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State Board · Class 10 · Mathematics Part 1

Lesson 5: Probability

3 concepts · 24 practice questions with answers · Free to read

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Random experiments and the sample space

From daily life

When we toss a coin we cannot say in advance whether it will fall head or tail, but we know there are only two possibilities. When we roll a die any number from 1 to 6 can come up.

Simple definition

An experiment whose possible results are all known in advance, but whose actual result cannot be predicted and in which every result is equally likely, is a random experiment. Each result is an outcome. The set of all outcomes is the sample space S and the number of outcomes is n(S). (H = head, T = tail.)

HTn(S) = 2123456n(S) = 66 × 6 = 36

In the figure: One coin: S = {H, T} so n(S) = 2; one die: S = {1, 2, 3, 4, 5, 6} so n(S) = 6; two dice: n(S) = 6 × 6 = 36.

Rule 1: Count the outcomes — n(S) is the total number of possible outcomes. Read the experiment, list every possibility and count them.

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Random experiments and the sample space

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  • Easy: One coin is tossed ⇒ S = {H, T}, n(S) = 2.
  • Medium: Cards numbered 1 to 50; one card is drawn ⇒ n(S) = 50.
  • Exam-type: A pack has 52 cards: 4 suits × 13 each = 52. Red 26, black 26 and face cards (king, queen, jack) 3 × 4 = 12.

Rule 2: More than one object: multiply — Multiply the number of possibilities of each object. A coin tossed twice and two coins tossed together have the same sample space.

  • Easy: Two coins ⇒ S = {HH, HT, TH, TT}, n(S) = 4.
  • Medium: Three coins ⇒ n(S) = 2 × 2 × 2 = 8.
  • Exam-type: One coin and one die ⇒ n(S) = 2 × 6 = 12. Two dice ⇒ n(S) = 6 × 6 = 36.

Rule 3: Committees: count the pairs — Choosing 2 people out of n: count each pair once. It can be counted as n × (n − 1) ÷ 2.

  • Easy: A committee of 2 from 4 children (1, 2, 3, 4) ⇒ S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}, n(S) = 6.
  • Medium: 2 out of 5 people ⇒ n(S) = 5 × 4 ÷ 2 = 10.
  • Exam-type: 3 boys and 2 girls, a committee of 2: boy-girl 3 × 2 = 6; 2 boys 3; 2 girls 1. Total n(S) = 6 + 3 + 1 = 10.
Remember

Always find n(S) by counting every outcome: two coins give 4 and two dice give 36.

Common mistake

HT and TH are different outcomes. So are (1, 2) and (2, 1). Do not count them as one.

Shortcut

For any number of objects, n(S) = the product of the possibilities of each object. Memory clue: "more objects, multiply" — coin 2, die 6.

Spot the difference

PointCoinsDice
One objectn(S) = 2n(S) = 6
Two objectsn(S) = 4n(S) = 36
Three objectsn(S) = 8n(S) = 216
Example outcomeHT(2, 5)

Practice questions

Easy
  1. 1Write the sample space of one die and give n(S).
  2. 2Two coins are tossed. What is n(S)?
  3. 3Cards are numbered 1 to 25. One card is drawn. What is n(S)?
Medium
  1. 4One coin and one die are thrown together. What is n(S)?
  2. 5Write the sample space for three coins.
  3. 6A committee of 2 is to be formed from 2 boys and 2 girls. What is n(S)?
Challenging
  1. 7Two dice are thrown. What is n(S), and how many outcomes are possible if the first die shows 4?
  2. 8How many face cards are in a pack of 52 cards, and how many of them are black?
Show answers (try them yourself first)
  1. 1S = {1, 2, 3, 4, 5, 6}, n(S) = 6.
  2. 22 × 2 = 4.
  3. 3n(S) = 25.
  4. 42 × 6 = 12.
  5. 5S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8.
  6. 64 × 3 ÷ 2 = 6.
  7. 7n(S) = 36. First die 4: (4, 1) to (4, 6), that is 6 outcomes.
  8. 8Face cards 3 × 4 = 12. Black ones 3 × 2 = 6.

