Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers finite equally likely outcomes, complements and small organised sample spaces. The coin-spinner examples state independence explicitly. Biased experiments and all textbook exercise contexts are not covered.
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LESSON 1
Count equally likely outcomes
Use favourable outcomes divided by all equally likely outcomes.
A sample space lists the outcomes of a random experiment.
Define the experiment
Roll a fair six-faced die once. Outcomes are 1, 2, 3, 4, 5, 6 and are equally likely.
Define the event
The event “greater than 4” contains outcomes 5 and 6.
Count both sets
There are 2 favourable outcomes among 6 equally likely outcomes, so P = 2/6 = 1/3.
Check the assumption
If the die is biased, simply counting faces is not enough to establish this probability. Equal likelihood is essential.
Worked example: A fair die event “multiple of 3” is {3, 6}. Its probability is 2/6 = 1/3. The event “at most 3” has three outcomes and probability 3/6 = 1/2.
Watch for this: Two named events need not be equally likely. “Roll a 6” and “do not roll a 6” are not a 1/2–1/2 split for a fair die.
Discuss: List an event on a fair die with probability 1/2, then create a different event with the same probability.
LESSON 2
Use complements and bounds
Find “not” probabilities and distinguish impossible from certain.
Every outcome either belongs to an event E or to its complement, but not both.
Partition all outcomes
A bag contains 3 blue and 5 red otherwise identical balls. A uniform random selection picks one ball.
Find the event probability
P(blue) = 3/8. The complementary event is not blue, which here means red.
Subtract from one
P(not blue) = 1 − 3/8 = 5/8. The two probabilities sum to 1.
Check endpoints
An impossible event has probability 0; a certain event has probability 1. A negative probability or one greater than 1 signals an error.
Worked example: For a fair die, P(not getting 1) = 1 − 1/6 = 5/6. P(getting a number above 6) = 0, while P(getting a number at most 6) = 1.
Watch for this: “Not blue” may include several colours if the bag contains more than two colours. The complement is everything outside the event.
Discuss: Make a three-colour bag model. Explain why not red includes both of the other colours, then check complementary probabilities add to 1.
LESSON 3
Organise a sample space without losing outcomes
Use a table or tree for a small combined experiment.
Ordered outcomes distinguish which result belongs to which action.
Define independent fair actions
Toss a fair coin and spin an independent fair three-sector spinner numbered 1, 2, 3.
List the six ordered outcomes
They are H1, H2, H3, T1, T2, T3. Independence and fairness make these six outcomes equally likely.
Select the event precisely
For “heads and an even number”, only H2 qualifies. Its probability is 1/6.
Compare a different event
For “heads or an even number”, H1, H2, H3 and T2 qualify, giving 4/6 = 2/3. H2 must be counted only once.
Worked example: With two independent fair coins, the ordered outcomes HH, HT, TH, TT are equally likely. Exactly one head occurs in HT and TH, so its probability is 2/4 = 1/2, not 1/3.
Watch for this: “And” requires both conditions; “or” includes outcomes satisfying either condition. Overlap is counted once.
Discuss: Make a 2-by-4 table for a coin and a four-sector spinner. Shade one “and” event and one “or” event, explaining any overlap.
Original learning companion, not full textbook exercise solutions. Covers finite equally likely outcomes, complements and small organised sample spaces. The coin-spinner examples state independence explicitly. Biased experiments and all textbook exercise contexts are not covered.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
