Cards In This Deck

• $\text{How do you calculate }P(atleastone)\text{ using the complement method?}$
  $$\begin{gathered} 1-P(none) \\ \text{OR} \\ 1-P(A^{\prime }\cap B^{\prime }) \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ P(A^{\prime }\cap B)\text{ represents the region inside }A\text{ but outside }B\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ P(A^{\prime }\cap B)\text{ means ’Not A AND B’. This represents the region inside B only (excluding the intersection).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ P(A\cup B{)}^{\prime }\text{ is the same region as }{A}^{\prime }\cup B^{\prime }\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ P(A\cup B{)}^{\prime }\text{ is the region outside both circles (everything except the union).} \\ {A}^{\prime }\cup B^{\prime }\text{ is everything except the intersection.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If two events }A\text{ and }B\text{ are Mutually Exclusive, then }P(A\cap B)=P(A)\times P(B)\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{Correct reasoning:} \\ \text{* Mutually Exclusive means }P(A\cap B)=0\text{.} \\ \text{* The formula }P(A)\times P(B)\text{ refers to Independent events.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If two events }A\text{ and }B\text{ are mutually exclusive, they are automatically independent.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{If }A\text{ and }B\text{ are mutually exclusive (}P(A\cap B)=0\text{) and have non-zero probabilities, the occurrence of }A\text{ makes the probability of }B\text{ occurring 0. Therefore, the outcome of }A\text{ does affect }B\text{, so they are not independent.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{In a conditional probability question asking for }P(A|B)\text{, the denominator of your fraction is always the Total Sample Space (e.g., }1\text{ or }n(\xi )\text{).} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{The denominator becomes the probability or count of the given event (}B\text{).} \\ P(A|B)=\frac{P(A\cap B)}{P(B)} \end{gathered}$$
• $\text{Formula for Conditional Probability }P(B|A)$
  $P(B|A)=\frac{P(B\cap A)}{P(A)}$
• $\text{When determining if events }A\text{ and }B\text{ are independent, which two mathematical tests can you perform?}$
  $$\begin{gathered} \text{1. Check if }P(A\cap B)=P(A)\times P(B) \\ \text{2. Check if }P(A|B)=P(A)\text{ (or }P(B|A)=P(B)\text{)} \end{gathered}$$
• $\text{What is the condition for }P(A\cup B)=P(A)+P(B)-P(A)P(B)\text{ to be valid?}$
  $$\begin{gathered} \text{Events }A\text{ and }B\text{ must be independent.} \\ \text{(This substitutes }P(A\cap B)\text{ with }P(A)P(B)\text{ in the addition formula).} \end{gathered}$$
• $\text{What condition must be met to use the formula }P(A\cup B)=P(A)+P(B)\text{?}$
  $\text{Events }A\text{ and }B\text{ must be Mutually Exclusive.}$
• $\text{You are given }P(A)\text{, }P(B)\text{, and }P(A\cup B)\text{. How do you calculate }P(A\cap B)\text{?}$
  $$\begin{gathered} \text{Rearrange the Addition Formula:} \\ P(A\cap B)=P(A)+P(B)-P(A\cup B) \end{gathered}$$
• $\text{When calculating }P(A\cup B)\text{, under what specific condition can you use the formula }P(A)+P(B)\text{?}$
  $$\begin{gathered} \text{Only when events }A\text{ and }B\text{ are mutually exclusive.} \\ \text{If they are not mutually exclusive, you must subtract the intersection: }P(A\cup B)=P(A)+P(B)-P(A\cap B)\text{.} \end{gathered}$$
• $\text{Formula for the Multiplication Rule (General)}$
  $P(A\cap B)=P(A)\times P(B|A)$
• $\text{What is the most reliable first step when filling in probabilities on a Venn Diagram?}$
  $$\begin{gathered} \text{Start from the center (intersection) and work outwards.} \\ \text{Fill in }P(A\cap B)\text{ first, then subtract this value from the total probabilities of }A\text{ and }B\text{ to find }P(Aonly)\text{ and }P(Bonly)\text{.} \end{gathered}$$
• $\text{When should you use a Tree Diagram instead of a Venn Diagram?}$
  $\text{When events happen sequentially (one after another) or when dealing with conditional probabilities where one outcome depends on the previous one.}$
• $\text{How do you calculate }P(B)\text{ if you are given }P(A)\text{, }P(B|A)\text{, and }P(B|A^{\prime })\text{?}$
  $$\begin{gathered} \text{Use the law of total probability (summing the branches):} \\ P(B)=P(A)\times P(B|A)+P(A^{\prime })\times P(B|A^{\prime }) \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If drawing two items without replacement, the probability on the second branch of the tree diagram remains the same as the first.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{Without replacement means the total number of items (denominator) decreases by 1, and the count of the specific item selected (numerator) might decrease by 1.} \end{gathered}$$