Binomial distribution: the Galton board

<p>In a computer game, a bug moves from left to right through a network of connected paths. The bug starts at $\mathrm{S}$ and, at each junction, randomly takes the left fork with probability $p$ or the right fork with probability $q$, where $q=1-p$. The forks taken at each junction are independent. The bug finishes its journey at one of the $9$ endpoints labelled $\mathrm{A}-\mathrm{I}$ (see diagram).</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that the probability that the bug finishes its journey at $\mathrm{D}$ is $56{{p}^{5}}{{q}^{3}}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Given that the probability that the bug finishes its journey at $\mathrm{D}$ is greater than the probability that the bug finishes its journey at any one of the other endpoints, find exactly the possible range of values of $p$.</span></div><p style="text-align: right;">[4]</p></li></ol>
<p>In another version of the game, the probability that, at each junction, the bug takes the left fork is $0.9p$, the probability that the bug takes the right fork is $0.9q$ and the probability that the bug is swallowed up by a ‘black hole’ is $0.1$.</p>
<ol class="i-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find the probability that, in this version of the game, the bug reaches one of the endpoints $\mathrm{A}-\mathrm{I}$, without being swallowed up by a black hole.</span></div><p style="text-align: right;">[1]</p></li></ol>

In a computer game, a bug moves from left to right through a network of connected paths. The bug starts at $\mathrm{S}$ and, at each junction, randomly takes the left fork with probability $p$ or the right fork with probability $q$, where $q=1-p$. The forks taken at each junction are independent. The bug finishes its journey at one of the $9$ endpoints labelled $\mathrm{A}-\mathrm{I}$ (see diagram).

Show that the probability that the bug finishes its journey at $\mathrm{D}$ is $56{{p}^{5}}{{q}^{3}}$.

Given that the probability that the bug finishes its journey at $\mathrm{D}$ is greater than the probability that the bug finishes its journey at any one of the other endpoints, find exactly the possible range of values of $p$.

In another version of the game, the probability that, at each junction, the bug takes the left fork is $0.9p$, the probability that the bug takes the right fork is $0.9q$ and the probability that the bug is swallowed up by a ‘black hole’ is $0.1$.

Find the probability that, in this version of the game, the bug reaches one of the endpoints $\mathrm{A}-\mathrm{I}$, without being swallowed up by a black hole.

Default solution

$\text{P}(L) = p$, $\text{P}(R) = q$. Starting at S, the bug reaches D by taking 5 left turns and 3 right turns out of 8 forks.

Probability of reaching D

$= \text{P}(LLLLLRRR + \cdots)$

$= \dfrac{8!}{5! \cdot 3!} \, p^5 q^3$

$= 56p^5 q^3$ (shown)

Let $X$ be the number of left forks taken by the bug out of 8 forks.

$X \sim \text{B}(8, p)$

$\text{P}(\text{finishes at D}) = \text{P}(X = 5)$

$\therefore X = 5$ is the mode.

$\text{P}(X=5) > \text{P}(X=6)$ and $\text{P}(X=5) > \text{P}(X=4)$

From $\text{P}(X=5) > \text{P}(X=6)$:

$2(1-p) > p$

$2 - 2p > p$

$\dfrac{2}{3} > p$

From $\text{P}(X=5) > \text{P}(X=4)$:

$\dfrac{4}{5} \cdot \dfrac{p}{1-p} > 1$

$4p > 5(1-p)$

$9p > 5$

$p > \dfrac{5}{9}$

$\therefore \dfrac{5}{9} < p < \dfrac{2}{3}$

At each fork, probability 0.1 of falling into a black hole.

Required probability $= \text{P}(B'B'B'\cdots B')$ (all 8 forks, no black hole)

$= (1 - 0.1)^8$

$= 0.9^8$

$= 0.43046 = 0.430$ (3 s.f.)

N2018 P2 Q6

N2018 P2 Q6

Video Solution:

Default solution