Events

From daily life

When rolling a die we may hope for "an even number". The results 2, 4, 6 are as expected and 1, 3, 5 are not.

Simple definition

The outcomes in the sample space that satisfy a given condition are favourable outcomes. The set of favourable outcomes is an event. An event is a subset of S. Events are named A, B, C and the number of outcomes is n(A).

112233445566777777636

In the figure: The 36 outcomes of two dice (row = first die, column = second): the 6 outcomes with sum 7 are coloured. n(E) = 6, n(S) = 36.

Rule 1: Write an event as a set — Read the condition, test each outcome of S and write the ones that satisfy it. Then count n(A).

  • Easy: One die, A = an even number ⇒ A = {2, 4, 6}, n(A) = 3.
  • Medium: One die, B = a prime number ⇒ B = {2, 3, 5}, n(B) = 3. (1 is not a prime.)
  • Exam-type: Cards numbered 1 to 30: A = divisible by 5 ⇒ {5, 10, 15, 20, 25, 30}, n(A) = 6. B = a perfect cube ⇒ {1, 8, 27}, n(B) = 3.

Rule 2: Coins: "at least" and "at most" — "At least" means not less than; "at most" means not more than. Pick such outcomes from the list.

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Events

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  • Easy: Two coins, A = at least one H ⇒ {HH, HT, TH}, n(A) = 3.
  • Medium: Three coins, A = at least two H ⇒ {HHH, HHT, HTH, THH}, n(A) = 4.
  • Exam-type: Three coins: B = no H ⇒ {TTT}, n(B) = 1. C = at most one T ⇒ {HHH, HHT, HTH, THH}, n(C) = 4.

Rule 3: Two dice: sums and conditions — Check the sums in the table of 36 outcomes. A sum of 7 comes most often (6 outcomes).

  • Easy: Sum 3 ⇒ {(1, 2), (2, 1)}, n = 2.
  • Medium: Sum 7 ⇒ 6 outcomes. Sum 12 ⇒ {(6, 6)}, n = 1.
  • Exam-type: Same number on both ⇒ {(1, 1), (2, 2), …, (6, 6)}, n = 6. Sum 13 ⇒ impossible event, n = 0. Sum 2 to 12 ⇒ sure event, n = 36.
Remember

In a sure event n(A) = n(S) and in an impossible event n(A) = 0.

Common mistake

"At least 2 H" includes both 2 and 3 heads. Taking only 2 heads is wrong.

Shortcut

For sums 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 the outcomes are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 in order. Memory clue: "count the outcomes that fit the condition" — a table prevents mistakes.

Spot the difference

PointSure eventImpossible event
n(A)n(S)0
One die: examplea number from 1 to 6 comes upthe number 7 comes up
Does it happenalwaysnever
Probability10

Practice questions

Easy
  1. 1One die; A = an odd number. Write A as a set.
  2. 2Two coins; A = no H. Write A.
  3. 3Cards numbered 1 to 20; how many cards are divisible by 4?
Medium
  1. 4Two dice; write the event that the sum is 5.
  2. 5Three coins; write the event of exactly two H.
  3. 6Two dice; 6 on the first and more than 3 on the second. Write the event.
Challenging
  1. 7Two dice; how many outcomes are in the event that the sum is more than 10?
  2. 84 boys and 2 girls; a committee of 2. How many outcomes have at least one girl?
Show answers (try them yourself first)
  1. 1A = {1, 3, 5}, n(A) = 3.
  2. 2A = {TT}, n(A) = 1.
  3. 3{4, 8, 12, 16, 20}, that is 5.
  4. 4{(1, 4), (2, 3), (3, 2), (4, 1)}, n = 4.
  5. 5{HHT, HTH, THH}, n = 3.
  6. 6{(6, 4), (6, 5), (6, 6)}, n = 3.
  7. 7{(5, 6), (6, 5), (6, 6)}, that is 3.
  8. 8n(S) = 6 × 5 ÷ 2 = 15; no girl 4 × 3 ÷ 2 = 6; at least one girl 15 − 6 = 9.

