same formula: P(A ∩ B) = P(A) + P(B) − P(A ∪ B). Memory clue : "add, then subtract the common part".
In CET "neither A nor B" is 1 − P(A ∪ B).
CET-style example : P(A) = 1/2, P(B) = 1/3, P(A ∩ B) = 1/6. P(A ∪ B) = ? (A) 2/3 (B) 5/6 (C) 1/6 (D) 3/4
Answer : (A). 3/6 + 2/6 − 1/6 = 4/6 = 2/3. Time-saving method : Common denominator 6: 3 + 2 − 1.
Spot the difference
| Point | Mutually exclusive | Not exclusive |
|---|---|---|
| P(A ∩ B) | 0 | not zero |
| P(A ∪ B) | P(A) + P(B) | P(A) + P(B) − P(A ∩ B) |
| Example | 2 or 5 on a die | even or greater than 4 on a die |
Practice questions
- 1For mutually exclusive A, B with P(A) = 0.2, P(B) = 0.5, find P(A ∪ B).
- 2For mutually exclusive A, B with P(A) = 3/10, P(B) = 1/5, find P(A ∪ B).
- 3If P(A ∪ B) = 0.6, find the probability that neither A nor B happens.
- 4A die is thrown. Probability of an even number or a number greater than 4?
- 5One card from 52; probability of a spade or an ace?
- 6P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Find P(A ∪ B).
- 7P(A ∪ B) = 0.9, P(A) = 0.6, P(B) = 0.7. Find P(A ∩ B).
- 8A die is thrown. Probability of a number divisible by 2 or by 3?
Show answers (try them yourself first)
- 10.2 + 0.5 = 0.7.
- 23/10 + 2/10 = 1/2.
- 31 − 0.6 = 0.4.
- 4{2, 4, 6, 5}: 4/6 = 2/3.
- 513 + 4 − 1 = 16; 16/52 = 4/13.
- 60.6 + 0.5 − 0.3 = 0.8.
- 70.6 + 0.7 − 0.9 = 0.4.
- 8{2, 3, 4, 6}: 4/6 = 2/3.
Conditional probability and multiplication theorem
A die is thrown and someone says "the number is even". Now only 2, 4, 6 are possible; the earlier 6 results no longer remain. When new information arrives, the probability changes. Building on classes 9–10 : In classes 9–10 we learnt the sample space and P = favourable ÷ total; here the list of total results becomes smaller.
The probability of A given that B has happened is the conditional probability P(A | B). Because B is known to have happened, the sample space shrinks to B.
Rule 1: Conditional probability — P(A | B) = P(A ∩ B) ÷ P(B) = n(A ∩ B) ÷ n(B) (when P(B) ≠ 0).
In the figure : B has happened: the new space is only B; the 3 points of A ∩ B (green) out of 5.
Rule 2: Multiplication theorem — P(A ∩ B) = P(B) × P(A | B) = P(A) × P(B | A). When balls are drawn without replacement the count is smaller the second time, so this formula is used.
How is the formula obtained? : In the conditional formula P(A | B) = P(A ∩ B)/P(B) multiply both sides by P(B): P(A ∩ B) = P(B) × P(A | B). Swapping A and B gives the second form.
- Easy : P(A ∩ B) = 0.2, P(B) = 0.5: P(A | B) = 0.2/0.5 = 0.4.
- Medium : A die; the number is known to be even. P(greater than 3 | even): even = {2, 4, 6}; of these {4, 6}. P = 2/3.
- Exam-type : A bag has 4 red and 6 blue balls; 2 are drawn without replacement; both red: 4/10 × 3/9 = 12/90 = 2/15.
In P(A | B) the event behind the "|" (B) is the denominator. P(A | B) and P(B | A) are usually different.
Taking 4/10 again the second time when there is no replacement. Correct: one ball fewer, so 3/9.
When a condition is given, make the new sample space in your mind and count its results directly; the formula may not even be needed. Memory clue : "condition given, space shrinks".
