Use arrays and tree diagrams to determine the outcomes and probabilities for multi-stage experiments, including two-way tables, multiplying probabilities along the branches of a tree diagram and adding the probabilities of mutually exclusive outcomes, with and without replacement

Set up the without-replacement branches: first draw $P(W) = \tfrac{4}{10}$, $P(B) = \tfrac{6}{10}$; second-draw denominators drop to $9$.

"Different colours" is white then black OR black then white: $P(WB) = \tfrac{4}{10} \times \tfrac{6}{9} = \tfrac{24}{90}$ and $P(BW) = \tfrac{6}{10} \times \tfrac{4}{9} = \tfrac{24}{90}$.

Add the two mutually exclusive paths: $\tfrac{24}{90} + \tfrac{24}{90} = \tfrac{48}{90} = \tfrac{8}{15}$.

Markers award one mark for a correct tree with changed second-draw denominators, one for the two path products, and one for adding the mutually exclusive paths to reach $\tfrac{8}{15}$. A common error is keeping the denominator at $10$ (treating it as with replacement).

Set up the without-replacement branches. First draw: $P(G) = \tfrac58$, $P(Y) = \tfrac38$. Once one lolly is removed, $7$ remain, so the second-draw denominators drop to $7$.

Part (a): both green is the single path $GG$. Multiply along it: $P(GG) = \tfrac58 \times \tfrac47 = \tfrac{20}{56} = \tfrac{5}{14}$.

Part (b): "different colours" is green then yellow OR yellow then green, two mutually exclusive paths. $P(GY) = \tfrac58 \times \tfrac37 = \tfrac{15}{56}$ and $P(YG) = \tfrac38 \times \tfrac57 = \tfrac{15}{56}$. Add them: $\tfrac{15}{56} + \tfrac{15}{56} = \tfrac{30}{56} = \tfrac{15}{28}$.

Markers award a mark for a correct tree with the second-draw denominators reduced to $7$, a mark for $P(GG) = \tfrac{5}{14}$, a mark for the two product paths in part (b), and a mark for adding the mutually exclusive paths to reach $\tfrac{15}{28}$. The classic error is leaving the second denominator at $8$ (treating it as with replacement).

Part (a): coffee is a single category, so use the coffee row marginal total of $160$ out of $250$: $P(\text{coffee}) = \tfrac{160}{250} = \tfrac{16}{25} = 0.64$.

Part (b): "tea and large" is the single inner cell where the tea row meets the large column, $36$ out of $250$: $P(\text{tea and large}) = \tfrac{36}{250} = \tfrac{18}{125} = 0.144$.

Part (c): regular is a single category, so use the regular column marginal total of $144$ out of $250$: $P(\text{regular}) = \tfrac{144}{250} = \tfrac{72}{125} = 0.576$.

Markers award a mark for each correct probability: parts (a) and (c) need a marginal total over the grand total $250$, and part (b) needs the single overlap cell, not a row or column total. The denominator is the grand total $250$ throughout.

Recognise this as a with-replacement (independent) two-stage event: each morning $P(R) = \tfrac25$, $P(G) = \tfrac35$.

Use the complement: "at least one red" is the opposite of "no red", i.e. green on both mornings. $P(GG) = \tfrac35 \times \tfrac35 = \tfrac{9}{25}$.

So $P(\text{at least one red}) = 1 - \tfrac{9}{25} = \tfrac{16}{25} = 0.64$.

Markers reward identifying the complement (or adding $P(RR) + P(RG) + P(GR) = \tfrac{4 + 6 + 6}{25}$), the correct branch products, and the final $\tfrac{16}{25}$. The efficient solution is the complement; the long way must list all three mutually exclusive paths.

Part (a) - list every outcome of the two-stage experiment. Each toss is a head ($H$) or a tail ($T$), and the first toss pairs with each second toss:

so there are $4$ equally likely outcomes.

Part (b) - find $P(HH)$. Multiply along the two branches that give a head each time. Each toss has $P(H) = \tfrac12$, so

so the probability of two heads is $\tfrac14$. (Check: the four outcomes are equally likely, and only one of them is $HH$, so $\tfrac14$ agrees.)

