Probability Class 10 Maths Notes

Probability is the second half of Unit VII Statistics and Probability, and the two chapters together carry 11 of the 80 theory marks. The retained content is deliberately narrow: the classical definition of probability, and simple problems in finding the probability of an event.There is no theory to prove and no new idea between one question and the next. You are counting, and the whole chapter is deciding what counts as one outcome.

• The classical definition, with the condition that all outcomes of the experiment are equally likely.
• Simple problems in finding the probability of an event — a single event, asked and answered in one step.
• The two boundary cases: the impossible event has probability 0 and the certain event has probability 1.
• The range of probability, 0 ≤ P(E) ≤ 1.

Favourable outcomes are the outcomes that satisfy the condition in the question. For example, if the question asks for a prime number on a die, the favourable outcomes are the faces showing 2, 3 and 5, so there are 3 of them out of the 6 possible faces.Always simplify the fraction before writing the answer. 3 out of 6 is written as 1/2 and not as 3/6, because the board wants the probability in its simplest form and an unsimplified fraction loses the final mark.

• Total outcomes for a fair die: 6.
• Total outcomes for a fair coin: 2, heads and tails.
• Total outcomes for a well-shuffled pack of 52 playing cards: 52.
• Total outcomes for a spinner with 8 equal sectors: 8.
• Total outcomes for n slips or n bottle caps each numbered from 1 to n: n.
• Favourable outcomes: count only those outcomes that satisfy the wording of the question, and nothing else.

The two ends of the scale are named, and a one-mark MCQ often tests them. An impossible event cannot happen under any outcome, so no outcome is favourable to it and P = 0. A certain event happens for sure, so every outcome is favourable and P = 1.These two values are exactly the ends of the range 0 ≤ P(E) ≤ 1. A probability is never negative and never greater than 1, because the number of favourable outcomes can never be less than none and never more than all of them.

• P(E) = 0 for an impossible event. Example: getting a 7 on a die, or drawing a king from a pack containing no kings.
• P(E) = 1 for a certain event. Example: getting a number from 1 to 6 on a die, or drawing a card from a pack of 52 cards.
• Any probability in the syllabus is a proper fraction between 0 and 1 inclusive, so an answer such as 3 is wrong at once.
• The complement of an event E is written E' and means not E, and P(E) = 1 − P(E') is a valid shortcut.

A fair die is thrown once. Find the probability of getting a prime number, a number greater than 4, and a number divisible by 2 or by 3.The total number of equally likely outcomes is 6, the six faces. Nothing changes between the three parts except the list of favourable faces.

• Total outcomes = 6, the faces 1, 2, 3, 4, 5 and 6.
• (a) Prime numbers on a die are 2, 3 and 5, so 3 favourable outcomes and P = 3/6 = 1/2.
• (b) Numbers greater than 4 are 5 and 6, so 2 favourable outcomes and P = 2/6 = 1/3.
• (c) Numbers divisible by 2 or by 3 are 2, 3, 4 and 6, so 4 favourable outcomes and P = 4/6 = 2/3.
• Note on (c): 6 has been counted once, not twice, because it is divisible by both. The word or means take the union, never add the two counts.

A card is drawn at random from a well-shuffled pack of 52 playing cards. Find the probability of getting a king, a face card and a red ace.A standard pack has 52 cards in 4 suits of 13 each. There are 4 kings, 12 face cards and 2 red aces, so the three answers come straight out of those counts.

• Total outcomes = 52 cards, all equally likely in a well-shuffled pack.
• (a) Kings are the king of each of the 4 suits, so 4 favourable outcomes and P = 4/52 = 1/13.
• (b) Face cards are the jack, queen and king of each suit, so 12 favourable outcomes and P = 12/52 = 3/13.
• (c) Red aces are the ace of hearts and the ace of diamonds, so 2 favourable outcomes and P = 2/52 = 1/26.
• Every part has the same denominator of 52, so the three answers differ only in the numerator.

Three short experiments on numbered objects, each with the same shape of answer. A fair coin is tossed twice. A bag holds 8 slips numbered 1 to 8. A spinner has 8 equal sectors numbered 1 to 8.In every case the total number of equally likely outcomes is written down first, then the favourable ones.

• Coin tossed twice: the outcomes are HH, HT, TH and TT, so 4 equally likely outcomes. Exactly one head is HT or TH, so P = 2/4 = 1/2.
• Bag with 8 slips: total outcomes = 8. Even numbers are 2, 4, 6 and 8, so P = 4/8 = 1/2. Numbers greater than 5 are 6, 7 and 8, so P = 3/8. Prime numbers are 2, 3, 5 and 7, so P = 4/8 = 1/2.
• Spinner with 8 equal sectors: total outcomes = 8. A number divisible by 3 is 3 or 6, so P = 2/8 = 1/4.
• A bag holds 12 bottle caps numbered 1 to 12. A multiple of 4 is 4, 8 or 12, so P = 3/12 = 1/4.
• A fair coin and a fair die are thrown together: list the 12 equally likely pairs, and only H with 6 is the single event asked for, so P = 1/12. Count the 12 pairs; do not multiply two probabilities, because the multiplication rule for independent events is outside this syllabus.

Once the total outcomes are counted, three mistakes account for nearly every lost mark. None of them is a conceptual error; all three are counting or presentation errors.

• Leaving the fraction unsimplified. 4 out of 8 is 1/2, and only the simplest form earns the final mark.
• Counting an outcome twice in an or question, such as adding the multiples of 2 and the multiples of 3 and counting 6 in both lists.
• Losing the line of working. State the total outcomes, list the favourable outcomes, write the fraction, then simplify — four short lines take one minute.

The 2024-25 paper is 80 marks in 3 hours with 38 questions. Section A has 18 MCQs and 2 assertion-reason questions of 1 mark each, Section B has 5 very short answer questions of 2 marks, Section C has 6 short answer questions of 3 marks, Section D has 4 long answer questions of 5 marks, and Section E has 3 case-study questions of 4 marks, with internal choice in two questions each of Sections B, C and D. Calculators are not allowed, so nothing in this chapter needs one.Probability is the natural source of the easy three-mark question, so expect at least one, and usually the definition itself turns up as a one-mark MCQ.

• State the classical definition of probability. One mark, one sentence.
• A fair die is thrown once. Find the probability of getting an even number. Answer 3/6 = 1/2.
• A card is drawn at random from a pack of 52 cards. Find the probability of getting a queen. Answer 4/52 = 1/13.
• A number is chosen at random from 1 to 20. Find the probability that it is a multiple of 5. Answer 4/20 = 1/5.
• Assertion-reason: an impossible event has probability 0 — assert, and give the reason that no outcome is favourable.
• MCQ trap: a certain event has probability 1 and not more than 1.
• MCQ trap: the probability of getting a head and the probability of getting a tail always add to 1, because the two events are complementary.