ATAR Notes: Forum

Uni Stuff => Faculties => Engineering => Topic started by: louiedaws on October 13, 2017, 09:57:56 pm

Title: Probability Question
Post by: louiedaws on October 13, 2017, 09:57:56 pm: Hey, I was never very good at probability in high school and was wondering if anyone knew how to do these three questions. Also don't if this is the right place to post so someone can move this to an appropriate section.
Title: Re: Probability Question
Post by: RuiAce on October 13, 2017, 10:18:36 pm: $\text{Because 1, 2 and 3 are connected in parallel to each other}\\ \text{but all of them are in series to 4,}\\ \text{Only one of 1, 2 or 3 need to work, but 4 must work}$
$\text{Using the independence assumption}\\ \begin{align*}\mathbb{P}(\text{circuit works})&= \mathbb{P}(\text{4 works}) \times \mathbb{P}(\textbf{at least}\text{ one of 1, 2 or 3 work})\\ &= \mathbb{P}(\text{4 works})\times \left[1- \mathbb{P}(\textbf{none of}\text{ 1, 2 or 3 work})\right]\end{align*}\\ \text{having taken advantage of the complement}$
$\text{Again, from the independence assumption}\\ \mathbb{P}(\text{circuit works})= \mathbb{P}(\text{4 works}) \times \left[1-\mathbb{P}(\text{1 doesn't work})\mathbb{P}(\text{2 doesn't work})\mathbb{P}(\text{3 doesn't work})\right]$
$\text{so since the probability each works is always }p\\ \text{our answer will be }p \left(1-(1-p)^3\right)$
Title: Re: Probability Question
Post by: RuiAce on October 13, 2017, 10:41:12 pm: We will assume that if Sam submits via an internet connection that works, then the assignment is guaranteed to be submitted successfully. Else, this question lacks information.
$\text{Denote the events}\\ \begin{align*}S&=\text{Successful submission}\\ W&=\text{Wi-Fi works}\\ G&=\text{4G is not out of data}\end{align*}\\ \text{Furthermore, for any event }A\text{, let }A^C\text{ denote the complementary event.}$
$\text{Observe that we are given the following information:}\\ \begin{align*}\mathbb{P}\left(S^C \mid W^C\right) &= 0.90\\ \mathbb{P}\left(S^C\mid G^C\right)&=0.95\\ \mathbb{P}\left(W^C\right)&=0.3\\ \mathbb{P}\left(G^C\right)&=0.25\end{align*}$
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Part 1) is not doable unless we make the assumption that the behaviour of the Wi-Fi and 4G are independent. So we will make this assumption.
(Reality check: This assumption may be viable, because Sam's Wi-Fi and 4G providers can be different.)
$\begin{align*}\mathbb{P}(W \cup G)&=1 - \mathbb{P}\left((W\cup G)^C\right)\\ &= 1 - \mathbb{P}\left(W^C \cap G^C\right)\\ &= 1- \mathbb{P}\left(W^C\right)\mathbb{P}\left(G^C\right)\tag{indep.}\\ &= 1-0.3\times0.25 = 0.925\end{align*}$

At this point, I don't want to do part b) because I can't verify if that 0.38 is extraneous information. Please provide the answers as a reference.