Lessons / Quantitative aptitude
Quantitative aptitude · Day 7

Probability

Probability = favourable / total, but only when every outcome is equally likely. List the sample space when in doubt; it is small more often than you think.

Practise (24 questions) Present as slidesCheat sheet: Factorials 0! to 15!Cheat sheet: Combinations nCrCheat sheet: Counting shortcuts (no listing by hand)

Why this topic matters

If I toss a coin, will it be heads? Nobody knows. But everyone in this room would bet it is heads about half the time. Probability is simply that 'about half', written as a number.

Placement tests (TCS, Infosys, Wipro, Accenture, AMCAT, CoCubes) ask 1 to 3 probability questions in almost every paper: coins, dice, cards, balls in a bag, and 'A and B try to solve a problem'. They look different but use the same four ideas: count the outcomes, use 'not' for 'at least one', add for OR, multiply for AND. Learn those four and this topic becomes free marks.

What probability means

Probability is a number between 0 and 1 that says how likely something is. 0 means it cannot happen. 1 means it is certain. 1/2 means it happens half the time.

To find it, count. Probability of an event = (number of ways the event can happen) / (total number of possible outcomes). This only works when every outcome is equally likely: a fair coin, a fair die, a well-shuffled deck, a ball drawn without looking.

Four words you will see in every question. Experiment: the action (toss a coin, draw a card). Outcome: one possible result (heads). Sample space: the list of ALL possible outcomes ({H, T}). Event: the outcomes we care about (getting heads).

Probabilities of all the outcomes add up to 1. That is why P(not E) = 1 - P(E): if there is a 1/6 chance of a six, there is a 5/6 chance of no six.

00.250.50.751impossibleeven chancecertaina six
Every probability sits somewhere between 0 and 1
P(E) = favourable outcomes / total outcomes
Count what you want, count everything, divide.
0 ≤ P(E) ≤ 1
If you get more than 1 or a negative number, you have made a counting mistake.
P(not E) = 1 - P(E)
Something either happens or it does not.
Example 1A bag has 3 red, 4 green and 5 blue balls. One ball is taken out without looking. What is the probability it is green?
  1. Total outcomes = 3 + 4 + 5 = 12 balls.Each ball is equally likely to be picked, so each ball is one outcome.
  2. Favourable outcomes = 4 green balls.These are the outcomes that make our event (green) happen.
  3. P(green) = 4/12 = 1/3.
Example 2A number is chosen at random from 1 to 20. What is the probability it is a multiple of 3 or 5?
  1. Multiples of 3: 3, 6, 9, 12, 15, 18 (6 numbers). Multiples of 5: 5, 10, 15, 20 (4 numbers).List them; the list is short.
  2. 15 is in both lists, so count it once: 6 + 4 - 1 = 9.Counting 15 twice is the most common mistake in 'or' questions.
  3. P = 9/20.
Exam tipBefore calculating, say the total out loud: '36 for two dice', '52 for cards', '8 for three coins'. Half of all wrong answers come from the wrong total.

Coins

One coin has 2 outcomes: H and T. Two coins have 2 × 2 = 4: HH, HT, TH, TT. Three coins have 2 × 2 × 2 = 8. In general n coins give 2n outcomes.

HT and TH are DIFFERENT outcomes. Think of the coins as a 1-rupee and a 2-rupee coin: 'first heads, second tails' is not the same as 'first tails, second heads'. If you merge them you get 3 outcomes that are not equally likely, and every answer comes out wrong.

For 'exactly k heads', you do not need to list anything: the number of ways is nCk (choose which k coins show heads). So P(exactly k heads) = nCk / 2n.

