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| + | [[Category:discrete math]] | ||
| + | [[Category:problem solving]] | ||
| + | =[[MA375]]: [[MA_375_Spring_2009_Walther_Week_5|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 | 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 | ||
| + | ---- | ||
| + | [[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
