Probability Applications Worked ExamplesIGCSE Maths 0580

(a) All the probabilities must add to $1$. Start by finding what is left over after $A$ and $B$.

Let $P(D) = d$. Then $P(C) = 2d$, so the two together make $3d$.

Solve for $d$.

So $P(D) = 0.15$ and $P(C)$ is double that.

Check the four probabilities total $1$.

(b) The expected number is the probability multiplied by the number of trials.

(a) Write the probability as favourable over total. The total is $5 + x$.

Cross-multiply.

Solve. The bag holds $5$ red and $15$ blue, $20$ in total.

(b) Let $y$ be the number of red counters added. Both the red count and the total increase by $y$.

Cross-multiply and expand.

Collect the $y$ terms.

Check: the bag would hold $15$ red and $15$ blue out of $30$.

(a) The relative frequency is successes divided by trials. This is the best estimate available for a biased die, since the theoretical value is unknown.

Simplify or convert to a decimal.

(b) Multiply the estimated probability by the new number of throws.

Evaluate.

(c) The probability itself came from a limited experiment, and results vary from one set of throws to the next.

"At least one" covers one, two or three failures — three separate cases. It is far quicker to find the opposite event: none fail.

The probability a single component does not fail is the complement.

The components are independent, so multiply for all three surviving.

Evaluate.

Subtract from $1$.

(a) Use the complement again: the opposite of "at least one green" is "both yellow".

There are $12$ counters. After one yellow is removed, $4$ yellows remain out of $11$ — the denominator drops too.

Simplify by dividing top and bottom by $4$.

Subtract from $1$.

(b) "Exactly one green" happens two ways: green then yellow, or yellow then green. Find the first.

The other order gives the same value — the numerators just swap places.

Add the two paths and simplify.

Check all three outcomes total $1$. $P(GG) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}$.

(a) Expected value is each prize multiplied by its probability, all added together.

Work out each product.

(b) Compare the expected winnings with the cost of playing.

The expected profit per game is winnings minus cost.

Multiply by the number of games for the total expected loss.

(c) A fair game is one where expected winnings equal the entry fee, so the expected profit is zero.

First find the total number of counters.

(a) Favourable outcomes over total outcomes.

(b) Use the complement: subtract the probability of green from $1$.

(c) Red and blue cannot both happen on one draw, so add the probabilities.

(a) Read the graph at $50$ throws. The small squares are $0.02$ apart vertically, so the cross is at $0.62$.

Relative frequency $= \dfrac{\text{number of successes}}{\text{number of trials}}$, so multiply back by the number of throws.

(b) Use the relative frequency after the most throws: $0.6$ at $100$ throws. More trials give a more reliable estimate — the graph also shows the values settling around $0.6$.

(c) The question asks about point down, so use the complement.

Expected number = probability $\times$ number of trials.

(a) For a fair spinner, each probability is the sector angle as a fraction of $360^{\circ}$.

(b) First find the other two probabilities, and check that all three sum to $1$.

"Different colours" has six orders, so work with the complement: the spins land on the same colour in just three ways.

Subtract from $1$.

(c) If the spinner is fair, the expected number of reds in $240$ spins is $\frac{1}{3} \times 240$.

Leila's $96$ reds give a relative frequency of $\frac{96}{240} = 0.4$, compared with $\frac{1}{3} \approx 0.333$ expected. With $240$ spins, $16$ more reds than expected suggests the spinner may be biased towards red — though it does not prove it.

(a) Read the three bars ($40$, $75$, $60$) and subtract their total from $250$.

(b) Turn each frequency into a relative frequency by dividing by $250$. These are the estimated probabilities.

A total of $5$ comes from $(1, 4)$, $(4, 1)$, $(2, 3)$ or $(3, 2)$. The spins are independent, so multiply along each pair and add — each unordered pair appears twice.

(c) In $500$ games the expected number of wins is $0.24 \times 500$.

Income from $500$ games at $50$ cents is $250. Paying $2 to each of $120$ winners costs $240.

Check per game: the expected payout is $0.24 \times 2 = 0.48$ dollars, so the fair expects to keep $2$ cents a game, and $2 \times 500 = 1000$ cents. ✓

(a) Each cell is the spinner number times the card number. The gaps are $1 \times 3 = 3$, $1 \times 6 = 6$, $3 \times 5 = 15$, $4 \times 2 = 8$ and $4 \times 5 = 20$.

(b) The spinner and the card are both fair, so all $3 \times 4 = 12$ cells are equally likely. The multiples of $6$ are $6$, $6$, $18$, $12$ and $24$ — five cells (the two $6$s are different outcomes, so count both).

(c) We are told the score is greater than $10$, so only those cells can have happened. They are $15$ and $18$ in the spinner-$3$ row, and $12$, $20$ and $24$ in the spinner-$4$ row.

Of these $5$ cells, the ones in the card-$5$ column are $15$ and $20$.

(a) The two picks are independent, so the probability that both are lemon is the product of the two probabilities. Form an equation.

Divide both sides by $0.35$.

(b) "Not lemon" is the complement for each bag. Find both complements first.

Neither is lemon means not lemon from $X$ and not lemon from $Y$ — multiply.

(c) Exactly one lemon can happen in two ways: lemon from $X$ only, or lemon from $Y$ only. Add the two products.

Check: the three outcomes (both, neither, exactly one) must total $1$.

(a) Find the number of members by adding all four regions.

(i) The overlap holds the members in both. (ii) The $9$ outside both circles do neither.

(b) The two members are chosen from the chess players only, so the sample space is $n(C) = 11 + 7 = 18$, not $40$. Of these, $7$ do drama and $11$ do not.

Exactly one does drama: drama then not, or not then drama. Without replacement, the second denominator is $17$.

Simplify, then convert to $3$ significant figures.

(a) Mark the odd numbers in the list. An odd number leaves remainder $1$ when divided by $2$ — this includes the negative number $-5$.

(b) Mark the positive integers: every card except $-5$ and $-2$.

(c) Expected frequency is probability times the number of trials. The answer is a number of times, not a fraction.