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probability:02_multiple_event_probability [2014/01/20 13:48]
marje [2.2. Mutually non exclusive events]
probability:02_multiple_event_probability [2014/01/20 15:44]
marje [2.3. Independent events]
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 $\text{Pr}[\text{"​we obtain two tails in a row"​}]=$ $\text{Pr}[\text{"​we obtain two tails in a row"​}]=$
  
-$\text{Pr}[\text{"​1st throw gives tails"​}\cap\text{"​2nd throw gives tails"​}]= +$\text{Pr}[\text{"​1st throw gives tails"​}\cap\text{"​2nd throw gives tails"​}]=$ 
-\text{Pr}[\text{"​1st throw gives tails"​}]\cdot\text{Pr}[\text{"​2nd throw gives tails"​}]=+ 
 +$\text{Pr}[\text{"​1st throw gives tails"​}]\cdot\text{Pr}[\text{"​2nd throw gives tails"​}]=
 \frac{1}{2}\cdot\frac{1}{2}= \frac{1}{2}\cdot\frac{1}{2}=
 \frac{1}{4}.$ \frac{1}{4}.$
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 Or, by tossing a coin and rolling a six-faced normal dice, the probability Or, by tossing a coin and rolling a six-faced normal dice, the probability
  
-$\text{Pr}[\text{"​we land on heads"​}\cap\text{"​we roll a six"​}]= +$\text{Pr}[\text{"​we land on heads"​}\cap\text{"​we roll a six"​}]=$ 
-\text{Pr}[\text{"​we land on heads"​}]\cdot\text{Pr}[\text{"​we roll a six"​}]=\frac{1}{2}\cdot\frac{1}{6}=+$\text{Pr}[\text{"​we land on heads"​}]\cdot\text{Pr}[\text{"​we roll a six"​}]=\frac{1}{2}\cdot\frac{1}{6}=
 \frac{1}{12}.$ \frac{1}{12}.$
  
probability/02_multiple_event_probability.txt · Last modified: 2014/03/06 13:22 by anna