There are 4 boxes colored red, yellow, green and blue. If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?

• lets fix that one box is green then no of selection of 1 box from red, yellow blue is 3C1......
  lets fix that one box is red then no of selection of 1 box from green, yellow blue is 3C1......
  so total combination=3c1+3c1=6
  but red and green occur in both cases which are similar so we reduce solution by 1 so total number of ways= 6-1= 5
  
• 9 years agoHelpfull: Yes(24) No(1)
• 
  ans-3/2
  if one green then probability 1/4 and one from others 3c1=3 =>(1/4)*3=3/4
  or
  if one red then probability 1/4 and one from others 3c1=3 =>(1/4)*3=3/4
  therefore (3/4)+(3/4)=3/2
• 9 years agoHelpfull: Yes(1) No(5)
• Ans is 5
  
  Combinations of the 4 boxes like
  
  red, yellow
  red,green
  red,blue
  yellow,green
  green,blue
• 9 years agoHelpfull: Yes(1) No(0)
• there are 8 couples...i.e. 16 persons...
  so total outcomes for choosing 4 persons is 16C4 = 1820
  p(atleast 1 couple) = p(1 couple or 2couples)
  for 1 couple -> we have 8 choices to choose a couple,
  for remaining 2 non couple persons, we can select any 1 person from 14 people in 14 ways
  now the 4th person can't be spouse of 3rd person, so we have only 12 persons to choose from
  so total choices for 3rd and 4th persons are 14*12/2 (divided by 2 as we are counting a pair twice) and total choices including 1st and 2nd persons are 8*14*12/2
  which is 672
  Now for 2 couples we have 8c2 = 28 choices
  so total favuorable outcomes = 672 + 28 = 700
  700/1820 = 5/13, or 15/39
  
  
• 9 years agoHelpfull: Yes(0) No(8)