In CET if the condition is "the first die shows 3", the 6 results of the second die are the new space; do not take 36.
CET-style example : Two dice; the first is known to show 3. The probability that the sum is 8? (A) 1/6 (B) 5/36 (C) 1/36 (D) 1/3
Answer : (A). New space: (3,1) to (3,6) = 6 results; only (3,5) has sum 8. P = 1/6. Time-saving method : The second die must show 5: 1 out of 6.
Spot the difference
| Point | P(A) | P(A ∣ B) |
|---|---|---|
| Space | whole S | only B |
| Denominator | n(S) | n(B) |
| Condition | none | B has happened |
Practice questions
- 1P(A ∩ B) = 0.3, P(B) = 0.6. Find P(A | B).
- 2A die; the number is known to be odd. Probability that it is 3?
- 3P(B) = 0.5, P(A | B) = 0.4. Find P(A ∩ B).
- 4Two dice; the sum is known to be 7. Probability that the first shows 4?
- 5A bag has 5 red and 3 blue balls; two are drawn without replacement. Probability that both are red?
- 6Two cards from 52 without replacement; probability that both are aces?
- 7P(A) = 0.6, P(B) = 0.5, P(A ∪ B) = 0.8. Find P(A | B).
- 8Two dice; the sum is known to be 10 or more. Probability that both numbers are equal?
Show answers (try them yourself first)
- 10.3/0.6 = 0.5.
- 2Of {1, 3, 5} the 3: 1/3.
- 30.5 × 0.4 = 0.2.
- 4Sum 7 has 6 results; only (4,3) has first 4. 1/6.
- 55/8 × 4/7 = 20/56 = 5/14.
- 64/52 × 3/51 = 12/2652 = 1/221.
- 7P(A ∩ B) = 0.6 + 0.5 − 0.8 = 0.3; P(A | B) = 0.3/0.5 = 0.6.
- 86 results; equal numbers (5,5), (6,6): 2/6 = 1/3.
Independent events
A coin is tossed and a die is thrown: the coin’s result has no effect on the die. Events that do not affect each other like this are independent events. Building on classes 9–10 : In classes 9–10 we learnt multiplying the possibilities (coin and die = 2 × 6); this is its formula form.
If whether B happens or not does not change the probability of A, then A and B are independent events: P(A | B) = P(A).
Rule 1: Formula for independent events — If A, B are independent then P(A ∩ B) = P(A) × P(B). Their complements are also independent: P(A′ ∩ B′) = P(A′) × P(B′).
In the figure : Tree diagram when two shooters hit with 0.6 and 0.7; the end numbers are the products.
Rule 2: Several events — "Both" = the product. "At least one" = 1 − (both fail). "Exactly one" = P(A) × P(B′) + P(A′) × P(B).
How is the formula obtained? : Put P(A | B) = P(A) in the multiplication theorem: P(A ∩ B) = P(B) × P(A | B) = P(B) × P(A). So P(A ∩ B) = P(A) × P(B). The same holds for complements, as P(A′) = 1 − P(A) and the conditions stay the same.
- Easy : Coin and die: P(head and 6) = 1/2 × 1/6 = 1/12.
- Medium : A, B independent; P(A) = 0.5, P(B) = 0.4. P(A ∩ B) = 0.2; P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7.
- Exam-type : Shooters 0.6 and 0.7: both hit 0.42; exactly one 0.18 + 0.28 = 0.46; at least one 1 − 0.4 × 0.3 = 0.88.
Balls drawn with replacement are independent; balls drawn without replacement are not (the first affects the second).
Treating mutually exclusive and independent as the same. For exclusive events P(A ∩ B) = 0; for independent events it is P(A) × P(B) (not zero when both are non-zero).
For "at least one" the shortest route in the tree is 1 − (all fail). Memory clue : "both = multiply; at least one = 1 − multiply (fails)".
When CET asks P(A′ ∩ B′), take the complement of each and multiply: (1 − P(A))(1 − P(B)).
