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| − | + | '''Conditional Probability''' | |
| + | One dice is rolled two separate times. Find the probability that the dice lands on an even number both times, and the sum of the two rolls is greater than 6 but the first roll must be larger than the second.<br> | ||
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| − | + | '''Solution:<br>''' | |
| − | <br> | + | The complete set would consist of the following:<br> |
| − | <br> | + | S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}<br> |
| − | + | We will let A="even number both times" and B="the sum of the two rolls is greater than 6 but the first roll must be larger than the second"<br> | |
| − | + | Therefore: A={(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)} and B={(4,3),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)}<br> | |
| − | + | There are 36 total outcomes, so P(AnB)={(6,2),(6,4)}=2/36 and P(B)=9/36<br> | |
| − | + | P(A|B)=(2/36)/(9/36)= 2/9<br> | |
| − | + | <br> | |
| − | + | [[Bonus point 1 ECE302 Spring2012 Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | |
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| − | + | [[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:Probability]] [[Category:Conditional_probability]] | |
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Revision as of 19:06, 27 January 2013
problem solving
Conditional Probability
One dice is rolled two separate times. Find the probability that the dice lands on an even number both times, and the sum of the two rolls is greater than 6 but the first roll must be larger than the second.
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Solution:
The complete set would consist of the following:
S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
We will let A="even number both times" and B="the sum of the two rolls is greater than 6 but the first roll must be larger than the second"
Therefore: A={(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)} and B={(4,3),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)}
There are 36 total outcomes, so P(AnB)={(6,2),(6,4)}=2/36 and P(B)=9/36
P(A|B)=(2/36)/(9/36)= 2/9
Back to first bonus point opportunity, ECE302 Spring 2013
