Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
Toss three fair coins. Let E be 'at least two heads' and F be 'the first coin shows a tail'. Then P(E) = 1/2. If we are told that F has occurred, the outcomes outside F are impossible, so F becomes the new sample space. Only THH of the four outcomes of F is in E, so the probability of E given F is 1/4.
If E and F are events of the same sample space and P(F) ≠ 0, the conditional probability of E given F is
Careful: the complement rule applies to the event before the bar, not after it. In general P(E|F′) ≠ 1 − P(E|F), and P(E|F) is not equal to P(F|E).
The same definition works when the outcomes are not equally likely. Toss a coin. If it shows heads, toss it again; if tails, throw a die. Let F be 'at least one tail' and E be 'the die shows more than 4'.
The probability of an outcome is the product of the probabilities along its branches.
Rearranging the definition of conditional probability gives the multiplication rule:
For three events, P(E ∩ F ∩ G) = P(E) P(F|E) P(G|E ∩ F), and similarly for more events.
Events E and F are independent if P(F|E) = P(F) (and P(E|F) = P(E)). Using the multiplication rule this is the same as
Events that are not independent are dependent.
These are different ideas. 'Mutually exclusive' is about the events (no common outcome); 'independent' is about their probabilities. Two mutually exclusive events with nonzero probabilities can never be independent, because P(E ∩ F) = 0 but P(E) P(F) > 0. Independent events with nonzero probabilities always have common outcomes.
Events E₁, E₂, …, Eₙ form a partition of S if they are pairwise disjoint (Eᵢ ∩ Eⱼ = φ), exhaustive (E₁ ∪ … ∪ Eₙ = S) and each has probability > 0. Any event E and its complement E′ form a partition.
If {E₁, …, Eₙ} is a partition and A is any event, then
Example. The probability of a strike is 0.65. The job is finished on time with probability 0.32 if there is a strike and 0.80 if there is none. P(on time) = 0.65 × 0.32 + 0.35 × 0.80 = 0.488.
With the same partition and P(A) > 0,
Bag I has 3 red and 4 black balls, Bag II has 5 red and 6 black. A bag is chosen at random and a red ball is drawn. Find the probability that it came from Bag II.
A test detects the disease in 90% of people who have it and wrongly flags 1% of healthy people. Only 0.1% of the population has the disease. For a person who tests positive,
Out of 100 000 people, about 100 are ill and 90 of them test positive, while about 999 healthy people also test positive. So only 90 out of 1089 positives are really ill.
A random variable is a real valued function whose domain is the sample space of a random experiment. For two tosses of a coin, S = {HH, HT, TH, TT}. Let X be the number of heads: X(HH) = 2, X(HT) = X(TH) = 1, X(TT) = 0. Let Y be heads minus tails: Y(HH) = 2, Y(HT) = Y(TH) = 0, Y(TT) = −2. Two different random variables can be defined on the same sample space.
| Situation | Formula |
|---|---|
| Conditional probability | P(E|F) = P(E ∩ F) / P(F) |
| Multiplication rule | P(E ∩ F) = P(E) P(F|E) |
| Independent events | P(E ∩ F) = P(E) P(F), P(E|F) = P(E) |
| At least one of independent A, B | 1 − P(A′) P(B′) |
| Total probability | P(A) = ΣP(Eⱼ) P(A|Eⱼ) |
| Bayes' theorem | P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / ΣP(Eⱼ) P(A|Eⱼ) |
Try each problem first, then open the solution to check your working.
If P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, find P(A|B) and P(B|A).
P(A|B) = P(A ∩ B)/P(B) = 0.2/0.4 = 1/2.
P(B|A) = P(A ∩ B)/P(A) = 0.2/0.5 = 2/5.
They differ because the conditions differ (P(A) ≠ P(B)).
Answer: P(A|B) = 1/2, P(B|A) = 2/5
Two dice are thrown and the sum is 8. Find the probability that the first die shows 3.
F: sum 8 = {(2,6), (3,5), (4,4), (5,3), (6,2)}, so n(F) = 5.
E ∩ F = {(3,5)}, so n(E ∩ F) = 1.
P(E|F) = 1/5.
Answer: 1/5
One card is drawn from a pack of 52 and is found to be a face card (jack, queen or king). Find the probability that it is a king.
