H2 Math Probability and Conditional Probability

JC2 · H2 Mathematics · Study guide

Solve conditional, independent-event, region-bound, tree, counting and repeated-game probability questions with complete event setups.

Concepts · worked methods · graphs10 min read · practice linked below

Translate the wording into events

“Given B” restricts the sample space to B. “A or B” includes the overlap unless the question says “but not both”. “At least one” often suggests a complement. Write the event before entering numbers: the main difficulty is usually deciding what belongs in the numerator and denominator.

Use the overlap to recover missing probabilities

Two conditional probabilities and “neither”

Events A and B satisfy $P(B\mid A)=1/4$ , $P(A\mid B)=1/2$ , and $P(A^{c}\cap B^{c})=0.3$ . Find $P(A\cap B)$ .

Let $x=P(A\cap B)$ . Turn the conditional statements into expressions for the whole sets:

\begin{aligned}\frac{x}{P(A)}&=\frac14\\P(A)&=4x,\\\frac{x}{P(B)}&=\frac12\\P(B)&=2x.\end{aligned}

The union is the complement of “neither”:

\begin{aligned}P(A\cup B)&=1-0.3=0.7,\\4x+2x-x&=0.7,\\5x&=0.7,\\x&=0.14.\end{aligned}

Check the four disjoint regions: A only is 0.42, both is 0.14, B only is 0.14, and neither is 0.30. They are non-negative and sum to one.

Independence needs a calculation

Mutually exclusive and independent are different tests

Two fair six-sided dice are thrown. Let A be “the sum is odd”, B be “at least one score exceeds 4”, and C be “the scores are equal”. Determine which pair is mutually exclusive and which is independent, then find $P(C\mid B)$ .

Equal scores give an even sum. Thus A and C cannot occur together: $P(A\cap C)=0$ . For independence, use all 36 equally likely ordered outcomes.

\begin{aligned}P(A)&=\frac{18}{36}=\frac12,\\P(B)&=1-\left(\frac46\right)^2\\&=\frac59,\\P(A\cap B)&=\frac{10}{36}=\frac5{18},\\P(A)P(B)&=\frac12\frac59=\frac5{18}.\end{aligned}

Therefore A and B are independent. The ten outcomes in their intersection are $(1,6),(3,6),(5,2),(5,4),(5,6)$ and their reverses. For B and C, only $(5,5),(6,6)$ satisfy both:

\begin{aligned}P(B\cap C)&=\frac2{36}=\frac1{18},\\P(B)P(C)&=\frac59\frac16=\frac5{54}.\end{aligned}

These are unequal, so B and C are not independent. Finally,

\begin{aligned}P(C\mid B)&=\frac{P(C\cap B)}{P(B)}\\&=\frac{1/18}{5/9}\\&=\frac1{10}.\end{aligned}

Positive-probability mutually exclusive events cannot be independent: their intersection is zero, while the product of their probabilities is positive.

Find bounds by making every region non-negative

An unknown overlap

Given $P(A)=0.7$ and $P(C)=0.6$ , find the least and greatest possible values of $P(A\cap C)$ and $P(A\mid C)$ .

Let $x=P(A\cap C)$ . Fill the four disjoint regions before forming inequalities:

\begin{aligned}P(A\cap C^{c})&=0.7-x,\\P(A^{c}\cap C)&=0.6-x,\\P(A^{c}\cap C^{c})&=1-(0.7+0.6-x)\\&=x-0.3.\end{aligned}

Each region must be at least zero:

\begin{aligned}x&\geq0,\\0.7-x&\geq0,\\0.6-x&\geq0,\\x-0.3&\geq0,\\\therefore\quad0.3&\leq x\leq0.6.\end{aligned}

\begin{aligned}P(A\mid C)&=\frac{x}{0.6},\\\frac12&\leq P(A\mid C)\leq1.\end{aligned}

Both endpoints are attainable: at $x=0.3$ , the “neither” region vanishes; at $x=0.6$ , C is contained in A. Independence is not assumed.

Move the overlap while keeping both totals fixed

Overlap x:0.40

x = 0.40: all four regions are non-negative; P(A | C) = 0.667.

Three events: a known subset removes regions

Let $P(A)=0.8$ , $P(B)=0.2$ , $P(C)=0.6$ , $P(A\cap B)=0.2$ and $P(B\cap C)=0.1$ . Since B is contained in A apart from probability-zero outcomes, B contributes 0.1 inside C and 0.1 outside C. Put $t=P(A\cap B^{c}\cap C)$ .

