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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 [2014/03/06 13:22] anna [2.4. Additional exercises] |
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| <WRAP task> | <WRAP task> | ||
| On Midsummer's Day, a probability of a person having a car accident is $0,08$. The probability of a person | On Midsummer's Day, a probability of a person having a car accident is $0,08$. The probability of a person | ||
| - | driving intoxicated is $0,35$ and probability that a person has a car accident while intoxicated is $0,2$. | + | driving intoxicated is $0,35$ and probability that a person has a car accident while intoxicated is $0,02$. |
| What is the probability of a person driving while intoxicated or having a car accident? | What is the probability of a person driving while intoxicated or having a car accident? | ||
| - | ++Answer| $\text{Pr}[\text{"a person driving while intoxicated or having a car accident"}]=0,35+0,08-0,2=0,23$++ | + | ++Answer| $\text{Pr}[\text{"a person driving while intoxicated or having a car accident"}]=0,35+0,08-0,02=0,41$++ |
| </WRAP> | </WRAP> | ||
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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}.$ | ||
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| ++Answer| Note, that $\text{"my new friend is female"}=$\\ | ++Answer| Note, that $\text{"my new friend is female"}=$\\ | ||
| $\text{"my new friend is female and at most $29$ years"}\cup | $\text{"my new friend is female and at most $29$ years"}\cup | ||
| - | \text{"my new friend is female and at least $30$ years"}$, and the events | + | \text{"my new friend is female and at least $30$ years"}$, and these events are mutually exclusive. |
| - | $\text{"my new friend is female and at most $29$ years"}$ and | + | Thus, $\text{Pr}[\text{"my new friend is female"}]=0,1+0,3=0,4$.++ |
| - | $\text{"my new friend is female and at least $30$ years"}$ are mutually exclusive. Thus, | + | |
| - | $\text{Pr}[\text{"my new friend is female"}]=0,1+0,3=0,4$.++ | + | |
| </WRAP> | </WRAP> | ||
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| <WRAP nl> | <WRAP nl> | ||
| - | [[probability:03_conditional_probability|3. Conditional probability]]\\ | + | [[probability:03_conditional_probability]]\\ |
| </WRAP> | </WRAP> | ||