. A drawer holds 4 red hats and 4 blue hats. what is probability of getting exactly 3 red hats or 3 blue hats when taking out 4 hats randomly out of drawer and immediately returning every hat to drawer before taking out next??

• Probability to draw exactly 3 red hats and 1 blue hat is= 1/2*1/2*1/2*1/2=1/16
  Probability to draw exactly 3 blue hats and 1 red hat is= 1/2*1/2*1/2*1/2=1/16
  Total Probability= 1/16+1/16=1/8
• 8 years agoHelpfull: Yes(8) No(1)
• The possibilities...
  3 red hats + 1 blue hat= 4C3 +4C1
  
  SImilarly
  3 blue hats + 1 red hat= 4C3 +4C1
  therfore
  porbablity = (4C3 *4C1)/8C4 + (4C3 *4C1)/8C4 =16/35
  
• 8 years agoHelpfull: Yes(3) No(8)
• Probablity of taking out a particular color hat = 4/8 = 1/2
  
  P(BBBR) = (1/2)^4
  P(BBRB) = (1/2)^4
  P(BRBB) = (1/2)^4
  P(RBBB) = (1/2)^4
  total = 4 * (1/2)^4
  
  similary for (RRRB) = 4 * (1/2)4
  
  answer = 8 * (1/2) ^4 = 1/2
  
• 8 years agoHelpfull: Yes(3) No(2)
• 2 choices for the first/second/third/fourth hat ==> Total number of possibilities : 2*2*2*2 = 16
  
  4 ways to have 3 blue hats (BBBR, BBRB, BRBB, RBBB)
  
  4 ways to have 3 red hats (RRRB, RRBR, RBRR, BRRR)
  
  ==> Answer is (4+4)/16 = 1/2
• 8 years agoHelpfull: Yes(2) No(0)
• can you please explain the problem properly
• 8 years agoHelpfull: Yes(1) No(0)
• Probability to draw exactly 3 red hats and 1 blue hat is= 1/2*1/2*1/2*1/2=1/16
  Probability to draw exactly 3 blue hats and 1 red hat is= 1/2*1/2*1/2*1/2=1/16
  (1/16*4)+(1/16*4) = 1/2
  [here 4 is the combination of red & blue hats ]
• 8 years agoHelpfull: Yes(1) No(1)
• As the objects are replaced, the probability of drawing red or blue is equal.
  Probability to draw 3 red hats consecutively = 12×12×12=1812×12×12=18
  Similarly probability to draw 3 blue hats consecutively = 12×12×12=1812×12×12=18
  Total probability = 12×12×12=1812×12×12=18+12×12×12=1812×12×12=18 = 14
• 8 years agoHelpfull: Yes(0) No(0)
• Probability to draw exactly 3 red hats and 1 blue hat is= 8C3*8C1
  Probability to draw exactly 3 blue hats and 1 red hat is=8C3*8C1
  Total Probability= (4C3*4C1+4C3*4C1)/8C2
• 4 years agoHelpfull: Yes(0) No(0)
• RRRB + BBBR
  = 4!/3! * (1/2*1/2*1/2*1/2*) + 4!/3! * (1/2*1/2*1/2*1/2*)
  = 1/4 +1/4 =1/2
• 4 years agoHelpfull: Yes(0) No(0)