H H HH H TH T HH T TT H HT H TT T HT T T
Three coins: 3 of the 8 outcomes have exactly two heads
n coins: 2n outcomes
1 coin 2, 2 coins 4, 3 coins 8, 4 coins 16.
P(exactly k heads) = nCk / 2n
Choose which k coins are heads.
P(at least one head) = 1 - 1/2n
The only way to fail is all tails.
Example 1Three coins are tossed. What is the probability of getting exactly two heads?
  1. Total = 23 = 8 outcomes.Each coin doubles the number of outcomes.
  2. Exactly two heads: HHT, HTH, THH = 3 outcomes (or 3C2 = 3).Choosing which 2 of the 3 coins are heads gives the count without listing.
  3. P = 3/8.
Example 2Four coins are tossed. What is the probability of getting at least one tail?
  1. Total = 24 = 16.Four coins.
  2. The opposite of 'at least one tail' is 'no tails', which is HHHH: only 1 outcome.Counting the opposite is quicker than counting 1, 2, 3 and 4 tails separately.
  3. P = 1 - 1/16 = 15/16.
Exam tip'At least one' with coins is always 1 - 1/2n. Four coins: 15/16. Five coins: 31/32.

Dice

A die has 6 faces, so one die gives 6 outcomes. Two dice give 6 × 6 = 36 outcomes, written as pairs: (1,1), (1,2) ... (6,6). Draw the 6 × 6 grid below once in your notebook. Every two-dice question is just shading the right boxes and counting them.

Sums follow a pattern. Sum 2 has 1 way, sum 3 has 2 ways, and so on up to sum 7, which has 6 ways. Then it falls: sum 8 has 5 ways, sum 12 has 1. So the number of ways for a sum s is (s - 1) when s is 7 or less, and (13 - s) when s is more than 7.

Doublets (both dice the same) are the diagonal of the grid: 6 of them.

second diefirst die112345672234567833456789445678910556789101166789101112
Sum 7: the 6 shaded boxes out of 36. P = 6/36 = 1/6
second diefirst die112345672234567833456789445678910556789101166789101112
Sum at least 10: 6 boxes. P = 6/36 = 1/6
Sum23456789101112
Ways out of 3612345654321
Example 1Two dice are thrown. What is the probability that the sum is 8?
  1. Total = 36.Two dice.
  2. Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 ways.Rule: for sums above 7, ways = 13 - sum = 13 - 8 = 5.
  3. P = 5/36.
Example 2Two dice are thrown. What is the probability of getting a doublet or a total of 4?
  1. Doublets: (1,1) ... (6,6) = 6 outcomes.The diagonal of the grid.
  2. Total 4: (1,3), (2,2), (3,1) = 3 outcomes.Sum 4 has 4 - 1 = 3 ways.
  3. (2,2) is in both lists, so the union is 6 + 3 - 1 = 8.OR means 'either', and the overlap must only be counted once.
  4. P = 8/36 = 2/9.
Example 3Three dice are thrown. What is the probability that all three show the same number?
  1. Total = 63 = 216.Each die multiplies the outcomes by 6.
  2. All the same: (1,1,1), (2,2,2) ... (6,6,6) = 6 outcomes.One outcome per number.
  3. P = 6/216 = 1/36.
Exam tipTwo dice, sum s: ways = 6 - |7 - s|. Sum 10 → 6 - 3 = 3 ways. You never need to list them.

Cards

A pack has 52 cards in 4 suits: spades and clubs (black), hearts and diamonds (red). Each suit has 13 cards: A, 2, 3 ... 10, J, Q, K.

So there are 4 of every rank (4 aces, 4 kings), 13 of every suit, and 26 of each colour. Face cards (also called court cards) are J, Q and K: 3 per suit, 12 in total.

When several cards are drawn together, count with combinations: the number of ways to draw 2 cards from 52 is 52C2 = 1326.

SpadesA2345678910JQKClubsA2345678910JQKHeartsA2345678910JQKDiamondsA2345678910JQK
A deck: 52 cards = 4 suits × 13 ranks. 26 red, 26 black. 12 face cards (J, Q, K), 4 aces
Example 1One card is drawn from a pack of 52. What is the probability that it is a king or a heart?
  1. Kings: 4. Hearts: 13.Count each group.
  2. The king of hearts is in both groups, so kings or hearts = 4 + 13 - 1 = 16.OR: add, then subtract the overlap.
  3. P = 16/52 = 4/13.
Example 2Two cards are drawn together from a pack of 52. What is the probability that both are red?
  1. Ways to draw any 2 cards: 52C2 = 52 × 51 / 2 = 1326.'Drawn together' means order does not matter, so use combinations.
  2. Ways to draw 2 red cards: 26C2 = 26 × 25 / 2 = 325.Choose both from the 26 red cards.
  3. P = 325/1326 = 25/102.
Example 3One card is drawn. What is the probability that it is a face card that is not red?
  1. Face cards: J, Q, K in each of 4 suits = 12.3 per suit.
  2. Not red means black: spades or clubs, so 2 suits × 3 = 6.Half of the face cards.
  3. P = 6/52 = 3/26.
Exam tipWrite '52 = 4 × 13, 26 red, 12 face, 4 of each' at the top of your rough sheet before a card question.

