Probability practice questions — GCSE Maths

Let $E$ be the expected number of flips to reach two heads in a row from the start. Condition on outcomes:

- With probability $\frac{1}{2}$, the first flip is T. We've used one flip and are back at the start: contributes $\frac{1}{2}(1+E)$. - With probability $\frac{1}{4}$, the first two flips are HT. We've used two flips and are back at the start: contributes $\frac{1}{4}(2+E)$. - With probability $\frac{1}{4}$, the first two flips are HH — done in two flips: contributes $\frac{1}{4}\cdot 2$.

So:

$$E=\frac{1}{2}(1+E)+\frac{1}{4}(2+E)+\frac{1}{4}\cdot 2$$

Expanding: $E=\frac{1}{2}+\frac{1}{2}E+\frac{1}{2}+\frac{1}{4}E+\frac{1}{2}=\frac{3}{2}+\frac{3}{4}E$. Therefore $\frac{1}{4}E=\frac{3}{2}$, i.e. $E=6$.

Sanity check: "flip until first head" has expected value $2$; requiring a *consecutive* second head triples it.

The pairs are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — six out of $36$ outcomes. Probability $={\displaystyle \frac{6}{36}}={\displaystyle \frac{1}{6}}$.

$P(\text{TTT})=\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{2}}\right)}^3={\displaystyle \frac{1}{8}}$, so $P(\ge 1H)=1-{\displaystyle \frac{1}{8}}={\displaystyle \frac{7}{8}}$.

Even outcomes: $\{2,4,6\}$. Among these, $\{4,6\}$ are $>3$: probability ${\displaystyle \frac{2}{3}}$.

$P={\displaystyle \frac{3}{8}}\cdot {\displaystyle \frac{2}{7}}={\displaystyle \frac{6}{56}}={\displaystyle \frac{3}{28}}$.

Let $D$ be the event "the part is defective". By the law of total probability over the branches $A$ and $B$: $$P(D)=P(A)P_A(D)+P(B)P_B(D)$$ $$P(D)=0.60\times 0.05+0.40\times 0.10$$ $$P(D)=0.03+0.04=0.07$$ So the probability that the part is defective is $0.07$.

There are $90$ two-digit numbers. Digit sums divisible by $11$ in range $[1,18]$: only $11$. Numbers whose digits sum to $11$: $29,38,47,56,65,74,83,92$ — nine numbers. Probability $={\displaystyle \frac{9}{90}}={\displaystyle \frac{1}{10}}$.

Multiply along each branch to get the joint probabilities: $$P(A\cap D)=0.6\times 0.05=0.03$$ $$P(B\cap D)=0.4\times 0.10=0.04$$ The total probability of a defective item is $$P(D)=0.03+0.04=0.07$$ Apply the conditional probability formula: $$P(A\mid D)=\frac{0.03}{0.07}=\frac{3}{7}$$

Even outcomes on a die: $\{2,4,6\}$ — that is $3$ outcomes.

Total outcomes: $6$.

Probability $={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$.

A fair coin has two equally likely outcomes; the probability of heads is ${\displaystyle \frac{1}{2}}$.