There are 12 face cards, so n(F) = 12.
The 4 kings are all face cards, so n(E ∩ F) = 4.
P(E|F) = 4/12 = 1/3.
Answer: 1/3
Three fair coins are tossed and the first shows a tail. Find the probability of at least two tails.
F = {THH, THT, TTH, TTT}, n(F) = 4.
At least two tails in F: THT, TTH, TTT, so 3 outcomes.
P = 3/4.
Answer: 3/4
In a class of 40, there are 24 girls and 16 boys. 6 girls and 4 boys play chess. A student who plays chess is chosen. Find the probability that it is a girl.
Chess players: 6 + 4 = 10, so n(F) = 10.
Girls who play chess: 6.
P(girl | chess) = 6/10 = 3/5.
Answer: 3/5
An urn has 6 red and 4 blue balls. Three balls are drawn one after another without replacement. Find the probability that all three are red.
P(1st red) = 6/10, P(2nd red | 1st red) = 5/9, P(3rd red | first two red) = 4/8.
P = (6/10)(5/9)(4/8) = 120/720 = 1/6.
Answer: 1/6
Two cards are drawn without replacement from a pack. Find the probability that the first is a heart and the second is a spade.
P(heart first) = 13/52.
Given that, there are still 13 spades among 51 cards: P(spade second | heart first) = 13/51.
P = (13/52)(13/51) = 13/204.
Answer: 13/204
If P(A) = 0.6 and P(B|A) = 0.5, find P(A ∩ B) and P(A ∩ B′).
P(A ∩ B) = P(A) P(B|A) = 0.6 × 0.5 = 0.3.
P(B′|A) = 1 − 0.5 = 0.5, so P(A ∩ B′) = 0.6 × 0.5 = 0.3.
Check: 0.3 + 0.3 = 0.6 = P(A).
Answer: P(A ∩ B) = 0.3, P(A ∩ B′) = 0.3
A box has 4 good and 2 defective bulbs. Three are drawn one after another without replacement. Find the probability that the first two are good and the third is defective.
P(good) = 4/6, then P(good | one good gone) = 3/5.
Now 2 good and 2 defective remain among 4: P(defective) = 2/4.
P = (4/6)(3/5)(2/4) = 1/5.
Answer: 1/5
A student passes Physics with probability 0.8. Given that he passes Physics, he passes Chemistry with probability 0.7. Find the probability that he passes both, and the probability that he passes Physics but fails Chemistry.
P(both) = 0.8 × 0.7 = 0.56.
P(Physics pass, Chemistry fail) = 0.8 × 0.3 = 0.24.
Together 0.56 + 0.24 = 0.80 = P(passes Physics).
Answer: 0.56 and 0.24
Two dice are thrown. E: the sum is 7, F: the first die shows 3. Are E and F independent?
P(E) = 6/36 = 1/6 and P(F) = 6/36 = 1/6.
E ∩ F = {(3, 4)}, so P(E ∩ F) = 1/36.
P(E) P(F) = 1/36 = P(E ∩ F), so they are independent.
Answer: Yes, E and F are independent
P(A) = 0.4, P(B) = 0.5 and P(A ∪ B) = 0.7. Are A and B independent?
P(A ∩ B) = 0.4 + 0.5 − 0.7 = 0.2.
P(A) P(B) = 0.4 × 0.5 = 0.2 = P(A ∩ B).
So A and B are independent.
Answer: Yes, independent
A, B and C solve a problem independently with probabilities 1/2, 1/3 and 1/4. Find the probability that the problem is solved.
The problem is solved if at least one succeeds.
P(none succeeds) = (1/2)(2/3)(3/4) = 1/4.
P(solved) = 1 − 1/4 = 3/4.
Answer: 3/4
A and B are independent with P(A) = 0.5 and P(A ∪ B) = 0.8. Find P(B).
Let P(B) = p. P(A ∪ B) = P(A) + P(B) − P(A)P(B) gives 0.8 = 0.5 + p − 0.5p = 0.5 + 0.5p.
So 0.5p = 0.3 and p = 0.6.
Check: 1 − (0.5)(0.4) = 0.8.
Answer: P(B) = 0.6
Three coins are tossed. E: the first is a head, F: the last two coins are alike. Are E and F independent?
