A&W assignment 3

T3.1 Probability spaces

a)

Probability space

The sample space is $\Omega =\{1,2,3,4,5,6{\}}^3$
An atomic event is one concrete outcome $(x_1,x_2,x_3)$, where $x_i$ is the result of die $i$.
$\mathrm{Pr}[(x_1,x_2,x_3)]=\frac{1}{|\Omega |}=\frac{1}{216}$

sum at least 16

We want all triples $(x_1,x_2,x_3)$ such that $x_1+x_2+x_3\ge 16$ Sum = 18

$$(6,6,6)$$

Sum = 17

$$(5,6,6),(6,5,6),(6,6,5)$$

Sum = 16

$$(4,6,6),(6,4,6),(6,6,4),(5,5,6),(5,6,5),(6,5,5)$$

Probability of $E$

There are 10 atomic events:

$$\mathrm{Pr}(E)=\frac{|E|}{|\Omega |}=\frac{10}{216}=\frac{5}{108}.$$

first die shows 6 given $E$

$$\mathrm{Pr}(x_1=6\mid E)=\frac{\mathrm{Pr}(x_1=6\cap E)}{\mathrm{Pr}(E)}.$$

There are 6 atomic events in $E$ whose first dice shows 6. Hence

$$\mathrm{Pr}(x_1=6\mid E)=\frac{6/216}{10/216}=\frac{6}{10}=\frac{3}{5}.$$

b)

Probability space

An atomic event is an ordered 4-tuple of distinct cards:

$$\Omega =\{(c_1,c_2,c_3,c_4):c_i\text{ distinct cards}\}.$$

with $c_1,c_2$ are Alice’s cards and $c_3,c_4$ are Bob’s cards
Number of atomic event $52\cdot 51\cdot 50\cdot 49=6,497,400$ $\mathrm{Pr}[(c_1,c_2,c_3,c_4)]=\frac{1}{|\Omega |}=\frac{1}{6497400}$

Computations

$$\mathrm{Pr}[A]=\frac{3}{51}=\boxed{\frac{1}{17}}$$ $$\mathrm{Pr}[B]=\mathrm{Pr}[A]=\boxed{\frac{1}{17}}$$

$\mathrm{Pr}[A\cap B]=\mathrm{Pr}[A]\cdot \mathrm{Pr}[B|A]$ We first calculate $\mathrm{Pr}[B|A]$ case 1: Bob’s first card has the same value as Alice’s pair contribution: $\frac{2}{50}\cdot \frac{1}{49}$ case 2: Bob’s first card does not have the same value as Alice’s pair contribution: $\frac{48}{50}\cdot \frac{3}{49}$ In total $\mathrm{Pr}[B|A]=\frac{2}{50}\cdot \frac{1}{49}+\frac{48}{50}\cdot \frac{3}{49}=\frac{146}{2450}=\frac{73}{1225}$ Therefore

$$\mathrm{Pr}[A\cap B]=\mathrm{Pr}[A]\cdot \mathrm{Pr}[B|A]=\frac{1}{17}\cdot \frac{73}{1225}=\boxed{\frac{73}{20825}}$$ $$\mathrm{Pr}[A|B]=\frac{\mathrm{Pr}[A\cap B]}{\mathrm{Pr}[B]}=\frac{\frac{73}{20825}}{\frac{1}{17}}=\boxed{\frac{73}{1225}}$$

c)

Probability space

$$\Omega =\{M,N\}\times \{H,T{\}}^k.$$

with $\mathrm{Pr}[N]=\mathrm{Pr}[M]=\frac{1}{2}$

Pr[N|E]

We first calculate $\mathrm{Pr}[E]$ : By total probability $\mathrm{Pr}[E]=\mathrm{Pr}[E\cap N]+\mathrm{Pr}[E\cap M]=\mathrm{Pr}[E|M]\cdot \mathrm{Pr}[M]+\mathrm{Pr}[E|N]\cdot \mathrm{Pr}[N]$

we know $\mathrm{Pr}[E\mid M]=1,\mathrm{Pr}[E\mid N]={\left(\frac{1}{2}\right)}^k$ Hence

$$\mathrm{Pr}[E]=\frac{1}{2}+\frac{1}{2^{k+1}}$$

By Bayes’ rule:

$$\mathrm{Pr}(N\mid E)=\frac{\mathrm{Pr}(E\mid N)\mathrm{Pr}(N)}{\mathrm{Pr}(E)}$$

By substitution:

$$\mathrm{Pr}[N\mid E]=\frac{{\left(\frac{1}{2}\right)}^k\cdot \frac{1}{2}}{\frac{1}{2}+\frac{1}{2^{k+1}}}=\frac{\frac{1}{2^{k+1}}}{\frac{1}{2}+\frac{1}{2^{k+1}}}=\boxed{\frac{1}{2^k+1}}$$

>99%

We want $\mathrm{Pr}[M|E]>0.99$

$$\mathrm{Pr}[M|E]=1-\mathrm{Pr}[N|E]=1-\frac{1}{2^k+1}=\frac{2^k}{2^k+1}>0.99\text{, k is integer}$$

By using calculator we get $\boxed{k\ge 7}$