CET-style example : A, B independent; P(A) = 1/3, P(B) = 1/4. P(A′ ∩ B′) = ? (A) 1/2 (B) 1/12 (C) 5/12 (D) 7/12
Answer : (A). 2/3 × 3/4 = 1/2. Time-saving method : The complements are 2/3 and 3/4; the 3 cancels, 2/4 = 1/2.
Spot the difference
| Point | Independent | Mutually exclusive |
|---|---|---|
| P(A ∩ B) | P(A) × P(B) | 0 |
| Meaning | no effect on each other | cannot happen together |
| Example | coin and die | 2 and 5 on a die |
Practice questions
- 1A coin is tossed twice. Probability of a head both times?
- 2A die is thrown twice. Probability of a 6 both times?
- 3A, B independent; P(A) = 0.3, P(B) = 0.5. Find P(A ∩ B).
- 4Two shooters hit with probabilities 0.8 and 0.5. Probability that both hit?
- 5For the same shooters, the probability that exactly one hits?
- 6For the same shooters, the probability that at least one hits?
- 7A, B independent; P(A) = 0.4 and P(A ∪ B) = 0.7. Find P(B).
- 8A die is thrown three times. Probability of no 6 at all?
Show answers (try them yourself first)
- 11/2 × 1/2 = 1/4.
- 21/6 × 1/6 = 1/36.
- 30.3 × 0.5 = 0.15.
- 40.8 × 0.5 = 0.4.
- 50.8 × 0.5 + 0.2 × 0.5 = 0.4 + 0.1 = 0.5.
- 61 − 0.2 × 0.5 = 0.9.
- 70.4 + p − 0.4p = 0.7, so 0.6p = 0.3 and p = 0.5.
- 8(5/6)³ = 125/216.
Total probability and Bayes' theorem
With eyes closed one of two bags is chosen and a ball is drawn; it turns out red. Now the reverse question arises: "which bag did it come from?" Finding the cause from the result is Bayes’ theorem. Building on classes 9–10 : In classes 9–10 we learnt adding and multiplying fractions; we also use P(A ∩ B) = P(B) × P(A | B) from the previous section again and again.
If E₁, E₂ are mutually exclusive and exhaustive (together S) and the event A occurs along with one of them, the overall probability of A is the total probability; the probability of E₁ given that A has occurred comes from Bayes’ theorem.
Rule 1: Two formulas — P(A) = P(E₁) × P(A | E₁) + P(E₂) × P(A | E₂). Bayes: P(E₁ | A) = P(E₁) × P(A | E₁) ÷ P(A).
In the figure : Bag I: 3 red 2 black; bag II: 2 red 4 black. The red paths are orange: 3/10 and 1/6.
Rule 2: The tree method — Multiply the probabilities along the branches; the sum of all paths ending in "red" = P(red). In Bayes the share of one path ÷ the sum of all red paths.
How is the formula obtained? : A = (A ∩ E₁) ∪ (A ∩ E₂), and these two parts are mutually exclusive, so P(A) = P(A ∩ E₁) + P(A ∩ E₂) = P(E₁)P(A | E₁) + P(E₂)P(A | E₂). In the conditional formula P(E₁ | A) = P(E₁ ∩ A) ÷ P(A) and P(E₁ ∩ A) = P(E₁)P(A | E₁); this is Bayes.
- Easy : P(E₁) = 0.4, P(A | E₁) = 0.5; P(E₂) = 0.6, P(A | E₂) = 0.25: P(A) = 0.2 + 0.15 = 0.35.
- Medium : Bag I (3 red 2 black) or II (2 red 4 black), each chosen with 1/2: P(red) = 1/2 × 3/5 + 1/2 × 2/6 = 3/10 + 1/6 = 7/15.
- Exam-type : The ball drawn is red; the probability that it came from bag I: (3/10) ÷ (7/15) = 45/70 = 9/14.