P(E) = 4/8 = 1/2. F = {HHH, HTT, THH, TTT}, so P(F) = 1/2.
E ∩ F = {HHH, HTT}, so P(E ∩ F) = 2/8 = 1/4.
P(E) P(F) = 1/4 = P(E ∩ F), so E and F are independent.
Answer: Yes, independent
Bag A has 2 red and 3 white balls, Bag B has 4 red and 1 white. A bag is chosen with probabilities 1/3 for A and 2/3 for B, and a ball is drawn. Find P(red), and the probability that a red ball came from Bag A.
P(red|A) = 2/5 and P(red|B) = 4/5.
P(red) = (1/3)(2/5) + (2/3)(4/5) = 2/15 + 8/15 = 2/3.
P(A|red) = (2/15) ÷ (2/3) = 1/5.
Answer: P(red) = 2/3; P(A | red) = 1/5
Factories I, II and III make 50%, 30% and 20% of the items, with 2%, 3% and 5% defective. An item is found defective. Find the probability that it came from factory III.
Joint probabilities: 0.5 × 0.02 = 0.010, 0.3 × 0.03 = 0.009, 0.2 × 0.05 = 0.010.
P(defective) = 0.029.
P(III | defective) = 0.010/0.029 = 10/29.
Answer: 10/29
2% of a population has a condition. A test is positive for 95% of those who have it and for 4% of those who do not. A person tests positive. Find the probability that the person has the condition.
P(has and positive) = 0.02 × 0.95 = 0.019. P(not has and positive) = 0.98 × 0.04 = 0.0392.
P(positive) = 0.0582.
P(has | positive) = 0.019/0.0582 = 190/582 = 95/291 ≈ 0.33.
Answer: 95/291 ≈ 0.33
60% of students prepared for a quiz. A prepared student answers a question correctly with probability 0.9, an unprepared one with probability 0.25. A student answers correctly. Find the probability that the student was prepared.
P(prepared and correct) = 0.6 × 0.9 = 0.54. P(unprepared and correct) = 0.4 × 0.25 = 0.10.
P(correct) = 0.64.
P(prepared | correct) = 0.54/0.64 = 27/32.
Answer: 27/32
Urn I has 3 red and 2 blue balls, Urn II has 2 red and 3 blue. A ball is moved from Urn I to Urn II and a ball is then drawn from Urn II. It is red. Find the probability that the moved ball was red.
If the moved ball is red (prob 3/5), Urn II has 3 red and 3 blue, so P(red) = 1/2. If it is blue (prob 2/5), Urn II has 2 red and 4 blue, so P(red) = 1/3.
P(moved red | drawn red) = (3/5)(1/2) ÷ [(3/5)(1/2) + (2/5)(1/3)] = (3/10) ÷ (3/10 + 2/15) = 9/13.
Answer: 9/13
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P(E|F) = P(E ∩ F) / P(F) = 0.2 / 0.3 = 2/3.
P(F|E) = P(E ∩ F) / P(E) = 0.2 / 0.6 = 1/3.
Answer: P(E|F) = 2/3 and P(F|E) = 1/3
P(A|B) = P(A ∩ B) / P(B) = 0.32 / 0.5 = 0.64.
Answer: 0.64 (= 16/25)
(i) P(A ∩ B) = P(A) P(B|A) = 0.8 × 0.4 = 0.32.
(ii) P(A|B) = P(A ∩ B) / P(B) = 0.32 / 0.5 = 0.64.
(iii) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.8 + 0.5 − 0.32 = 0.98.
Answer: (i) 0.32 (ii) 0.64 (iii) 0.98
P(B) = 5/13 and P(A) = 5/26.
P(A ∩ B) = P(A|B) P(B) = (2/5)(5/13) = 2/13.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 5/26 + 10/26 − 4/26 = 11/26.
Answer: 11/26
(i) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 6/11 + 5/11 − 7/11 = 4/11.
(ii) P(A|B) = (4/11) / (5/11) = 4/5.
(iii) P(B|A) = (4/11) / (6/11) = 2/3.
Answer: (i) 4/11 (ii) 4/5 (iii) 2/3
The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, 8 equally likely outcomes.
