Counting tools in probability

What you will be able to do

• count favourable and total outcomes with $^nP_r$ and $^nC_r$;
• solve questions on cards, committees, arrangements and seating;
• keep the favourable count and the total count in the same style, both ordered or both unordered.

From counting to chance

In the chapter on permutations and combinations you counted ways. Here we turn ways into chances. When all outcomes are equally likely,

and each number of ways is counted with the multiplication principle, $^nP_r$ or $^nC_r$. Remember: $^nP_r = \dfrac{n!}{(n - r)!}$ counts arrangements, and $^nC_r = \dfrac{n!}{r!(n - r)!}$ counts selections. A-Level and IB texts write $\binom{n}{r}$ for $^nC_r$ and read it as "$n$ choose $r$".

The golden rule. Count the favourable and the total in the same style. If the total counts hands without order, the favourable count must also be without order. If the total counts arrangements, count the favourable arrangements.

A hand of five cards

A hand of $5$ cards is dealt from a pack of $52$. The order of the cards in a hand does not matter, so the number of hands is $^{52}C_5 = 2\,598\,960$.

What is the probability that the hand has exactly two aces? Choose the two aces from the $4$ aces in $^4C_2 = 6$ ways. Choose the other three cards from the $48$ non-aces in $^{48}C_3 = 17\,296$ ways. So

Choosing a team or a committee

A selection of $r$ people from a group of $N$, with $K$ of them of one kind, has exactly $k$ of that kind in

of the cases. This is the model of every committee, sample from a batch and hand of cards.

A committee of $4$ is chosen from $5$ men and $4$ women. The number of committees is $^9C_4 = 126$. Committees with two men and two women: $^5C_2 \times {}^4C_2 = 10 \times 6 = 60$. So the probability is $\dfrac{60}{126} = \dfrac{10}{21}$.

Arrangements

The letters of a word are arranged at random. Take the word EQUATION. It has $8$ different letters, $5$ vowels (E, U, A, I, O) and $3$ consonants. The number of arrangements is $8! = 40\,320$.

What is the chance that all the vowels are together? Glue the vowels into one block. The block and the three consonants make $4$ objects, which can be arranged in $4!$ ways. The vowels can be arranged inside the block in $5!$ ways. So

Seating

Six friends sit in a row. What is the chance that two particular friends sit next to each other? Glue the two into one block. The block and four other friends make $5$ objects, arranged in $5!$ ways. The two inside can swap in $2!$ ways. So $P = \dfrac{5! \times 2!}{6!} = \dfrac{1}{3}$.

At a round table, rotating everybody does not give a new seating. Fix one friend, and arrange the other $n - 1$ in $(n - 1)!$ ways. For six friends, two particular friends sit next to each other with probability $\dfrac{4! \times 2!}{5!} = \dfrac{2}{5}$. In general, the other friend takes one of the $n - 1$ seats, and $2$ of them are next to you, so the chance is $\dfrac{2}{n - 1}$.

With and without putting back

A bag holds $6$ red and $4$ blue balls. Three balls are drawn.

• Together, or one by one without putting back. Exactly two red: $\dfrac{^6C_2 \times {}^4C_1}{^{10}C_3} = \dfrac{60}{120} = \dfrac{1}{2}$.
• One by one with putting back. Choose which draw is blue in $3$ ways, then count $6 \times 6 \times 4$ outcomes out of $10^3$. So $P = \dfrac{3 \times 144}{1000} = 0.432$.

The two answers differ, because putting a ball back changes the bag. Read the question for the words "with replacement".

A worked example

Five letters are written on cards: A, B, C, D and E. Three cards are drawn together. Find the probability that the cards A and B are both drawn.

1. All outcomes. $^5C_3 = 10$ selections.
2. Favourable. A and B are in, so only one more card is chosen from the other $3$: $^3C_1 = 3$.
3. The probability. $\dfrac{3}{10}$.
4. Check by ordering. Count ordered draws instead. All: $5 \times 4 \times 3 = 60$. Favourable: choose the places of A and B in $3 \times 2 = 6$ ways, and the third card in $3$ ways, so $18$. $\dfrac{18}{60} = \dfrac{3}{10}$. The answers agree.

Practise with new numbers

Try it in Python

The program counts the hands of five cards with math.comb and finds the chance of exactly two aces. It then does the committee of four from five men and four women.

Common mistakes

• Counting the favourable outcomes in order and the total without order. Both must be in the same style.
• Forgetting the arrangement factor. When a block of friends or letters sits together, multiply by the arrangements inside the block.
• Using a table formula for a row. A row of $n$ friends has $n!$ arrangements. A round table has $(n - 1)!$.
• Mixing replacement. A draw with replacement is not a combination.

Remember these

• $P(E) = \dfrac{\text{favourable ways}}{\text{all ways}}$, both counted in the same style.
• Committees and hands: $\dfrac{^KC_k \times {}^{N-K}C_{r-k}}{^NC_r}$.
• Glue objects that must be together into a block, then multiply by the arrangements inside it.
• A row of $n$ has $n!$ arrangements. A round table has $(n - 1)!$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A hand of $5$ cards is dealt. What is the probability that all five are hearts? Leave the answer as a fraction of combinations.

2. A committee of $3$ is chosen from $6$ men and $4$ women. Find the probability that it has no woman.

3. The letters of the word PLANET are arranged in a row. Find the probability that the vowels are together.

4. Five friends sit at a round table. What is the chance that two particular friends sit side by side?

5. Why do the answers to "three balls together" and "three balls with replacement" differ?

References

• NCERT, Mathematics for Class XI (earlier edition), Permutations and Combinations and Probability, and the chapter of this book on permutations and combinations.
• OpenStax, Precalculus 2e, Chapter 11, sections on counting principles and on probability.