Probability

From daily life

When there are clouds we say "it may rain". In maths we give "may" a number. That number is the probability.

Simple definition

The probability of an event A is P(A) = favourable outcomes ÷ all outcomes = n(A) ÷ n(S). It can be written as a fraction, a decimal or a percentage.

Formula: P(A) = n(A) / n(S)

01/211/65/61/4P(A) + P(A′) = 1

In the figure: Probability always lies between 0 and 1: 0 is impossible, 1/2 is equal chance, 1 is sure. P(A) + P(A′) = 1.

Rule 1: Count n(A) and n(S), then divide — First find n(S), then n(A), and write P(A) = n(A) ÷ n(S). Simplify the answer.

  • Easy: One die; A = getting 5 ⇒ P(A) = 1/6.
  • Medium: A bag has 3 white and 5 black balls; one ball is drawn ⇒ n(S) = 8; P(white) = 3/8; P(black) = 5/8.
  • Exam-type: Cards numbered 1 to 25: multiples of 3 are {3, 6, 9, 12, 15, 18, 21, 24}, n(A) = 8 ⇒ P(A) = 8/25 = 32 %.

Rule 2: Playing cards and two dice — For playing cards n(S) = 52 and for two dice n(S) = 36. Count the favourable outcomes from the table.

Live figure

Probability

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  • Easy: One from 52 cards: P(a spade) = 13/52 = 1/4.
  • Medium: P(a king) = 4/52 = 1/13. P(a face card) = 12/52 = 3/13.
  • Exam-type: Two dice: P(sum 7) = 6/36 = 1/6. P(sum 9 or more) = (4 + 3 + 2 + 1)/36 = 10/36 = 5/18.

Rule 3: Limits of probability and "not A" — 0 ≤ P(A) ≤ 1. The probability that A does not happen is 1 − P(A).

  • Easy: Impossible event: P = 0 ÷ n(S) = 0. Sure event: P = n(S) ÷ n(S) = 1.
  • Medium: One die: P(a number less than 7) = 6/6 = 1 (sure). P(the number 7) = 0 (impossible).
  • Exam-type: If the probability of rain is 3/10, the probability of no rain is 1 − 3/10 = 7/10. One die: P(not 6) = 1 − 1/6 = 5/6.
Remember

0 ≤ P(A) ≤ 1. A probability is never negative or greater than 1.

Common mistake

Taking n(S) = 6 for two dice is wrong. n(S) = 36.

Shortcut

For "at least one": find 1 − P(none). Memory clue: probability = favourable ÷ total; "favourable on top, total below".

Spot the difference

PointEvent AA does not happen
Favourable outcomesn(A)n(S) − n(A)
Probabilityn(A) / n(S)1 − P(A)
One die, A = getting 61/65/6
Range0 to 10 to 1

Practice questions

Easy
  1. 1A coin is tossed. What is P(H)?
  2. 2A die is rolled. What is P(an even number)?
  3. 3One card is drawn from 52. What is P(a red card)?
Medium
  1. 4A bag has 4 red and 6 blue balls. What is P(blue)?
  2. 5Two coins are tossed. What is P(at least one H)?
  3. 6Two dice are rolled. What is P(sum 8)?
Challenging
  1. 7Cards numbered 1 to 40. What is P(a prime number)?
  2. 8One die; find P(a number greater than 4) and P(4 or less).
Show answers (try them yourself first)
  1. 1P = 1/2.
  2. 23/6 = 1/2.
  3. 326/52 = 1/2.
  4. 46/10 = 3/5.
  5. 53/4.
  6. 6Sum 8 has 5 outcomes ⇒ 5/36.
  7. 7Primes 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 = 12 ⇒ 12/40 = 3/10.
  8. 8{5, 6} ⇒ 2/6 = 1/3; the other is 1 − 1/3 = 2/3.