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Cameron rolls a number cube and spins a 5-part spinner, labeled 1, 2, 3, 4, and 5. What is the probability that he rolls a 5 or spins a 5?

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3 Answers

These are independent events.
The probability of rolling a 5 is 1/6.
The probability of spinning a 5 is 1/5.
The probability of rolling a 5 or spinning a 5 is 1/5+1/6=11/30 but you have to subtract 1/30 because you can roll a 5 and spin a 5 at the same time. Therefore 11/30-1/30=10/30=1/3.
There are 30 outcomes. Roll first then spin to get the following outcomes:
11,12,13,14,15,21,22,23,24,25,31,32,33,34,35,41,42,43,44,45,51,52,53,54,55,61,62,63,64,65
You want to roll a 5 OR spin a 5. How many combinations have a 5 ? 10 combinations have a 5.
(you don't count 55 twice) There are 10 combinations with a 5 out of a possible 30 combinations.
Another way of looking at the problem is the following:What is the probability of NOT rolling a 5 AND NOT spinning a 5 ?
The probability of not rolling a 5 AND not spinning a 5 is (5/6)*(4/5)=20/30. Therefore the probability of rolling a 5 or spinning a 5 is 1-20/30=10/30.
 
There are six outcomes for the cube and five outcomes for the spinner, for a total of 6*5=30 outcomes (1-1, 1-2,..., 6-5).
 
In six of those outcomes the spinner spins a 5 (1-5, 2-5,...,6-5), and in five outcomes the cube rolls a 5 (5-1, 5-2, ..., 5-5). There is exactly one outcome where the spinner spins 5 AND the cube rolls 5 (5-5), which should only be counted once, so there are 6+5-1=10 outcomes where the spinner spins 5 OR the cube rolls 5.
 
Therefore, the probability that he rolls 5 or spins a 5 is 10/30=1/3.
Hi Chris;
The cube must have six sides, six potential results.  I assume each side is denoted the numbers 1-6 and that each denotation occurs once.
The spinner has five parts, five potential results.  You have informed me that each part is denoted the numbers 1-5 and that each denotation occurs once.
6 sides + 5 parts=11 parts
The number 5 appears twice.
2/11