Probability · Lesson 4 of 8
Equally Likely Outcomes
“Derive the counting formula and use elementary outcomes to calculate probabilities accurately.”
• Derive the probability of a single equally likely outcome. • Use P(E)=n(E)/n(S) only when the elementary outcomes are equally likely. • Enumerate ordered coin and dice outcomes without losing multiplicities. • Count favourable card and letter-position outcomes. • Translate exactly, at least, and at most into favourable-outcome counts.
Why equal chances produce a counting fraction
If every elementary outcome has the same weight, the probability model has extra symmetry. We do not have to write a different weight beside every point. Instead, find the common weight from the fact that all the weights together total one.
Suppose the finite sample space has n outcomes and the common probability is p. Adding p once for every point gives np=1. Dividing by n gives p=1/n. Now an event containing m of those points collects m copies of the same weight, so its probability is m/n.
Favourable means satisfying the event, not necessarily desirable to a person. A defective item can be a favourable outcome for the event “the selected item is defective”. The word is about the question being asked.
Before using the fraction, state what an elementary outcome is and why the model gives those outcomes equal weights. A fair die provides six equally likely faces. A well-shuffled standard deck selected uniformly provides 52 equally likely cards. Two equally likely labels do not by themselves guarantee that all sequences of labels are equally likely; for fair coin-toss problems here, the model specifies the equally likely ordered sequences.
Coins: count sequences before totals
For two fair tosses, the four ordered sequences HH, HT, TH, TT are equally likely. The numbers of heads 2,1,1,0 come from those sequences. Thus one head has two ways to occur, while zero or two heads each have one. Collapsing the sample space to the three counts and giving them equal weights loses this multiplicity.
Problem
A fair coin is tossed twice. Find the probability that at least one tail appears.
- 1.Use S={HH,HT,TH,TT}, four equally likely ordered sequences.
- 2.At least one tail occurs in E={HT,TH,TT}; HH is the only excluded sequence.
- 3.The favourable count is 3 and the total count is 4, so P(E)=3/4.
| Three-toss head count | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Sequences | TTT | HTT,THT,TTH | HHT,HTH,THH | HHH |
| Number of sequences | 1 | 3 | 3 | 1 |
| Probability | 1/8 | 3/8 | 3/8 | 1/8 |
The middle counts have more elementary sequences behind them. This explains the pattern 1,3,3,1 without requiring a new probability distribution formula. Add the counts corresponding to a verbal condition: at least two heads means the categories 2 and 3, while at most two heads means 0,1,2.
Problem
Find the probabilities of exactly two heads, at least two heads, at most two heads, and no head in three fair tosses.
- 1.The full sample space has eight equally likely sequences.
- 2.Exactly two heads gives three sequences, so the probability is 3/8.
- 3.At least two heads allows exactly two or exactly three: (3+1)/8=1/2.
- 4.At most two heads allows zero, one, or two: (1+3+3)/8=7/8.
- 5.No head allows only TTT, giving 1/8. It is the same event as three tails, while no tail means HHH.
Dice: preserve which die gives which score
For one fair die, an event can be counted directly among the six faces. For two labelled dice, an outcome records the first and second scores as an ordered pair. A six-by-six grid helps show all 36 possibilities and how different pairs produce the same sum.
Problem
Find the probabilities of a prime, a number at least 3, and a number greater than 6 on one fair die.
- 1.The six elementary outcomes are 1,2,3,4,5,6, each with probability 1/6.
- 2.The primes are 2,3,5, so P(prime)=3/6=1/2. The number 1 is not prime.
- 3.At least 3 gives {3,4,5,6}, so its probability is 4/6=2/3.
- 4.Greater than 6 is the empty event, so its favourable count is zero and its probability is 0/6=0.
The possible totals 2 through 12 are not eleven equally likely outcomes. Probability must account for how many equally likely ordered pairs produce each total.
Problem
A fair coin has faces labelled 1 and 6. In a model with twelve equally likely coin–die pairs, find the chances of totals 3 and 12.
- 1.Record (coin label,die score). S consists of (1,1) through (1,6) and (6,1) through (6,6), so n(S)=12.