(i) F = {HHH, HHT}, E ∩ F = {HHH}, so P(E|F) = 1/2.
(ii) F = all outcomes except HHH, so n(F) = 7. E ∩ F = {HHT, HTH, THH}, so P(E|F) = 3/7.
(iii) F = all outcomes except HHH, n(F) = 7. E (at least one head) ∩ F = all outcomes with both a head and a tail = 6 outcomes, so P(E|F) = 6/7.
Answer: (i) 1/2 (ii) 3/7 (iii) 6/7
Sample space {HH, HT, TH, TT}.
(i) F: one coin shows head = {HT, TH}. E: a tail appears = {HT, TH, TT}. E ∩ F = {HT, TH} = F, so P(E|F) = 1.
(ii) E: no tail = {HH}. F: no head = {TT}. E ∩ F = ∅, so P(E|F) = 0.
Answer: (i) 1 (ii) 0
F = {(6, 5, 1), (6, 5, 2), …, (6, 5, 6)} has 6 outcomes out of 216, so P(F) = 6/216.
E ∩ F = {(6, 5, 4)}, so P(E ∩ F) = 1/216.
P(E|F) = (1/216) / (6/216) = 1/6.
Answer: 1/6
There are 3! = 6 arrangements. F: father in the middle has 2 arrangements: (M, Fa, S) and (S, Fa, M).
In both, the son is at an end, so E ∩ F = F.
P(E|F) = P(E ∩ F) / P(F) = (2/6) / (2/6) = 1.
Answer: 1
Write outcomes as (black, red).
(a) F: black = 5 has 6 outcomes. Sum > 9 with black 5: (5, 5), (5, 6), so 2 outcomes. P = 2/6 = 1/3.
(b) F: red ∈ {1, 2, 3} has 18 outcomes. Sum 8 with red < 4: (5, 3), (6, 2), so 2 outcomes. P = 2/18 = 1/9.
Answer: (a) 1/3 (b) 1/9
(i) E ∩ F = {3}. P(E|F) = (1/6)/(2/6) = 1/2. P(F|E) = (1/6)/(3/6) = 1/3.
(ii) E ∩ G = {3, 5}. P(E|G) = (2/6)/(4/6) = 1/2. P(G|E) = (2/6)/(3/6) = 2/3.
(iii) E ∪ F = {1, 2, 3, 5}, and (E ∪ F) ∩ G = {2, 3, 5}, so P((E ∪ F)|G) = 3/4. (E ∩ F) ∩ G = {3}, so P((E ∩ F)|G) = 1/4.
Answer: (i) 1/2, 1/3 (ii) 1/2, 2/3 (iii) 3/4, 1/4
Sample space (elder, younger): {(b, b), (b, g), (g, b), (g, g)}. Let E = both girls = {(g, g)}.
(i) F = youngest is a girl = {(b, g), (g, g)}. P(E|F) = (1/4)/(2/4) = 1/2.
(ii) F = at least one girl = {(b, g), (g, b), (g, g)}. P(E|F) = (1/4)/(3/4) = 1/3.
Answer: (i) 1/2 (ii) 1/3
Total = 300 + 200 + 500 + 400 = 1400. Let E: easy, F: multiple choice.
P(F) = 900/1400 and P(E ∩ F) = 500/1400.
P(E|F) = 500/900 = 5/9.
Answer: 5/9
F: the numbers are different has 36 − 6 = 30 outcomes.
E: sum 4 = {(1, 3), (2, 2), (3, 1)}, and E ∩ F = {(1, 3), (3, 1)} has 2 outcomes.
P(E|F) = 2/30 = 1/15.
Answer: 1/15
Outcomes: (3, k) and (6, k) for k = 1, …, 6, each of probability 1/36; and (n, H) and (n, T) for n = 1, 2, 4, 5, each of probability 1/12.
F: at least one die shows 3 = {(3, 1), …, (3, 6), (6, 3)}, so P(F) = 7/36.
A coin is tossed only when the first number is not a multiple of 3, and then no die is thrown again. So no outcome of F contains a coin, and E ∩ F = ∅.
P(E|F) = 0 / (7/36) = 0.
Answer: 0
P(A|B) = P(A ∩ B) / P(B) is defined only when P(B) ≠ 0.
Here P(B) = 0, so the conditional probability is not defined.
