W williamrobertsuk New member Joined May 18, 2019 Messages 32 May 18, 2019 #1 Find the problem attached! Last edited by a moderator: May 18, 2019

Dr.Peterson Elite Member Joined Nov 12, 2017 Messages 5,010 May 19, 2019 #2 Have you tried anything? I would first try using De Morgan's laws to simplify \(\displaystyle A\cup(B^c\cap C^c)^c\). Then question 1 will be fairly easy. Please show us your work, and ask any specific questions it raises. Also, please reread the submission guidelines.

Have you tried anything? I would first try using De Morgan's laws to simplify \(\displaystyle A\cup(B^c\cap C^c)^c\). Then question 1 will be fairly easy. Please show us your work, and ask any specific questions it raises. Also, please reread the submission guidelines.

pka Elite Member Joined Jan 29, 2005 Messages 8,670 May 19, 2019 #3 williamrobertsuk said: Find the problem attached! View attachment 12190 Click to expand... Here are HINTS only. You must apply them and post results. For 1. \(\displaystyle {\left( {{B^c} \cup {C^c}} \right)^c} = B \cap C\) For 2. If \(\displaystyle H~\&~G\) are disjoint events then \(\displaystyle \mathcal{P}(H\cup G)=\mathcal{P}(H)+\mathcal{P}(G)\) For 3. \(\displaystyle {\left( {{A^c} \cap \left( {{B^c} \cup {C^c}} \right)} \right)^c} = A \cup {\left( {{B^c} \cup {C^c}} \right)^c}\)

williamrobertsuk said: Find the problem attached! View attachment 12190 Click to expand... Here are HINTS only. You must apply them and post results. For 1. \(\displaystyle {\left( {{B^c} \cup {C^c}} \right)^c} = B \cap C\) For 2. If \(\displaystyle H~\&~G\) are disjoint events then \(\displaystyle \mathcal{P}(H\cup G)=\mathcal{P}(H)+\mathcal{P}(G)\) For 3. \(\displaystyle {\left( {{A^c} \cap \left( {{B^c} \cup {C^c}} \right)} \right)^c} = A \cup {\left( {{B^c} \cup {C^c}} \right)^c}\)