The other non-zero regions are A only $0.6-t$ , C only $0.5-t$ , B outside C 0.1, B inside C 0.1, and outside A and C $t-0.3$ . Hence

\begin{aligned}t&\geq0,\\t&\leq0.6,\\t&\leq0.5,\\t&\geq0.3,\\\therefore\quad0.3&\leq t\leq0.5.\end{aligned}

\begin{aligned}P(A\cap B^{c}\cap C\mid A\cup C)&=\frac{t}{1.3-t}.\end{aligned}

This fraction increases with t: its numerator increases and its positive denominator decreases. The least value is $0.3/1=0.3$ , and the greatest is $0.5/0.8=0.625$ . If A and C are independent, $0.1+t=0.8(0.6)$ , giving $t=0.38$ .

Multiply along paths, add the required paths

Reverse a conditional probability

A factory uses machine A for 60% of items and machine B for 40%. Defect probabilities are 0.02 and 0.05 respectively. An item is defective. Find the probability that B made it.

The first split is the machine, then defect/no defect. The defective event has two disjoint paths:

\begin{aligned}P(D)&=0.6(0.02)+0.4(0.05)\\&=0.012+0.020\\&=0.032,\\P(B\mid D)&=\frac{P(B\cap D)}{P(D)}\\&=\frac{0.4(0.05)}{0.032}\\&=0.625.\end{aligned}

The answer is not 0.05: that is $P(D\mid B)$ , with the conditioning reversed. The more defective machine can account for most defects despite making fewer items.

When the next result depends on the previous result

A player hits on the first throw with probability 0.7. After a hit the next hit probability is 0.8; after a miss the next hit probability is 0.6. Find the probability of a hit on throw 2, then the probability of exactly two hits in three throws given that throw 2 was a miss.

\begin{aligned}P(H_2)&=P(H_1H_2)+P(M_1H_2)\\&=0.7(0.8)+0.3(0.6)\\&=0.74.\end{aligned}

Given a miss on throw 2, two hits in total require the single path HMH. Its last hit follows a miss, so use 0.6:

\begin{aligned}P(H_1M_2H_3)&=0.7(0.2)(0.6)\\&=0.084,\\P(\text{two hits}\mid M_2)&=\frac{0.084}{1-0.74}\\&=\frac{21}{65}\\&\approx0.323.\end{aligned}

Do not reuse the first-throw probability on later branches. These successive throws are not independent.

Conditional counting without replacement

Restrict the denominator to the given selections

A bag contains four red and six blue balls. Five are selected without replacement. Given that at least three are red, find the probability that exactly two are blue.

The condition allows three red or four red; five red is impossible. In the numerator, exactly two blue means three red.

Let q be the required conditional probability.

\begin{aligned}N_{3R,2B}&=\binom43\binom62=60,\\N_{4R,1B}&=\binom44\binom61=6,\\q&=\frac{60}{60+6}\\&=\frac{10}{11}.\end{aligned}

The denominator is 66 eligible selections, not all $\binom{10}{5}$ selections. The common unrestricted denominator cancels in the conditional ratio.

A repeated game can give a geometric series

Alternating turns until the first success

A and B take turns, A first. Each attempt succeeds independently with probability $1/3$ , and the first success wins. Find the probability A wins. Given that B wins, find the probability B wins on B’s second turn.

A wins immediately, or after one complete failed round, or after two, and so on. Each failed round has probability $(2/3)^2=4/9$ :

\begin{aligned}P(A\text{ wins})&=\frac13+\frac49\frac13+\cdots\\&=\frac{1/3}{1-4/9}\\&=\frac35.\end{aligned}

There is no positive probability of endless play, since $(4/9)^n\to0$ . Hence $P(B\text{ wins})=2/5$ . B wins on B’s second turn on the path A fails, B fails, A fails, B succeeds:

\begin{aligned}P(B\text{ wins on turn }2)&=\left(\frac23\right)^3\frac13\\&=\frac8{81},\\P(\text{turn }2\mid B\text{ wins})&=\frac{8/81}{2/5}\\&=\frac{20}{81}.\end{aligned}

A turn number is not a total throw number. B’s second turn is throw 4.

Before you finish

Check that the conditioning event has positive probability, tree branches at each split add to one, and all disjoint regions are non-negative. Test independence with an equality; do not infer it from different event names.