Balls in a bag: with and without replacement

'With replacement' means the ball goes back before the next draw, so every draw looks exactly like the first. 'Without replacement' means it stays out, so the second draw has one ball fewer, and one fewer of whatever colour came out.

If balls are drawn 'together' or 'at the same time', that is the same as without replacement, and combinations are the fastest way to count.

Two ways to solve a without-replacement question, and both give the same answer: multiply step by step (5/8 × 4/7), or count with combinations (5C2 / 8C2). Use whichever you find easier.

5/8R4/7RR R3/7BR B3/8B5/7RB R2/7BB B
Without replacement: the second branch depends on the first draw
Example 1A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. What is the probability both are red?
  1. First draw red: 5/8.5 red out of 8 balls.
  2. Second draw red, given the first was red: 4/7.One red ball is already out, so 4 red remain out of 7.
  3. Both: 5/8 × 4/7 = 20/56 = 5/14.AND means multiply.
  4. Check with combinations: 5C2 / 8C2 = 10/28 = 5/14. Same.
Example 2Same bag, but the first ball is put back before the second draw. Probability both are red?
  1. First red: 5/8. Second red: 5/8 again.The ball went back, so the bag is unchanged.
  2. P = 5/8 × 5/8 = 25/64.
Example 3A bag has 3 red and 5 black balls. Two are drawn together. What is the probability of getting one of each colour?
  1. Total ways: 8C2 = 28.Drawn together, so combinations.
  2. One red AND one black: 3C1 × 5C1 = 3 × 5 = 15.Choose the red one (3 ways) and the black one (5 ways); multiply.
  3. P = 15/28.
Exam tip'Together' = combinations. 'One after another without replacement' = multiply fractions that shrink by one.

At least one: use the complement

'At least one' means one, or two, or three, or more. Counting all of those cases separately is slow and easy to get wrong.

The opposite of 'at least one' is 'none'. And 'none' is usually one simple case. So P(at least one) = 1 - P(none).

This is exactly how to solve 'A and B try to solve a problem' questions. The problem gets solved if at least one of them solves it. So find the probability that NEITHER solves it, and subtract from 1.

1/2A✓1/3B✓A✓ B✓2/3B✗A✓ B✗1/2A✗1/3B✓A✗ B✓2/3B✗A✗ B✗
Only the A✗ B✗ path leaves the problem unsolved: 1/2 × 2/3 = 1/3
P(at least one) = 1 - P(none)
Works for coins, dice, people, machines, shots at a target.
P(none) for independent events = (1 - p1)(1 - p2)(1 - p3)...
Each person fails independently, so multiply the failure chances.
Example 1A and B solve a problem independently with probabilities 1/2 and 1/3. What is the probability that the problem gets solved?
  1. 'The problem gets solved' means at least one of A and B solves it.Rewrite the question in these words first; it tells you which rule to use.
  2. P(A fails) = 1 - 1/2 = 1/2. P(B fails) = 1 - 1/3 = 2/3.Failure is the complement of success for each person.
  3. P(both fail) = 1/2 × 2/3 = 1/3.They work independently, so multiply.
  4. P(solved) = 1 - 1/3 = 2/3.Solved is the opposite of 'both fail'.
  5. Check: 1/2 + 1/3 would be 5/6, which counts the case where both solve it twice. That is why 5/6 is the trap option.
Example 2A die is thrown 3 times. What is the probability of getting at least one six?
  1. P(no six on one throw) = 5/6.5 of the 6 faces are not a six.
  2. P(no six in 3 throws) = (5/6)3 = 125/216.The throws are independent, so multiply.
  3. P(at least one six) = 1 - 125/216 = 91/216.
Example 3Three students solve a problem with probabilities 1/2, 1/3 and 1/4, independently. What is the probability that it is solved?
  1. Failure chances: 1/2, 2/3, 3/4.1 minus each success chance.
  2. All three fail: 1/2 × 2/3 × 3/4 = 6/24 = 1/4.Independent, so multiply.
  3. P(solved) = 1 - 1/4 = 3/4.
Exam tipRead 'at least', 'not all', 'the problem is solved', 'the target is hit' as: 1 minus P(nobody).

