EN6 - Counting and probability problems

EN6 - Counting and probability problems

A counting problem becomes manageable when you decide what counts as one outcome, whether order matters and which cases satisfy the restriction. You will combine combinations, ordered selections and binomial probabilities in problems where the method is yours to choose. Sometimes the shortest route counts what you do not want first.

From counts to probabilities

Roll five ordinary fair dice. Would you expect no sixes or exactly one six to be more likely? Surprisingly, their probabilities are equal. We will find out why, then use that fact to calculate the chance of at least two sixes.

First, build the habit that makes these problems reliable: define one outcome before counting it.

You already know

Use combinations for unordered selections and permutations or the product rule for specified roles. In a probability tree, multiply along a path and add mutually exclusive paths. A binomial model counts successes in fixed independent trials with unchanged success probability.

Suppose one team is chosen at random from all possible teams of a fixed size. Each team is equally likely, so counting favourable teams and dividing by all teams works. If a captain and deputy are chosen instead, switching their names creates a different outcome. Both the top and bottom of the fraction must use that same meaning of an outcome.

Probability from equally likely outcomes

P(A)=number of outcomes satisfying Atotal number of outcomes\mathrm{P}(A)=\frac{\text{number of outcomes satisfying }A}{\text{total number of outcomes}}

The possible outcomes counted in the denominator must be equally likely. Count numerator and denominator using the same ordered or unordered sample space.
Not on the formulae sheet: learn it.

A count and a probability answer different questions. A count can be 2020; a probability must be between 00 and 11. Read “in how many ways” and “find the probability” carefully before deciding where to stop.

Worked example 1: an unordered pair

Question. A club has five Year 10 members and four Year 11 members. Two members are chosen at random to form a team. Find the probability that the team has one member from each year.

Plan. Count mixed-year pairs, then divide by all unordered pairs of club members.

N(all pairs)=9C2=36choose two of nineN(mixed pairs)=5C1×4C1=5×4=20one from each yearP(mixed pair)=2036=59favourable divided by total\begin{aligned} N(\text{all pairs})&={}^9C_2=36 && \text{choose two of nine}\\ N(\text{mixed pairs})&={}^5C_1\times{}^4C_1=5\times4=20 && \text{one from each year}\\ \mathrm{P}(\text{mixed pair})&=\frac{20}{36}=\frac59 && \text{favourable divided by total} \end{aligned}

Why not multiply the numerator by two? Choosing a Year 10 member and a Year 11 member already specifies the pair once. Their order of selection is not part of the outcome.

Answer. 59\frac59.

Check. The tree method includes two orders: 59×48+49×58=4072=59\frac59\times\frac48+\frac49\times\frac58=\frac{40}{72}=\frac59.

The next example changes one feature: the two people now have specified roles. The desired outcome requires the Year 10 member to have one particular role.

Worked example 2: specified roles

Question. From the same club, a captain and then a deputy are chosen at random, with no person holding both roles. Find the probability that the captain is in Year 10 and the deputy is in Year 11.

Plan. Count ordered assignments, since captain and deputy are distinct roles.

N(all assignments)=9P2=9×8=72two different rolesN(required assignments)=5×4=20Year 10 then Year 11P(required roles)=2072=518divide within the same sample space\begin{aligned} N(\text{all assignments})&={}^9P_2=9\times8=72 && \text{two different roles}\\ N(\text{required assignments})&=5\times4=20 && \text{Year 10 then Year 11}\\ \mathrm{P}(\text{required roles})&=\frac{20}{72}=\frac5{18} && \text{divide within the same sample space} \end{aligned}

Why is the answer half the previous probability? Each mixed pair has two possible role assignments, but only one puts the Year 10 member in charge.

Answer. 518\frac5{18}.

Check. The tree path gives 59×48=518\frac59\times\frac48=\frac5{18}. This selection is without replacement, so the second denominator is 88.

An ordered sample space is fine even for an unordered event, provided you count both orders in its numerator.

Watch out

Wrong for the mixed-team question: 2072\frac{20}{72}. This uses unordered favourable pairs over ordered outcomes. Right: use 2036\frac{20}{36}, or count both orders and use 4072\frac{40}{72}.

In the next problem, the first and second draws have different requirements. Keep that order when you count.

Restricted selections and probability

A restriction such as “at least two” usually allows several different group compositions; use your combination counts to find their total probability in one consistent sample space.

Inside one composition, multiply choices: choosing experienced members and choosing new members must both happen. Between different compositions, add counts: a team with two experienced members and a team with three experienced members are different cases. This multiply-within, add-between strategy is not printed as a formula on the formulae sheet; learn when each operation applies. The combination formula itself is given there.

Worked example 3: at least two from a group

Question. A team of three is chosen at random from five experienced members and four new members. Find the probability that it contains at least two experienced members.

Plan. Count the two allowed compositions separately, add their counts and divide by all three-person teams.

N(2 experienced)=5C2×4C1=10×4=40choose both groupsN(3 experienced)=5C3=10no new membersN(at least 2)=40+10=50add distinct cases\begin{aligned} N(2\text{ experienced})&={}^5C_2\times{}^4C_1=10\times4=40 && \text{choose both groups}\\ N(3\text{ experienced})&={}^5C_3=10 && \text{no new members}\\ N(\text{at least }2)&=40+10=50 && \text{add distinct cases} \end{aligned} P(at least 2)=509C3=5084=2542all teams are equally likely\mathrm{P}(\text{at least }2)=\frac{50}{{}^9C_3}=\frac{50}{84}=\frac{25}{42} \qquad \text{all teams are equally likely}

Why add rather than multiply the two rows? One team cannot have both exactly two and exactly three experienced members. The rows are alternatives, not simultaneous choices.

Answer. 2542\frac{25}{42}.

Check. The excluded cases give 4C3+5C14C2=4+30=34{}^4C_3+{}^5C_1{}^4C_2=4+30=34, and 5084+3484=1\frac{50}{84}+\frac{34}{84}=1.

The next example changes only “at least two” to “at most two”. Exactly two belongs to both descriptions, so these two questions are not complements.

Worked example 4: at most two from a group

Question. A team of three is chosen at random from the same group. Find the probability that it contains at most two experienced members.

Plan. Include zero, one and two experienced members, then divide their total count by the same 8484 possible teams.

N(0 experienced)=4C3=4three new membersN(1 experienced)=5C14C2=5×6=30one experienced, two newN(2 experienced)=5C24C1=10×4=40two experienced, one newN(at most 2)=4+30+40=74add the allowed cases\begin{aligned} N(0\text{ experienced})&={}^4C_3=4 && \text{three new members}\\ N(1\text{ experienced})&={}^5C_1{}^4C_2=5\times6=30 && \text{one experienced, two new}\\ N(2\text{ experienced})&={}^5C_2{}^4C_1=10\times4=40 && \text{two experienced, one new}\\ N(\text{at most }2)&=4+30+40=74 && \text{add the allowed cases} \end{aligned} P(at most 2)=7484=3742use the same total\mathrm{P}(\text{at most }2)=\frac{74}{84}=\frac{37}{42} \qquad \text{use the same total}

Why include zero? “At most two” places an upper limit; it does not require an experienced member at all.

Answer. 3742\frac{37}{42}.

Check. Only the 5C3=10{}^5C_3=10 all-experienced teams are excluded, and 1−1084=37421-\frac{10}{84}=\frac{37}{42}.

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