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Rafael spins the pointers of the two spinners shown at the right. Find the probability of each possible sum.

Rafael spins the pointers of the two spinners shown at the right. Find the probability-example-1
User Pufos
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1 Answer

5 votes

Answer:

P(sum 2)=1/6

P(sum 3)=1/3

P(sum 4)=1/3

P(sum 5)=1/6

Explanation:


\left|\begin{array}ccc&1&2&3\\---&---&---&----\\1&2&3&4\\2&3&4&5\end{array}\right|

On the table, there are a a total of 6 outcomes.

  • Number of sum that equals 2=1
  • Number of sum that equals 3=2
  • Number of sum that equals 4=2
  • Number of sum that equals 5=1

Therefore:

P(sum 2)=1/6

P(sum 3)=2/6=1/3

P(sum 4)=2/6=1/3

P(sum 5)=1/6

User Hejdav
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