OR: the addition rule

P(A or B) means the chance that A happens, or B happens, or both. If you just add P(A) + P(B), the outcomes where both happen get counted twice. So subtract them once: P(A or B) = P(A) + P(B) - P(A and B).

If A and B can never happen together (mutually exclusive), the overlap is empty and you simply add. Example: one die shows 2 or shows 5. It cannot show both, so 1/6 + 1/6 = 1/3.

Kings (4)Hearts (13)311252 cards
King or heart: 4 + 13 - 1 = 16 cards
P(A or B) = P(A) + P(B) - P(A and B)
Add, then remove the double-counted overlap.
Mutually exclusive: P(A or B) = P(A) + P(B)
No overlap, nothing to remove.
Example 1From the numbers 1 to 30, one is picked at random. What is the probability it is divisible by 2 or 3?
  1. Divisible by 2: 15 numbers. By 3: 10 numbers.30/2 and 30/3.
  2. Divisible by both means divisible by 6: 5 numbers.The overlap of multiples of 2 and 3 is multiples of 6.
  3. 15 + 10 - 5 = 20. P = 20/30 = 2/3.
Exam tipSee the word OR? Add, then ask 'can both happen at once?' If yes, subtract the overlap.

AND: the multiplication rule

P(A and B) is the chance that both happen. If the two events do not affect each other (independent), multiply: P(A) × P(B). Coins, dice and draws WITH replacement are independent.

If the first event changes the second (dependent), multiply by the second probability AFTER the first has happened. Draws WITHOUT replacement are dependent.

A probability tree shows this: multiply along a branch to get AND, add different branches to get OR.

1/2H1/66H 65/6not 6H not 61/2T1/66T 65/6not 6T not 6
Coin then die: heads AND six = 1/2 × 1/6 = 1/12
Example 1A coin is tossed and a die is thrown. What is the probability of heads and an even number?
  1. P(heads) = 1/2. P(even) = 3/6 = 1/2.Three even faces: 2, 4, 6.
  2. Independent, so P = 1/2 × 1/2 = 1/4.The coin does not affect the die.
Example 2Two cards are drawn one after the other without replacement. What is the probability that both are aces?
  1. First ace: 4/52.4 aces in 52 cards.
  2. Second ace: 3/51.One ace and one card are gone.
  3. P = 4/52 × 3/51 = 12/2652 = 1/221.
Example 3A speaks the truth 3/4 of the time and B 4/5 of the time. What is the probability that they contradict each other on the same statement?
  1. They contradict when exactly one of them tells the truth.If both tell the truth or both lie, they agree.
  2. A true and B lies: 3/4 × 1/5 = 3/20.AND, independent: multiply.
  3. A lies and B true: 1/4 × 4/5 = 4/20.The other way round.
  4. Add the two cases: 3/20 + 4/20 = 7/20.The two cases cannot happen together, so plain addition.
Exam tipSay it in words: 'this AND that' → multiply. 'this case OR that case' → add the cases.

Conditional probability: 'given that'

Sometimes the question tells you something has already happened: 'given that the sum is even', 'if it is known that one child is a boy'. That information shrinks the sample space. Throw away every outcome that does not fit, and count only among the ones left.