The end probabilities of all branches always add up to 1: 3/10 + 1/5 + 1/6 + 1/3 = 1. If not, something is wrong.
Treating P(E₁ | A) and P(A | E₁) as the same. "Bag I given red" and "red given bag I" are different.
To avoid fractions bring all probabilities to a common denominator; in Bayes the denominators cancel and only the numerators remain. Memory clue : "one path ÷ all paths".
When CET says "it was found defective, probability that it is from machine A" this is Bayes; find P(defective) and divide immediately.
CET-style example : Machine A makes 60% of the items (2% defective) and machine B 40% (5% defective). One item is chosen. Probability that it is defective? (A) 0.032 (B) 0.07 (C) 0.035 (D) more than 0.07
Answer : (A). 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.02 = 0.032. Time-saving method : Add two products; (B) and (D) are clearly wrong because the result cannot exceed 0.05.
Spot the difference
| Point | Total probability | Bayes |
|---|---|---|
| Finds | P(A) | P(E₁ ∣ A) |
| Method | sum of all paths | one path ÷ sum |
| Question | how likely? | which cause? |
Practice questions
- 1P(E₁) = P(E₂) = 0.5; P(A | E₁) = 0.2, P(A | E₂) = 0.4. Find P(A).
- 2P(E) = 0.3, P(A | E) = 0.5, P(A | E′) = 0.1. Find P(A).
- 3Machine A makes 50% of the items (4% defective), machine B 50% (2% defective). Find the probability that a chosen item is defective.
- 4Bag I: 2 white 3 black; bag II: 4 white 1 black. A bag is chosen at random. Probability that the ball is white?
- 5The same bags: the ball is white; probability that it came from bag II?
- 6In the earlier machine question (A 50% and 4%, B 50% and 2%) the item is defective. Probability that it is from machine A?
- 7Three machines: 20% (5% defective), 30% (4%), 50% (2%). Probability that a defective item is from the first machine?
- 8Two coins: one fair, one with heads on both sides. One is chosen at random and tossed; a head comes. Probability that it is the two-headed coin?
Show answers (try them yourself first)
- 10.1 + 0.2 = 0.3.
- 20.3 × 0.5 + 0.7 × 0.1 = 0.15 + 0.07 = 0.22.
- 30.5 × 0.04 + 0.5 × 0.02 = 0.03.
- 41/2 × 2/5 + 1/2 × 4/5 = 1/5 + 2/5 = 3/5.
- 5(2/5) ÷ (3/5) = 2/3.
- 60.02 ÷ 0.03 = 2/3.
- 7P(defective) = 0.01 + 0.012 + 0.01 = 0.032; 0.01 ÷ 0.032 = 5/16.
- 8P(head) = 1/2 × 1/2 + 1/2 × 1 = 3/4; (1/2) ÷ (3/4) = 2/3.
Odds
Before a match people say "our team’s chance is 2 against 3"; that is 2 wins and 3 losses out of 5. The ratio of favourable to unfavourable is the odds. Building on classes 9–10 : In classes 9–10 we learnt ratios; odds are also a ratio, only probability comes from a division.
The odds in favour of an event = favourable results : unfavourable results. The odds against the event = unfavourable : favourable.
Rule 1: Odds and probability — If the odds in favour are a : b then P = a ÷ (a + b). Conversely, if P(A) is known the odds in favour are P(A) : P(A′) = P(A) : (1 − P(A)).
In the figure : Die: a number greater than 4 (5, 6) is favourable 2; unfavourable 4. Odds 2 : 4 = 1 : 2; P = 2/6 = 1/3.
Rule 2: Both ways — In odds the sum (a + b) is the total number of results. If the odds in favour are a : b, the odds against are b : a.
How is the formula obtained? : With a favourable and b unfavourable results the total is a + b. So P = a/(a + b) and P(A′) = b/(a + b). The ratio P : P(A′) = a/(a + b) : b/(a + b) = a : b.
- Easy : P(A) = 2/5: odds in favour 2/5 : 3/5 = 2 : 3.
