A dice has one of the first 6 prime number on each its six sides ,with no two sides having the same number .the dice is rolled 10 times and the results added.the addition is most likely to be closet to
a-41 b-48 c-60 d- 68

• if die is rolled then avg score = (2+3+5+7+11+13 )/6 = 41/6
  
  so, most likely sum for 10 times = 41/6 * 10 = 68 approax
• 8 years agoHelpfull: Yes(36) No(0)
• yeah take avg=41/6=7*10=68 something ,no other option to calculate
• 8 years agoHelpfull: Yes(3) No(1)
• If the dice is rolled only once then average or expected value=(2+3+5+7+11+13)/6
  But the dice is rolled 10 times ,then: 41*10/6=68 approx.
• 6 years agoHelpfull: Yes(0) No(0)
• is this really an elitmus question
• 4 years agoHelpfull: Yes(0) No(0)