As a formula: P(A given B) = P(A and B) / P(B). But the shrink-and-count method is easier to understand and gives the same answer.

second diefirst die112345672234567833456789445678910556789101166789101112
Given the sum is even (18 boxes), sum 8 is 5 of them: 5/18
P(A given B) = P(A and B) / P(B)
Among the outcomes where B happened, what fraction also have A?
Example 1Two dice are thrown. Given that the sum is even, what is the probability that the sum is 8?
  1. Outcomes with an even sum: 18 of the 36.Half the grid; this is our new, smaller sample space.
  2. Among those, sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5.All of them have an even sum, so they are all inside the new space.
  3. P = 5/18.
Example 2A family has two children. Given that at least one is a boy, what is the probability that both are boys?
  1. All families: BB, BG, GB, GG.Older child first; BG and GB are different families.
  2. 'At least one boy' removes GG. Left: BB, BG, GB.The given information shrinks the sample space to 3.
  3. Both boys = BB, 1 of 3. P = 1/3.
Exam tip'Given that' → cross out every outcome that breaks the given fact, then count again.

Odds, and how questions are worded

Odds are another way of saying probability. 'Odds in favour of E are 3 : 5' means for every 3 ways E happens there are 5 ways it does not. So there are 8 equally likely ways in total and P(E) = 3/8. 'Odds against E are 3 : 5' means P(E) = 5/8.

The table below turns the usual phrases into the rule you need.

The question saysIt meansUse
at least oneone or more1 - P(none)
at most onezero or oneP(0) + P(1)
exactly oneone and only oneadd the cases (first only, second only...)
neithernone of themmultiply the failure chances
either ... orORaddition rule
both / andANDmultiplication rule
given that / if it is knownconditionalshrink the sample space
drawn together / at random at onceno ordercombinations nC r
odds in favour a : bP = a/(a + b)
Example 1The odds against an event are 7 : 2. What is its probability?
  1. Odds against 7 : 2 means 7 ways it fails for every 2 ways it happens.Against puts failure first.
  2. Total 9 ways. P(event) = 2/9.

Formula summary

P(E) = favourable / total
P(not E) = 1 - P(E)
'At least one' is almost always 1 - P(none).
Addition
P(A or B) = P(A) + P(B) - P(A and B).
Independent
P(A and B) = P(A) × P(B).
Conditional
P(A | B) = P(A and B) / P(B).
Two dice
36 outcomes. Sum 7 has 6 ways, the most of any sum.
Cards
52 = 4 suits × 13. 12 face cards. 4 of each rank.

Shortcuts and tricks

At least one = 1 minus none

Use it when: 'At least one six', 'the problem is solved', 'the target is hit'.

  1. Find the chance that it fails every time.
  2. Subtract from 1.
ExampleAt least one six in three throws of a die?
  1. P(no six) in one throw = 5/6, so in three throws (5/6)3 = 125/216.Independent throws multiply.
  2. 1 - 125/216 = 91/216.

Sum of two dice without listing

Use it when: Any 'sum is s' question with two dice.

  1. Ways = 6 - |7 - s|.
  2. Divide by 36.
ExampleSum of 10 with two dice?
  1. 6 - |7 - 10| = 6 - 3 = 3 ways: (4,6), (5,5), (6,4).Sums are symmetric around 7.
  2. 3/36 = 1/12.

Exactly k heads

Use it when: Any coin question asking for an exact count.

  1. Ways = nCk. Total = 2n.
ExampleExactly 3 heads in 5 tosses?
  1. 5C3 = 10 ways.Choose which 3 tosses are heads.
  2. Total 25 = 32. P = 10/32 = 5/16.

Common mistakes

Treating HT and TH as the same outcome, so two coins seem to have 3 outcomes.
Label the coins (first, second). Two coins always have 4 outcomes.
Adding success chances: 1/2 + 1/3 = 5/6 for 'A or B solves it'.
Use 1 - P(both fail) = 1 - 1/2 × 2/3 = 2/3.
Keeping the bag the same size on the second draw without replacement.
Take one ball out of the total and out of its colour: 5/8 then 4/7.
Counting the king of hearts twice in 'king or heart'.
OR: add, then subtract what is in both groups.
Forgetting that 'drawn together' is a combination question.
Use nCr for both the favourable and the total count.

Before you move on, you should be able to...

