a drawer holds 4 red hats and for blue hats. What is the probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats at random and return every hat to drawer before taking out another one

• It is quite simple.
  
  When you take out hats one by one after replacement, there are equal chances of getting red or blue hat.
  
  So possible outcomes are
  RRRB, RRBR,RBRR,BRRR, BBBR,BBRB,BRBB,RBBB.. with exactly three red or blue hats.
  RRBB,BBRR,RBRB,BRBR,BRRB,RBBR, BBBB,RRRR.. with other combinations.
  
  so out of 16 possible combinations , eight are desired combinations.
  so probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats at random and return every hat to drawer before taking out another one = 8/16 =1/2
• 12 years agoHelpfull: Yes(29) No(11)
• Let we denote the red hats as : R1,R2,R3,R4
  and blue hats as : B1,B2,B3,B4.
  Now, reading the given problem carefully,we observe that one hat is drawn at a time and after recording its color the drawn hat is again returned to the drawer.So, it,therefore,follows from the condition of the prob. that at every step of drawing a hat, we draw a hat from a set of 8 hats,namely:R1, R2, R3, R4, B1, B2, B3, B4.
  So, 1st hat can be drawn in 8 ways and the color is recorded.
  Similarly,2nd hat can also be drawn in 8 ways & the color is recorded.
  3rd hat can also be drawn in 8 ways & the color is recorded.
  Lastly 4th hat can also be drawn in 8 ways & the color is recorded.
  So, total no. of ways of drawing 4 hats one by one with replacement in the drawer = 8*8*8*8 = (8^4).
  
  Now, 1st we consider the case of drawing 3 red hats and 1 blue hat.
  It is obvious that the event of drawing 1 blue hat can happen at any of the 1st draw ,2nd draw ,3rd draw & 4th draw,i.e,this event can happen in 4 ways and also a blue hat can be chosen from among 4 blue hats namely :B1,B2,B3,B4 , in 4 ways.
  So,as a whole a blue hat may get included in our drawing in (4*4)=(4^2) ways.
  And for each of these (4^2)cases,3 red hats can be drawn 1 by 1 with replacement in the drawer before drawing another hat in (4^3)ways.
  exactly 3 red hats and a blue hat can be drawn in :(4^2)*(4^3)=(4^5)ways.
  
  So, similarly,exactly 3 blue and a red hats can also be drawn in (4^5)ways.
  
  So, the given event can happen in 2*(4^5)ways.
  
  So, the required probability = {2*(4^5)}/(8^4) = (1/2).
  
  So, the required probability = (1/2)..
  
• 12 years agoHelpfull: Yes(19) No(3)
• RRRB has 4 possible arrangements and BBBR too. So a total of 8 desired arrangements out of possible 16 arrangements.
  
  so probability = 8/16 = 1/2
• 12 years agoHelpfull: Yes(4) No(8)
• There are 4 red and 4 blue hats.
  Probability of drawing a red hat is 4/8 = 1/2 so for the blue hat.
  Required probability for RRRB +BBBR,
  There will be 4 ways to get exactly 3 reds - BRRR, RBRR, RRBR, RRRB
  Same way there will be 4ways to get exactly blue.
  
  So probability of getting exactly 3 reds and 3 blue = 4*1/2*1/2*1/2*1/2 + 4*1/2*1/2*1/2*1/2 = 1/2
• 12 years agoHelpfull: Yes(4) No(2)
• plz explain clearly..
• 12 years agoHelpfull: Yes(2) No(3)
• thnks sdeep
• 12 years agoHelpfull: Yes(2) No(3)