Two finals are scheduled - The Wimbledon match and the World Cup Cricket at the same time. Anu wants to watch the Wimbledon finals and her brother Vinu wants to watch WCC final. They decide to roll a tetrahedral die twice. The tetrahedral is numbered 1,2,3,4 on its four sides and all numbers are equally likely to appear. Anu rolls first and then Vinu rolls. If the number on the first roll is strictly greater than the number on the second roll. Anu wins and gets to watch Wimbledon. What is the probability that Anu will get to watch Wimbledon?
1)7/16 2)9/16 3)3/8 4)1/2

• Answer:
  
  The probability that Anu will get to watch the Wimbledon match is 3/8
  
  Step-by-step explanation:
  
  Probability of any number appearing = 1/4
  
  Anu wins in the following condition
  
  1) If 1 occus on the first roll then there is no chance of her winning because in the next roll the number appeaing will be either 1 or greater than 1
  
  2) If 2 appears on 1st roll then Anu will win if 1 appears on the second roll
  
  3) If 3 appears on the 1st roll then Anu will win if 1 or 2 appears on the second roll
  
  4) If 4 appears on the first roll then Anu will win if 1, 2, 3 appears on the second roll
  
  Therefore, the probability of Anu winning
  1/4x1/4+1/4x2/4+1/4x3/4=3/8
• 4 years agoHelpfull: Yes(3) No(0)
• The answer is 3/8
• 4 years agoHelpfull: Yes(0) No(1)
• The probability that Anu will get to watch the Wimbledon match is 3/8
  
  Step-by-step explanation:
  
  Probability of any number appearing = 1/4
  
  Anu wins in the following condition
  
  1) If 1 occus on the first roll then there is no chance of her winning because in the next roll the number appeaing will be either 1 or greater than 1
  
  2) If 2 appears on 1st roll then Anu will win if 1 appears on the second roll
  
  3) If 3 appears on the 1st roll then Anu will win if 1 or 2 appears on the second roll
  
  4) If 4 appears on the first roll then Anu will win if 1, 2, 3 appears on the second roll
  
  Therefore, the probability of Anu winning
  
  frac{1}{4}times frac{1}{4} + frac{1}{4}times frac{2}{4} + frac{1}{4}times frac{3}{4} = frac{3}{8}[latex]
• 4 years agoHelpfull: Yes(0) No(0)