Tree Diagrams Worked ExamplesIGCSE Maths 0580

Work out the second-stage probabilities. In box $A$ there are $3$ reds out of $5$ counters.

In box $B$ there is $1$ red out of $5$.

Red can be reached by two different routes through the tree. Multiply along the first branch.

Multiply along the second branch.

Add the two routes, because either one produces a red counter.

This is a conditional probability. Write the definition: the wanted outcome over everything that satisfies the condition.

The numerator is the single branch through box $A$, found earlier.

The denominator is the total probability of red — both routes, not just this one.

Divide.

Simplify by dividing top and bottom by $4$.

(a) The spins are independent, so multiply along the branch. Both spins red gives $p \times p$.

Take the square root. A probability cannot be negative, so reject $-0.3$.

So the probability of not red on a single spin is the complement.

(b) "Exactly once" has two branches: red then not-red, or not-red then red.

The other order gives the same product.

Add the two branches.

Check all three outcomes sum to $1$: both red, exactly one, neither $(0.7^{2} = 0.49)$.

(a) There are $7$ counters. After each white is removed, both the white count and the total fall by one.

Multiply the numerators and the denominators.

Simplify by dividing top and bottom by $6$.

(b) "Exactly two white" means one black, which can appear in any of three positions. Take them in turn.

Now with the black counter second. The numerators are the same three numbers, just reordered.

And with the black counter first.

Add all three orders and simplify.

"At least twice" means exactly two or exactly three. Handle the two cases separately and add.

For exactly three, multiply along the single branch where every shot scores.

For exactly two, one shot is missed. A single such branch has probability $0.4 \times 0.4 \times 0.6$.

The miss can be the first, second or third shot — three branches, each with the same probability.

Add the two cases.

Check against the other outcomes: exactly one is $3 \times 0.4 \times 0.6^{2} = 0.432$, and none is $0.6^{3} = 0.216$.

Write the total number of counters, then the probability of two reds in a row.

Note the second denominator is $5 + n$, one less than the first — a counter has been removed.

Cross-multiply to clear both fractions.

Expand the brackets.

Bring everything to one side to form a quadratic.

Factorise: two numbers multiplying to $-60$ and adding to $11$ are $15$ and $-4$.

A number of counters cannot be negative, so reject $n = -15$.

Check: the bag holds $6$ red out of $10$, then $5$ red out of $9$.

The spins are independent, so the probabilities do not change between stages.

(a) Multiply along the red–red branch.

(b) Multiply along the not-red–not-red branch.

(c) Two branches give exactly one red. Find one of them.

Double it for the other order, then simplify.

Check the three answers total $1$.

(a) The branches from each point must add up to $1$. First match:

Second match — each pair of branches has its own total of $1$, because the probabilities change with the first result.

(b) Multiply along the Win–Win path.

(c) "Exactly one" has two paths: Win then Not win, and Not win then Win. Multiply along each path.

Add the two paths.

Check: the remaining path (Not win, Not win) is $0.4 \times 0.55 = 0.22$, and $0.42 + 0.36 + 0.22 = 1$. ✓

(a) To pass on the second attempt she must fail the first and then pass. Multiply along that path.

(b) Adding the three "pass" paths works, but it is quicker to use the complement: she fails to pass only by failing all three times.

Subtract from $1$.

Check by adding the paths: $0.6 + 0.3 + 0.4 \times 0.25 \times 0.75 = 0.6 + 0.3 + 0.075 = 0.975$. ✓

(c) A learner needs a third attempt if they fail the first two — whatever then happens on the third.

(d) "Given that she passes" restricts attention to the three pass paths, which total $0.975$. The first-attempt pass is $0.6$ of that.

(a) Multiply along each of Jin's four paths, using a common denominator of $648$.

The outcomes of a complete tree must total exactly $1$, so Jin's tree cannot be right.

(b) Mistake 1 — after an orange sweet, Jin still uses ninths. That treats the draw as with replacement. Once an orange is taken, $8$ sweets remain: $4$ orange and $4$ lime.

Mistake 2 — after a lime sweet, Jin has reduced the total to $8$ but not the number of limes, so his branches add up to $\frac{9}{8}$. $5$ orange and $3$ lime remain.

(c) Same flavour means Orange–Orange or Lime–Lime, using the corrected tree.

Check the corrected tree: the other two paths are $\frac{20}{72}$ each, and $\frac{20 + 20 + 20 + 12}{72} = 1$. ✓

(a) After a blue pen is taken, $9$ pens are left: $3$ blue and $6$ black. The pair must sum to $1$.

After a black pen is taken, $9$ pens are left: $4$ blue and $5$ black. The number of blue pens has not changed.

(b) Exactly one blue happens on two paths: blue then black, or black then blue. Multiply along each path.

Add the two paths and simplify.

(a) Add the Failed column and divide by the total number of cars.

(b) The cars are chosen from the electric cars only, so the sample space is $14 + 4 = 18$ cars, not $75$. The two cars are different cars, so this is without replacement.

Exactly one failed: passed then failed, or failed then passed. The second denominator is $17$.

Simplify, then round to $3$ significant figures.