Answer: (C) not defined
P(A ∩ B)/P(B) = P(A ∩ B)/P(A). If P(A ∩ B) ≠ 0 we can cancel it and get P(A) = P(B).
Options (A) and (B) are not forced, since two different events can have equal probability, and (C) is a special case in which both sides are 0.
So the statement that must hold is P(A) = P(B).
Answer: (D) P(A) = P(B)
For independent events P(A ∩ B) = P(A) P(B) = (3/5)(1/5) = 3/25.
Answer: 3/25
There are 26 black cards. P(first black) = 26/52. After one black card is removed, P(second black | first black) = 25/51.
P(both black) = (26/52)(25/51) = 25/102.
Answer: 25/102
The box is approved if all three drawn oranges are good.
P = (12/15)(11/14)(10/13) = 1320/2730 = 44/91.
Answer: 44/91
P(A) = 1/2, P(B) = 1/6. The 12 outcomes are equally likely, and A ∩ B = {(H, 3)}, so P(A ∩ B) = 1/12.
P(A) P(B) = 1/12 = P(A ∩ B).
Answer: Yes, A and B are independent
A = {2, 4, 6}, B = {1, 2, 3}, A ∩ B = {2}.
P(A) = 1/2, P(B) = 1/2, P(A ∩ B) = 1/6.
P(A) P(B) = 1/4 ≠ 1/6.
Answer: No, A and B are not independent
P(E) P(F) = (3/5)(3/10) = 9/50.
P(E ∩ F) = 1/5 = 10/50, which is not equal to 9/50.
Answer: No, E and F are not independent
(i) Mutually exclusive: P(A ∪ B) = P(A) + P(B), so 3/5 = 1/2 + p and p = 1/10.
(ii) Independent: P(A ∪ B) = P(A) + P(B) − P(A) P(B), so 3/5 = 1/2 + p − p/2 = 1/2 + p/2. Hence p/2 = 1/10 and p = 1/5.
Answer: (i) p = 1/10 (ii) p = 1/5
(i) P(A ∩ B) = 0.3 × 0.4 = 0.12.
(ii) P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58.
(iii) For independent events P(A|B) = P(A) = 0.3.
(iv) P(B|A) = P(B) = 0.4.
Answer: (i) 0.12 (ii) 0.58 (iii) 0.3 (iv) 0.4
P(A ∪ B) = 1/4 + 1/2 − 1/8 = 5/8.
P(not A and not B) = P((A ∪ B)′) = 1 − 5/8 = 3/8.
Answer: 3/8
P(not A or not B) = P((A ∩ B)′) = 1 − P(A ∩ B) = 1/4, so P(A ∩ B) = 3/4.
P(A) P(B) = (1/2)(7/12) = 7/24, which is not equal to 3/4.
So A and B are not independent. (The printed data is in fact inconsistent, because P(A ∩ B) = 3/4 is larger than P(A) = 1/2, which is impossible. The textbook answer, based on P(A ∩ B) ≠ P(A) P(B), is 'not independent'.)
Answer: Not independent
(i) P(A and B) = 0.3 × 0.6 = 0.18.
(ii) P(A and not B) = P(A) P(B′) = 0.3 × 0.4 = 0.12.
(iii) P(A or B) = 0.3 + 0.6 − 0.18 = 0.72.
(iv) P(neither) = P(A′) P(B′) = 0.7 × 0.4 = 0.28.
Answer: (i) 0.18 (ii) 0.12 (iii) 0.72 (iv) 0.28
P(no odd number in a throw) = 1/2, so P(no odd number in three throws) = (1/2)³ = 1/8.
P(at least one odd) = 1 − 1/8 = 7/8.
Answer: 7/8
With replacement the draws are independent. P(red) = 8/18 = 4/9, P(black) = 10/18 = 5/9.
(i) (4/9)(4/9) = 16/81.
(ii) (5/9)(4/9) = 20/81.
(iii) The two orders (black, red) and (red, black) are exclusive: 20/81 + 20/81 = 40/81.
Answer: (i) 16/81 (ii) 20/81 (iii) 40/81
(i) The problem is solved if at least one succeeds: 1 − P(A fails) P(B fails) = 1 − (1/2)(2/3) = 2/3.