  • say what total to use for coins (2n), dice (6n) and cards (52), before calculating
  • turn 'at least one' into 1 - P(none) and solve the 'A and B solve a problem' question
  • add for OR (and subtract the overlap), multiply for AND
  • tell with replacement from without, and adjust the second fraction
  • shrink the sample space for 'given that' questions

Practice questions (24)

Try each one before opening the solution. Or practise them one by one so your score is saved.

easy A die is thrown once. What is the probability of getting a prime number?
  1. Primes on a die: 2, 3, 5.1 is not prime.
  2. 3/6 = 1/2.

Answer: 1/2

easy Three coins are tossed. What is the probability of getting at least two heads?
  1. Total 8.23.
  2. Two heads: 3 ways (3C2). Three heads: 1 way.'At least two' = two or three.
  3. 4/8 = 1/2.

Answer: 1/2

easy Two dice are thrown. What is the probability that the sum is at least 10?
  1. Sum 10: 3 ways, 11: 2 ways, 12: 1 way.Ways = 13 - sum above 7.
  2. 6/36 = 1/6.

Answer: 1/6

easy One card is drawn from a pack of 52. What is the probability that it is a king or a heart?
  1. Kings 4, hearts 13, king of hearts counted in both.Overlap of 1.
  2. 4 + 13 - 1 = 16. 16/52 = 4/13.

Answer: 4/13

easy A bag has 4 white and 6 black balls. One is drawn. What is the probability it is not white?
  1. Not white = black: 6 of 10.Or 1 - P(white) = 1 - 4/10.
  2. 6/10 = 3/5.

Answer: 3/5

easy A number is chosen from 1 to 50. What is the probability that it is a perfect square?
  1. Squares up to 50: 1, 4, 9, 16, 25, 36, 49.72 = 49 is the last one below 50.
  2. 7/50.

Answer: 7/50

easy Two dice are thrown. What is the probability of a doublet?
  1. Doublets: (1,1) to (6,6), 6 outcomes.The diagonal of the 6 × 6 grid.
  2. 6/36 = 1/6.

Answer: 1/6

medium A bag has 5 red and 4 blue balls. Two are drawn together. What is the probability both are red?
  1. Total: 9C2 = 36.Drawn together, so combinations.
  2. Both red: 5C2 = 10.Choose 2 of the 5 red.
  3. 10/36 = 5/18.

Answer: 5/18

medium A family has two children and at least one is a boy. What is the probability both are boys?
  1. Families: BB, BG, GB, GG. 'At least one boy' removes GG.The given fact shrinks the sample space to 3.
  2. BB is 1 of 3.

Answer: 1/3

medium Three dice are thrown. What is the probability that all three show different numbers?
  1. Total 216.63.
  2. All different: 6 × 5 × 4 = 120.First die any, second avoids one number, third avoids two.
  3. 120/216 = 5/9.

Answer: 5/9

medium Two cards are drawn together from a pack of 52. What is the probability that one is a spade and one is a heart?
  1. Total: 52C2 = 1326.Drawn together.
  2. One spade and one heart: 13 × 13 = 169.Choose one from each suit and multiply.
  3. 169/1326 = 13/102.

Answer: 13/102

medium Two dice are thrown. What is the probability that the product of the numbers is even?
  1. The product is odd only when both numbers are odd: 3 × 3 = 9 outcomes.One even number makes the whole product even.
  2. P(even) = 1 - 9/36 = 3/4.

Answer: 3/4

medium A bag has 3 red and 5 black balls. Two are drawn together. What is the probability of one of each colour?
  1. Total 8C2 = 28.Combinations.
  2. One red AND one black: 3 × 5 = 15.Multiply the choices.
  3. 15/28.

Answer: 15/28

medium The odds in favour of an event are 3 : 5. What is the probability that it does NOT happen?
  1. 3 ways for, 5 ways against, 8 in total.Odds in favour a : b means P = a/(a + b).
  2. P(not) = 5/8.

Answer: 5/8

medium A die is thrown twice. What is the probability that the second number is greater than the first?
  1. Equal numbers: 6 outcomes. The other 30 split evenly between 'first bigger' and 'second bigger'.By symmetry, the two cases are equally likely.
  2. 15/36 = 5/12.