- 2.A total of 3 occurs only at (1,2); (6,−3) is not a possible outcome. Therefore P(total 3)=1/12.
- 3.A total of 12 occurs only at (6,6), so P(total 12)=1/12.
- 4.Although the die has six faces, the complete experiment also includes the coin. Using a denominator of 6 would omit that part of the outcome.
Cards and repeated letters
Uniform selection makes individual physical objects the elementary outcomes. We can group those objects by suit, colour, or letter, but repeated labels must retain the weights of all the objects carrying them. Counting distinct labels instead of selectable positions is a common source of errors.
| Feature of a standard deck without jokers | Count |
|---|---|
| Suits: clubs, diamonds, hearts, spades | 4 |
| Cards in each suit | 13 |
| Total cards | 52 |
| Aces, one per suit | 4 |
| Black cards: clubs and spades | 26 |
Problem
A card is drawn uniformly from a shuffled standard deck. Find the chance of the ace of spades, any ace, and a black card.
- 1.Every individual card is an outcome, so n(S)=52.
- 2.The ace of spades is one particular card, giving probability 1/52.
- 3.There are four aces, so P(ace)=4/52=1/13.
- 4.The two black suits contain 13+13=26 cards, so P(black)=26/52=1/2.
Problem
One of the thirteen positions in ASSASSINATION is chosen uniformly. Find the chance that its letter is a vowel or a consonant.
- 1.Read the positions A,S,S,A,S,S,I,N,A,T,I,O,N. The multiplicities are A:3, S:4, I:2, N:2, T:1, O:1.
- 2.The vowel positions consist of three A’s, two I’s, and one O, making 6 positions.
- 3.Thus P(vowel)=6/13. The remaining seven positions are consonants, giving P(consonant)=7/13.
- 4.The six distinct letter names are not equally likely when a position is selected: S occurs four times while O occurs once.
The same method works for a city council with four men and six women. Uniformly selecting one of its ten members gives six favourable outcomes for “woman”, so the probability is 6/10=3/5. The relevant equality is between individual members, not between the two group labels.
After calculating a probability, check its size. A favourable set cannot contain more points than S, so the fraction must be between 0 and 1. The event S itself gives n(S)/n(S)=1; the empty event gives zero. These special cases agree with the axiomatic approach rather than forming a separate theory.
Quiz
Three fair tosses give how many equally likely ordered sequences?
What is the probability of exactly one head in two fair tosses?
Why is the total 7 more likely than 2 on two fair labelled dice?
Selecting a position in ASSASSINATION uniformly gives what probability of a vowel?
For a uniform card draw, what is the probability of any ace?
Which assumption is required for a favourable-count/total-count fraction?
Practice Problems
- Two fair coin tosses are modelled by four equally likely sequences. Find the probability of at least one tail by listing the favourable outcomes.
- For a fair die find the probabilities of: a prime; at least 3; at most 1; greater than 6; and less than 6. Give the event sets as well as the probabilities.
- For a uniformly selected card from 52 cards, state the size of S and find the chances of the ace of spades, any ace, and a black card.
- A fair coin labelled 1 and 6 and a fair die give twelve equally likely pairs. Find the probabilities that the sum is 3 and that it is 12.
- One of four men and six women on a council is selected uniformly. Find the probability of selecting a woman and explain your denominator.
- For three fair tosses find the probabilities of: three heads, exactly two heads, at least two heads, at most two heads, no head, three tails, exactly two tails, no tail, and at most two tails.
- A position in ASSASSINATION is selected uniformly. Find the probabilities of a vowel and a consonant. Explain how repeated letters affect the count.
- On the two-dice grid, count the pairs with totals 6 and 8. Compare their probabilities and explain why total values themselves should not be treated uniformly.
Key Takeaways
• Equal outcome weights and total probability one imply a weight of 1/n per point. • P(E)=n(E)/n(S) is a special case of summing outcome weights. • Count elementary sequences, ordered pairs, individual cards, or letter positions consistently. • Aggregated labels such as head counts and dice totals need not be equally likely. • Translate the event condition first, then count and check that the answer lies in [0,1].