(ii) Exactly one solves it: (1/2)(2/3) + (1/2)(1/3) = 1/3 + 1/6 = 1/2.
Answer: (i) 2/3 (ii) 1/2
(i) P(E) = 13/52 = 1/4, P(F) = 4/52 = 1/13, P(E ∩ F) = P(ace of spades) = 1/52 = P(E) P(F). Independent.
(ii) P(E) = 26/52 = 1/2, P(F) = 4/52 = 1/13, P(E ∩ F) = P(black king) = 2/52 = 1/26 = P(E) P(F). Independent.
(iii) P(E) = 8/52 = 2/13, P(F) = 8/52 = 2/13, E ∩ F = {queens}, so P(E ∩ F) = 4/52 = 1/13. But P(E) P(F) = 4/169 ≠ 1/13. Not independent.
Answer: Independent in (i) and (ii); not independent in (iii)
Let H: reads Hindi, E: reads English. P(H) = 0.6, P(E) = 0.4, P(H ∩ E) = 0.2.
(a) P(H ∪ E) = 0.6 + 0.4 − 0.2 = 0.8, so P(neither) = 1 − 0.8 = 0.2.
(b) P(E|H) = 0.2 / 0.6 = 1/3.
(c) P(H|E) = 0.2 / 0.4 = 1/2.
Answer: (a) 0.2 (b) 1/3 (c) 1/2
The only even prime is 2, so on each die P(2) = 1/6.
The dice are independent, so P(2 on both) = (1/6)(1/6) = 1/36.
Answer: (D) 1/36
If A and B are independent, so are A′ and B′, so P(A′ ∩ B′) = P(A′) P(B′) = [1 − P(A)][1 − P(B)].
Mutually exclusive events with nonzero probabilities are never independent, and equal or complementary probabilities say nothing about independence.
Answer: (B)
Let R1, B1 be the colour of the first ball. P(R1) = P(B1) = 1/2.
If R1: the urn has 7 red and 5 black (12 balls), so P(red second | R1) = 7/12. If B1: 5 red and 7 black, so P(red second | B1) = 5/12.
By total probability, P(second red) = (1/2)(7/12) + (1/2)(5/12) = 12/24 = 1/2.
Answer: 1/2
Let E1, E2 be the first and second bag, A: red ball. P(E1) = P(E2) = 1/2, P(A|E1) = 4/8 = 1/2, P(A|E2) = 2/8 = 1/4.
P(E1|A) = (1/2)(1/2) / [(1/2)(1/2) + (1/2)(1/4)] = (1/4)/(3/8) = 2/3.
Answer: 2/3
Let H: hostel, D: day scholar, A: A grade. P(H) = 0.6, P(D) = 0.4, P(A|H) = 0.3, P(A|D) = 0.2.
P(H|A) = (0.6 × 0.3) / (0.6 × 0.3 + 0.4 × 0.2) = 0.18 / 0.26 = 9/13.
Answer: 9/13
Let K: knows, G: guesses, C: correct. P(K) = 3/4, P(G) = 1/4, P(C|K) = 1, P(C|G) = 1/4.
P(K|C) = (3/4)(1) / [(3/4)(1) + (1/4)(1/4)] = (3/4)/(13/16) = 12/13.
Answer: 12/13
Let E: has the disease, A: test positive. P(E) = 0.001, P(E′) = 0.999, P(A|E) = 0.99, P(A|E′) = 0.005.
P(E|A) = (0.001 × 0.99) / (0.001 × 0.99 + 0.999 × 0.005) = 0.00099 / 0.005985 = 22/133.
That is about 0.165, so even after a positive test the chance is only about 16.5%, because the disease is so rare.
Answer: 22/133 (≈ 0.165)
Let E1: two headed, E2: biased, E3: fair; each has probability 1/3. A: heads. P(A|E1) = 1, P(A|E2) = 3/4, P(A|E3) = 1/2.
P(E1|A) = (1/3)(1) / [(1/3)(1) + (1/3)(3/4) + (1/3)(1/2)] = 1 / (1 + 3/4 + 1/2) = 1/(9/4) = 4/9.
Answer: 4/9
Total insured = 12000. P(scooter) = 1/6, P(car) = 1/3, P(truck) = 1/2.