Answer: 5/12

hard A and B solve a problem independently with probabilities 1/2 and 1/3. What is the probability that the problem gets solved?
  1. P(both fail) = 1/2 × 2/3 = 1/3.Solved = at least one solves it = 1 - P(both fail).
  2. 1 - 1/3 = 2/3.

Answer: 2/3

hard Three students solve a problem independently with probabilities 1/2, 1/3 and 1/4. What is the probability that the problem gets solved?
  1. P(all fail) = 1/2 × 2/3 × 3/4 = 1/4.Multiply the failure chances.
  2. 1 - 1/4 = 3/4.

Answer: 3/4

hard Two cards are drawn one after another without replacement. What is the probability both are aces?
  1. 4/52 for the first ace, 3/51 for the second.One ace and one card are gone after the first draw.
  2. 12/2652 = 1/221.

Answer: 1/221

hard A speaks the truth 3/4 of the time and B 4/5 of the time. What is the probability they contradict each other on a statement?
  1. Contradict = exactly one tells the truth.Both true or both false means they agree.
  2. 3/4 × 1/5 + 1/4 × 4/5 = 3/20 + 4/20.Two separate cases, so add.
  3. 7/20.

Answer: 7/20

hard The letters of ASSASSIN are arranged at random. What is the probability that all four S come together?
  1. Total arrangements: 8!/(4! 2!) = 840.4 S alike and 2 A alike.
  2. Glue the four S into one block. Now arrange 5 items (SSSS, A, A, I, N): 5!/2! = 60.The block counts as one item; the two A are alike.
  3. 60/840 = 1/14.

Answer: 1/14

hard Two dice are thrown. Given that the sum is even, what is the probability that it is 8?
  1. Even sums: 18 outcomes. Sum 8: 5 of them.Shrink the sample space to the given fact.
  2. 5/18.

Answer: 5/18

hard A box has 6 good and 4 defective bulbs. 3 are picked at random. What is the probability that at least one is defective?
  1. P(none defective) = 6C3 / 10C3 = 20/120 = 1/6.'At least one' → complement.
  2. 1 - 1/6 = 5/6.

Answer: 5/6

hard A coin is tossed 6 times. What is the probability of getting exactly 4 heads?
  1. Ways: 6C4 = 15. Total: 26 = 64.Choose which 4 tosses are heads.
  2. 15/64.

Answer: 15/64

hard Four people are chosen from 5 men and 4 women. What is the probability that exactly 2 are women?
  1. Total: 9C4 = 126.Choose any 4 of 9.
  2. 2 women and 2 men: 4C2 × 5C2 = 6 × 10 = 60.Choose each group separately and multiply.
  3. 60/126 = 10/21.

Answer: 10/21

Class plan (2 hours, for trainers)

A tested order for teaching this topic in one 2-hour session. Present mode follows the same order.

10 minSpeed maths warm-up: five 2-digit criss-cross multiplications on the board, students race. (Keeps the daily habit.)
10 minHook: the coin question. Build the 0-to-1 line together; students place 'rain tomorrow', 'sun rises', 'a six' on it. Define outcome, sample space, event.
15 minCoins and dice. Draw the 6 × 6 dice grid on the board; students shade sum 7, sum 8, doublets and count. Show HT and TH are different.
10 minCards: 52 = 4 × 13. Students fill the deck breakdown from memory, then the king-or-heart example.
15 minAt least one = 1 - none. Do the A and B problem (the one students got wrong in the test) with the tree, then the three-students version.
15 minOR and AND. Venn diagram for the addition rule, tree for the multiplication rule. With vs without replacement with the bag example.
10 minConditional: shrink the sample space. Dice given even sum; the two-children puzzle (let the class argue 1/2 vs 1/3 first).
20 minIndividual practice: the easy and medium questions on the practice page, timed. Walk the room.
10 minGo through the three hard questions together on the projector (present mode).
5 minRecap: each student says one rule in their own words. Exit ticket: at least one six in two throws (answer 11/36).
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