Recognise this is with replacement. The first counter goes back, so the bag is unchanged for the second draw and the two draws are independent. Each draw has $P(\text{red}) = \tfrac35$.

Multiply along the two red branches.

so the probability that both are red is $\tfrac{9}{25}$. (Check: $\tfrac{9}{25} = 0.36$, a sensible chance since red is the more common colour.)

Read the totals from the margins of the table. The "Total" row and column already count the categories for you.

Part (a) - $P(\text{wears glasses})$. The glasses column totals $48$ out of $200$:

Part (b) - $P(\text{female})$. The female row totals $120$ out of $200$:

so $P(\text{wears glasses}) = 0.24$ and $P(\text{female}) = 0.6$. (Check: the table total is $200$, which matches the denominator used each time.)

Recognise this is without replacement. The first counter is kept, so the bag has only $4$ counters for the second draw and the second probability changes.

First draw. There are $3$ red out of $5$:

Second draw, after a red is removed. One red has gone, leaving $2$ red out of the remaining $4$ counters, so the second red branch is:

Multiply along the branch.

so the probability that both are red is $\tfrac{3}{10}$. (Check: $\tfrac{3}{10} = 0.3$, lower than the with-replacement answer of $0.36$, which makes sense - removing a red leaves fewer reds for the second draw.)

Set up the four paths. Drawing without replacement, the second denominator drops to $4$. After the first counter is removed, $4$ counters remain, split as whatever the first draw left behind. The four outcomes and their probabilities are

For $P(RR)$ the second factor $\tfrac24$ is the $2$ red left among $4$; for $P(RB)$ a red has gone, so the bag is $2$ red and $2$ blue and the blue branch is $\tfrac24$ (the $2$ blue among $4$). They give the same number here, but for different reasons. These four add to $\tfrac{20}{20} = 1$, a good check.

Part (a) - one of each colour. "One of each" means red then blue OR blue then red, two mutually exclusive paths, so add them:

Part (b) - at least one red. The complement of "at least one red" is "no red", which is $BB$:

so $P(\text{one of each}) = \tfrac35$ and $P(\text{at least one red}) = \tfrac{9}{10}$. (Check via the long way: $P(RR) + P(RB) + P(BR) = \tfrac{6+6+6}{20} = \tfrac{18}{20}$, which agrees.)

Set up the three-stage experiment. Each spin is green ($G$) with probability $\tfrac34$ or red ($R$) with probability $\tfrac14$. A three-stage tree has $2 \times 2 \times 2 = 8$ paths, and you multiply the three branch probabilities along each path.

Part (a) - three greens. Multiply three green branches:

Part (b) - exactly two greens. Exactly two greens means one red, and the red can fall on spin $1$, $2$ or $3$: the paths $RGG$, $GRG$, $GGR$. Each has the same probability:

There are $3$ such mutually exclusive paths, so add them:

Part (c) - at least one red. The complement is "no red", which is $GGG$ from part (a):

so $P(GGG) = \tfrac{27}{64}$, $P(\text{exactly two greens}) = \tfrac{27}{64}$ and $P(\text{at least one red}) = \tfrac{37}{64}$. (Check: $\tfrac{37}{64} \approx 0.58$, a believable chance of seeing red at least once in three spins.)

Part (a) - female AND wears glasses. Read the single cell where the female row meets the glasses column: $30$ students, out of $200$:

Part (b) - male. This is a single category, so use the marginal total from the bottom of the male row: $80$ out of $200$:

Part (c) - does not wear glasses, via the complement. The glasses column totals $48$, so $P(\text{wears glasses}) = \tfrac{48}{200} = \tfrac{6}{25}$. "Does not wear glasses" is the complement:

so the answers are $0.15$, $0.4$ and $0.76$. (Check on (c): the no-glasses column totals $152$, and $\tfrac{152}{200} = \tfrac{19}{25}$, which agrees.)