P(scooter|accident) = (1/6)(0.01) / [(1/6)(0.01) + (1/3)(0.03) + (1/2)(0.15)].
Multiply numerator and denominator by 12000: 20 / (20 + 120 + 900) = 20/1040 = 1/52.
Answer: 1/52
P(A) = 0.6, P(B) = 0.4, P(D|A) = 0.02, P(D|B) = 0.01.
P(B|D) = (0.4 × 0.01) / (0.6 × 0.02 + 0.4 × 0.01) = 0.004 / 0.016 = 1/4.
Answer: 1/4
P(G1) = 0.6, P(G2) = 0.4, P(N|G1) = 0.7, P(N|G2) = 0.3.
P(G2|N) = (0.4 × 0.3) / (0.6 × 0.7 + 0.4 × 0.3) = 0.12 / 0.54 = 2/9.
Answer: 2/9
Let E1: die shows 1, 2, 3 or 4, with P(E1) = 4/6 = 2/3. Let E2: die shows 5 or 6, with P(E2) = 1/3. A: exactly one head.
P(A|E1) = 1/2 (one toss gives a head). P(A|E2) = 3/8 (three tosses give exactly one head in 3 of 8 outcomes).
P(E1|A) = (2/3)(1/2) / [(2/3)(1/2) + (1/3)(3/8)] = (1/3) / (1/3 + 1/8) = (8/24)/(11/24) = 8/11.
Answer: 8/11
P(A) = 0.5, P(B) = 0.3, P(C) = 0.2; P(D|A) = 0.01, P(D|B) = 0.05, P(D|C) = 0.07.
P(A|D) = 0.005 / (0.005 + 0.015 + 0.014) = 0.005 / 0.034 = 5/34.
Answer: 5/34
Let E1: lost card is a diamond, P(E1) = 13/52 = 1/4. E2: lost card is not a diamond, P(E2) = 3/4. A: two diamonds drawn from the remaining 51 cards.
P(A|E1): 12 diamonds are left, so P = (12/51)(11/50) = 132/2550.
P(A|E2): 13 diamonds are left, so P = (13/51)(12/50) = 156/2550.
P(E1|A) = (1/4)(132) / [(1/4)(132) + (3/4)(156)] = 132 / (132 + 468) = 132/600 = 11/50.
Answer: 11/50
Let E1: a head occurs, E2: a tail occurs, each of probability 1/2. R: A reports a head. P(R|E1) = 4/5 (tells the truth), P(R|E2) = 1/5 (lies).
P(E1|R) = (1/2)(4/5) / [(1/2)(4/5) + (1/2)(1/5)] = (4/10)/(5/10) = 4/5.
Answer: (A) 4/5
Since A ⊂ B, A ∩ B = A, so P(A|B) = P(A) / P(B).
As P(B) ≤ 1, we have 1/P(B) ≥ 1, so P(A|B) = P(A)/P(B) ≥ P(A).
(A) is the reciprocal of the correct ratio, and (B) says the opposite.
Answer: (C) P(A|B) ≥ P(A)
(i) If A ⊂ B then A ∩ B = A, so P(B|A) = P(A)/P(A) = 1.
(ii) If A ∩ B = φ then P(A ∩ B) = 0, so P(B|A) = 0.
Answer: (i) 1 (ii) 0
Sample space {(b, b), (b, g), (g, b), (g, g)}.
(i) F = at least one male = {(b, b), (b, g), (g, b)}. P(both male | F) = (1/4)/(3/4) = 1/3.
(ii) F = elder is female = {(g, b), (g, g)}. P(both female | F) = (1/4)/(2/4) = 1/2.
Answer: (i) 1/3 (ii) 1/2
Let M: male, W: female, G: grey hair. P(M) = P(W) = 1/2, P(G|M) = 0.05, P(G|W) = 0.0025.
P(M|G) = (0.5)(0.05) / [(0.5)(0.05) + (0.5)(0.0025)] = 0.05 / 0.0525 = 20/21.
Answer: 20/21
Each person is right-handed with probability 9/10, independently. The number X of right-handed people among 10 has P(X = k) = ¹⁰Cₖ (9/10)ᵏ (1/10)¹⁰⁻ᵏ.
