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=[[MA375]]: [[MA_375_Spring_2009_Walther_Week_5|Solution to a homework problem from this week or last week's homework]]=
====What is the probability of each outcome when a loaded die is rolled, if 3 is twice as likely to appear as each of the other five numbers on the die====
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Spring 2009, Prof. Walther
 
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==Question==  
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What is the probability of each outcome when a loaded die is rolled, if 3 is twice as likely to appear as each of the other five numbers on the die
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==Answer==
 
Well the probability of 3 is twice as likely.  So we set up an equation that is equal to 1 because that is the probability of any answer.  So, 2x +5x = 1 is the equation.  So solve for a and you get 1/7 probability of rolling 1,2,4,5,6 and since 3 is twice as likely you get 2*1/7  which equals 2/7
 
Well the probability of 3 is twice as likely.  So we set up an equation that is equal to 1 because that is the probability of any answer.  So, 2x +5x = 1 is the equation.  So solve for a and you get 1/7 probability of rolling 1,2,4,5,6 and since 3 is twice as likely you get 2*1/7  which equals 2/7
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[[MA375_%28WaltherSpring2009%29|Back to MA375, Spring 2009, Prof. Walther]]

Latest revision as of 08:20, 20 May 2013


MA375: Solution to a homework problem from this week or last week's homework

Spring 2009, Prof. Walther


Question

What is the probability of each outcome when a loaded die is rolled, if 3 is twice as likely to appear as each of the other five numbers on the die


Answer

Well the probability of 3 is twice as likely. So we set up an equation that is equal to 1 because that is the probability of any answer. So, 2x +5x = 1 is the equation. So solve for a and you get 1/7 probability of rolling 1,2,4,5,6 and since 3 is twice as likely you get 2*1/7 which equals 2/7


Back to MA375, Spring 2009, Prof. Walther

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