Ina single throw of two dice, what is the probability that neither a doublet nor a total of 8 will appear?
a)7/15
b)5/18
c)13/18
d)3/16

• doublets (1,1)(2,2)...(4,4),..(6,6)=> n(A)=6
  
  ans: 13/18
  total 8=> (2,6)(3,5) (4,4) (5,3)(6,2)=>n(B)=5
  A I B=>(4,4) n(A I B)=1
  n(A U B)=n(A)+n(B)-n(A I B)=6+5-1=10
  due to neither a doublet nor a total of 8
  p=(36-10)/36=26/36
  =13/18
• 11 years agoHelpfull: Yes(5) No(0)
• An. c) 13/18
  doublet {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
  total 8 will be from the combination of {(2,6),(6,2),(3,5),(5,3),(4,4)}
  
  total no. of combination 6+5-1=10;(-1 because (4,4) taken two times)
  
  so, no. of possibilities that neither a doublet nor a total of 8 will be
  =36-10=26
  total no. possibilities =6*6=36
  so,
  probability = 26/36=13/18
  
• 11 years agoHelpfull: Yes(3) No(0)
• let pa(A) be probability of doublet & p(B) be the probability of total of 8.now we need to find p(A^c intersection B^c).
  now p(A^c intersection B^c)=p(AUB)^c=1-p(AUB)=1-p(A)-p(B)+p(A intersection B)
  again,p(A)=6/36, p(B)=5/36,p(A intersection B)=1/36.
  putting all terms in above equation we get=13/18 ans.
• 11 years agoHelpfull: Yes(2) No(0)
• (11,22,33,44,55,66,26,35,53,62)
  10/36=5/18
• 11 years agoHelpfull: Yes(0) No(4)