P(X ≤ 6) = Σ (k = 0 to 6) ¹⁰Cₖ (9/10)ᵏ (1/10)¹⁰⁻ᵏ = 1 − P(X ≥ 7).
P(X = 7) = 120 (0.9)⁷(0.1)³ ≈ 0.0574, P(X = 8) = 45 (0.9)⁸(0.1)² ≈ 0.1937, P(X = 9) = 10 (0.9)⁹(0.1) ≈ 0.3874, P(X = 10) = (0.9)¹⁰ ≈ 0.3487.
These add to about 0.9872, so P(X ≤ 6) ≈ 1 − 0.9872 = 0.0128.
Answer: Σ (k = 0 to 6) ¹⁰Cₖ (9/10)ᵏ (1/10)¹⁰⁻ᵏ ≈ 0.0128
A leap year has 366 = 52 × 7 + 2 days, so it has 52 full weeks and 2 extra days. These 2 days are consecutive days of the week.
The 7 possible pairs are (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon), all equally likely.
There are 53 Tuesdays if a Tuesday is among the two extra days: (Mon, Tue) and (Tue, Wed), which is 2 pairs.
Answer: 2/7
Each box has 10 marbles and is chosen with probability 1/4. P(red|A) = 1/10, P(red|B) = 6/10, P(red|C) = 8/10, P(red|D) = 0.
The common factor 1/4 cancels, so P(box | red) is proportional to 1, 6, 8, 0, whose total is 15.
P(A|red) = 1/15, P(B|red) = 6/15 = 2/5, P(C|red) = 8/15.
Answer: A: 1/15, B: 2/5, C: 8/15
Let E1: meditation and yoga, E2: drug, each of probability 1/2. A: heart attack.
With yoga the risk falls by 30%: P(A|E1) = 0.4 × 0.70 = 0.28. With the drug it falls by 25%: P(A|E2) = 0.4 × 0.75 = 0.30.
P(E1|A) = (1/2)(0.28) / [(1/2)(0.28) + (1/2)(0.30)] = 0.28 / 0.58 = 14/29.
Answer: 14/29
A determinant |a b; c d| has value ad − bc. There are 2⁴ = 16 equally likely arrays.
ad − bc > 0 only when ad = 1 and bc = 0. ad = 1 needs a = d = 1 (probability 1/4). bc = 0 means b and c are not both 1, which has probability 3/4.
So the favourable arrays are a = d = 1 with (b, c) ∈ {(0, 0), (0, 1), (1, 0)}: 3 arrays. Probability = 3/16.
Answer: 3/16
P(B fails) = P(B alone) + P(A and B fail) = 0.15 + 0.15 = 0.30.
(i) P(A fails | B has failed) = 0.15 / 0.30 = 1/2.
(ii) P(A fails alone) = P(A fails) − P(A and B fail) = 0.20 − 0.15 = 0.05 = 1/20.
Answer: (i) 1/2 (ii) 1/20 (= 0.05)
Let E1: transferred ball is red, with P(E1) = 3/7. E2: transferred ball is black, with P(E2) = 4/7. A: ball drawn from Bag II is red.
If E1, Bag II has 5 red and 5 black, so P(A|E1) = 5/10. If E2, Bag II has 4 red and 6 black, so P(A|E2) = 4/10.
P(E2|A) = (4/7)(4/10) / [(3/7)(5/10) + (4/7)(4/10)] = 16 / (15 + 16) = 16/31.
Answer: 16/31
P(B|A) = P(A ∩ B)/P(A) = 1 gives P(A ∩ B) = P(A).
So every outcome of A is an outcome of B, which means A ⊂ B.
Answer: (A) A ⊂ B
P(A|B) > P(A) means P(A ∩ B)/P(B) > P(A), so P(A ∩ B) > P(A) P(B).
Dividing by P(A): P(B|A) = P(A ∩ B)/P(A) > P(B).
Answer: (C) P(B|A) > P(B)
The left side is P(A ∪ B), so P(A ∪ B) = P(A). From P(A ∪ B) = P(A) + P(B) − P(A ∩ B) we get P(B) = P(A ∩ B).
Hence P(A|B) = P(A ∩ B)/P(B) = 1 (for P(B) ≠ 0). This says B ⊂ A. P(B|A) need not equal 1.
Answer: (B) P(A|B) = 1