- Medium : Odds against 7 : 3: in favour 3 : 7, P = 3/10.
- Exam-type : A king from 52 cards: favourable 4, unfavourable 48. Odds in favour 4 : 48 = 1 : 12; P = 4/52 = 1/13.
In odds keep the part and the whole apart: odds 1 : 2 means P = 1/3 (not 1/2).
Taking P = a/b from odds a : b. The correct formula is P = a/(a + b).
From P to odds: the numerator and (denominator − numerator); for P = 2/5 these are 2 and 5 − 2 = 3, so odds 2 : 3. Memory clue : "numerator : (denominator − numerator)".
When CET gives two sets of odds, convert each to P first and then use the addition theorem.
CET-style example : The odds in favour of A are 3 : 4 and of B are 2 : 5; A, B are mutually exclusive. P(A ∪ B) = ? (A) 5/7 (B) 5/11 (C) 1/2 (D) 6/7
Answer : (A). P(A) = 3/7, P(B) = 2/7; the sum is 5/7. Time-saving method : Both have denominator 7: 3 + 2.
Spot the difference
| Point | Probability | Odds |
|---|---|---|
| Form | fraction 0 to 1 | ratio a : b |
| Denominator | total results | unfavourable results |
| Example | 2/6 | 2 : 4 |
Practice questions
- 1If P(A) = 3/7, find the odds in favour of A.
- 2If the odds in favour are 4 : 5, find P.
- 3A die is thrown; find the odds in favour of getting 6.
- 4If the odds against are 7 : 3, find the probability of the event.
- 5If P(A) = 0.6, find the odds against A.
- 6Two coins; find the odds in favour of both heads.
- 7The odds in favour of A are 2 : 3 and of B are 1 : 4; A, B are mutually exclusive. Find the odds in favour of A ∪ B.
- 8The odds in favour of A are 1 : 2 and of B are 1 : 3; A, B are independent. Find P(A ∪ B).
Show answers (try them yourself first)
- 13 : (7 − 3) = 3 : 4.
- 24/(4 + 5) = 4/9.
- 3Favourable 1, unfavourable 5: 1 : 5.
- 4In favour 3 : 7; P = 3/10.
- 5P(A′) = 0.4; odds against 0.4 : 0.6 = 2 : 3.
- 6Favourable 1 (HH), unfavourable 3: 1 : 3.
- 7P(A) = 2/5, P(B) = 1/5, P(A ∪ B) = 3/5; odds 3 : 2.
- 8P(A) = 1/3, P(B) = 1/4; P(A ∩ B) = 1/12; P(A ∪ B) = 1/3 + 1/4 − 1/12 = 1/2.
Lesson 9: Probability — quick revision
| Topic | Formula / point |
|---|---|
| Probability | P(A) = n(A)/n(S); 0 ≤ P ≤ 1; P(A′) = 1 − P(A) |
| Counting results | n coins 2ⁿ; two dice 36; cards 52 |
| Addition theorem | P(A ∪ B) = P(A) + P(B) − P(A ∩ B); exclusive: P(A) + P(B) |
| Conditional | P(A ∣ B) = P(A ∩ B)/P(B); P(A ∩ B) = P(B) P(A ∣ B) |
| Independent | P(A ∩ B) = P(A) P(B); at least one = 1 − P(all fail) |
| Total probability / Bayes | P(A) = Σ P(Eᵢ) P(A ∣ Eᵢ); P(E₁ ∣ A) = P(E₁) P(A ∣ E₁)/P(A) |
| Odds | in favour a : b then P = a/(a + b); against b : a |
🎯 Most important points for the exam
- Probability is never below 0 or above 1.
- "Or" means addition minus the common part; "and" means multiplication.
- A condition shrinks the space; the denominator is n(B).
- Without replacement the events are not independent; the count is smaller the second time.
- In odds the denominator is a + b, not a : b.
"favourable ÷ total; hard? then 1 − complement".