. in a single throw with two dice find the probability that their sum is a multiple either of 3 or 4 a)1/3 b)5/9 c)17/36 d)1/2

• Their sum can be 3, 4 , 6, 8 ,9 ,12
  
  The probability of getting 3 as sum of two die is = 2/36
  The probability of getting 4 as sum of two die is = 3/36
  The probability of getting 6 as sum of two die is = 5/36
  The probability of getting 8 as sum of two die is = 5/36
  The probability of getting 9 as sum of two die is = 4/36
  The probability of getting 12 as sum of two die is = 1/36
  
  So, the probability that their sum is a multiple either of 3 or 4 will be = (2+3+5+5+4+1)/36
  = 20/36
  =5/9
  
• 9 years agoHelpfull: Yes(15) No(0)
• number of event =6*6= 36
  the number which are multiple of 3 or 4 is =3 4 6 8 9 12
  the number of ways of 3 to 7 is (n-1)
  that is n=3 value is 2
  when n=4 value is 3
  when n=6 value is 5
  similarly from number 8 to 12 (13-n)
  so when n=8 value is 5
  n=9 value is 4
  n=12 value is 1
  add all the value and we get 2 +3+5+5+4+1 =20
  prob=20/36
  5/9 answer
• 9 years agoHelpfull: Yes(4) No(0)
• Their sum can be 3,4,6,8,9,12
  For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 - n) ways.
  Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
  So probability is (20/36)=(5/9)
  so ans z b)5/9
• 9 years agoHelpfull: Yes(1) No(2)
• To get the sum which is multiple either of 3 or 4,
  favourable cases n(E) :- { (1,2), (1,3),(1,5),(2,1),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(3,6),(4,2),(4,4),(4,5),(5,1),(5,3),(5,4),(6,2),(6,3),(6,6)} = 20
  n(S) :- 36(for two dice = 6*6)
  Therefore Probability :- 20/36 = 5/9
  
• 9 years agoHelpfull: Yes(1) No(0)
• B)5/9
  Put the sample space when 2 dice is tossed from(1,1) to (6,6).
  check the sum of numbers that are multiples of 4 or 3.
• 7 years agoHelpfull: Yes(0) No(0)
• b
  two dice sum multiple of 3 are 12 pairs
  and multiple of 4 are 9pairs
  in both multiples pair(6,6) is in common excluding it for one time we get total 20 pairs
  probability in getting all pairs is 36 therefore----20/36=5/9
• 6 years agoHelpfull: Yes(0) No(0)
• 1 2 3 4 5 6
  1 1 2 3 4 5 6
  2 2 4 5 6 7 8
  3 3 5 6 7 8 9
  4 4 6 7 8 9 11
  5 5 7 8 9 10 11
  6 6 8 9 10 11 12
  in above matrics find multiple of 3 or 4 value(for or + is required)
• 5 years agoHelpfull: Yes(0) No(0)
• total outcomes=36
  favorable outcome =18
  probability =18/36=1/2
• 4 years agoHelpfull: Yes(0) No(0)
• Answer is 20/36=5/9
• 4 years agoHelpfull: Yes(0) No(0)