EXN.2.AHL.TZ0.9

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

METHOD 1

$E_n$ is the event “the first tail occurs on the $2$nd, $4$th, $6$th, …, $2n$th toss”

$\text{P}\left(E\right)=\underset{n=1}{\overset{\infty }{\text{Σ}}}\text{P}\left(E_n\right)$         (A1)

 

Note: Award A1 for deducing that either $1$ head before a tail or $3$ heads before a tail or $5$ heads before a tail etc. is required. In other words, deduces $\left(2n-1\right)$ heads before a tail.

 

$\text{P}\left(E\right)=0.4\times 0.6+{\left(0.4\right)}^3\times 0.6+{\left(0.4\right)}^5\times 0.6+\dots$         M1A1

 

Note: Award M1 for attempting to form an infinite geometric series.

Note: Award A1 for $\text{P}\left(E\right)=\underset{n=1}{\overset{\infty }{\text{Σ}}}{\left(0.4\right)}^{2n-1}\left(0.6\right)$.

 

uses $S_{\infty }=\frac{u_1}{1-r}$ with $u_1=0.6\times 0.4$ and $r={\left(0.4\right)}^2$         (M1)

 

Note: Award M1 for using $S_{\infty }=\frac{u_1}{1-r}$ with $u_1=0.4$ and $r={\left(0.4\right)}^2$

 

$=\frac{0.6\times 0.4}{1-{\left(0.4\right)}^2}$        A1

$=0.286 \left(=\frac{2}{7}\right)$        A1

 

METHOD 2

let $T_1$ be the event “tail occurs on the first toss”

uses $\text{P}\left(E\right)=\text{P}\left(E \overline{)T_1}\right)\text{P}\left(T_1\right)+\text{P}\left(E \overline{)T_1\text{'}}\right)\text{P}\left(T_1\text{'}\right)$         M1

concludes that $\text{P}\left(E \overline{)T_1}\right)=0$ and so $\text{P}\left(E\right)=\text{P}\left(E \overline{)T_1\text{'}}\right)\text{P}\left(T_1\text{'}\right)$         R1

$\text{P}\left(E \overline{)T_1\text{'}}\right)=\text{P}\left(E\text{'}\right)\left(=1-\text{P}\left(E\right)\right)$         A1

 

Note: Award A1 for concluding: given that a tail is not obtained on the first toss, then $\text{P}\left(E \overline{)T_1\text{'}}\right)$ is the probability that the first tail is obtained after a further odd number of tosses, $\text{P}\left(E\text{'}\right)$.

 

$\text{P}\left(T_1\text{'}\right)=0.4$

$\text{P}\left(E\right)=0.4\left(1-\text{P}\left(E\right)\right)$         A1

attempts to solve for $\text{P}\left(E\right)$         (M1)

$=0.286 \left(=\frac{2}{7}\right)$         A1

 

[6 marks]

[N/A]