Venn Diagrams and Tables Worked ExamplesIGCSE Maths 0580

(a) Fill in what you are given, then subtract to find each remaining cell. Boys without glasses:

Girls without glasses:

Now the column totals. Glasses altogether, and no glasses altogether:

The completed table is shown below. Both the rows and the columns must total $120$.

(b) Read the glasses total over the grand total, then simplify.

(c) The student is known to wear glasses, so the denominator is the glasses total, not $120$.

Two events are independent exactly when $P(A \cap B) = P(A) \times P(B)$. Compute each piece and compare.

The left-hand side is the single cell for boys with glasses.

Now the two separate probabilities.

Multiply them.

The two values differ, so the condition fails and the events are not independent.

You can see it directly too: $P(\text{glasses} \mid \text{boy}) = \frac{45}{70} = 0.643$, well above the overall $P(\text{glasses}) = 0.542$. Knowing the student is a boy changes the probability.

(a) Add all eight regions, including the $17$ outside the circles.

$n(M)$ is every region inside the Music circle — four of them, not just "$M$ only".

(b) The condition is "in Music", so the denominator becomes $n(M) = 33$.

$n(A \cap M)$ is both the "$M \cap A$ only" region and the centre — anyone in both $A$ and $M$.

Divide and simplify.

(c) "At least two" is the three pairwise regions plus the centre.

"All three" is the centre alone. Divide by the conditioning total.

(a) Add each row across.

Now add each column down.

Check both ways: the row totals and the column totals must each give $200$.

(b) Car total over the grand total, then simplify.

(c) The condition restricts us to the $50$ walkers, of whom $40$ are under $30$.

(a) If $12$ study neither, the rest are inside at least one circle.

Now apply the inclusion–exclusion formula.

Substitute and solve for the overlap.

(b) Find each "only" region by removing the overlap.

Add them for "exactly one", then check all four regions total $60$.

Write the probability.

(c) Given the student studies French, the denominator is $n(F) = 30$.

(a) Fill the two given cells, then subtract within each row.

Total the columns.

The completed table, checking that everything totals $1000$.

(b) The condition given is "has the condition", so the denominator is that row total, $80$.

(c) Now the condition given is "tests positive", so the denominator is the column total, $118$.

Simplify by dividing by $2$, then evaluate.

(d) The same cell $(72)$ sits over two different denominators, giving very different answers. The test catches $90\%$ of true cases, but because so many more people are healthy, $46$ of the $118$ positives are false alarms.

(a) The Year 10 row must total $40$. Subtract the two known entries.

(b) Add each column down.

And the orange column, using the entry just found.

Check the column totals add to $80$.

(c) Banana total over the grand total, then simplify.

(a) Each empty cell is the row number plus the column number: $X = 2, Y = 4$; $X = 3, Y = 2$; $X = 3, Y = 5$; $X = 4, Y = 1$; $X = 4, Y = 3$.

The diagram has $4 \times 5 = 20$ cells. Both spinners are fair, so all $20$ outcomes are equally likely — that is what lets you count cells.

(b) Count the cells showing $6$: $(1, 5)$, $(2, 4)$, $(3, 3)$, $(4, 2)$ — one of them is a cell you filled in.

(c) "Greater than $7$" means $8$ or $9$: the cells $(3, 5)$, $(4, 4)$ and $(4, 5)$. The $7$s are not included.

(d) Expected frequency = probability $\times$ number of trials.

(a) The diagram shows probabilities, not numbers of members, so the four regions add up to $1$.

Collect terms and solve.

(b) "Exactly one" is $W$ only plus $S$ only; the overlap is excluded. $S$ only is $2x = 0.34$.

(c) This is conditional: the member is known to use the pool, so work inside circle $S$ only.

Divide the overlap by $P(S)$.

(d) Weights but not swimming is the region $W$ only, with probability $0.35$.

(a) Count the multiples of $4$ systematically. Every cell in the row or column for a $4$ qualifies: $6 + 6 - 1 = 11$ cells (the cell $4 \times 4$ is shared).

Outside that row and column, a product is a multiple of $4$ only when both dice are even: $2 \times 2 = 4$, $2 \times 6 = 12$, $6 \times 2 = 12$, $6 \times 6 = 36$. That is $4$ more cells.

(b) The condition shrinks the sample space to those $15$ cells. Both dice even means both scores are from $\{2, 4, 6\}$: $3 \times 3 = 9$ cells, and each of these products is a multiple of $4$, so all $9$ are among the $15$.

(c) The product is odd only when both scores are odd: $3 \times 3 = 9$ cells.

Expected wins in $72$ games for each player, then the difference.