Probability practice questions — A-Level Maths

$P(\text{TTT}) = \!\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$, so $P(\ge 1 H) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$.

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 $= \dfrac{9}{90} = \dfrac{1}{10}$.

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 $0.07$.

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: $$P(A\mid D)=\frac{0.03}{0.07}=\frac{3}{7}$$

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

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

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

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

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

So:

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

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

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

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

Total outcomes: $6$.

Probability $= \dfrac{3}{6} = \dfrac{1}{2}$.

$P(\bar{A}) = 1 - P(A) = 1 - 0.3 = 0.7$.