a drawer holds 4 red hats and 4 blue hats.what is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediatly returning every hat to the drawer before taking out the next?
a)1/2
b)3/8
c)1/4
d)1/8

• 4 red,4 blue (asked exactly 3r or 3b)
  probability for drawing red hat is 4/8 note: he returning every hat to drawer
  firstly exactly 3 red is= 4/8*4/8*4/8*(4/8)=1/16
  exactly 3 blue= 4/8*4/8*4/8*(4/8)=1/16
  OR =1/16+1/16=1/8
  
• 12 years agoHelpfull: Yes(165) No(59)
• The order is also important here. RRRB or BBBR.
  RRRB has 4 possible arrangements and BBBR too. So a total of 8 arrangements. For each of these arrangements p=1/16 so
  Ans = 8/16 = 1/2
• 12 years agoHelpfull: Yes(55) No(67)
• yes i forgot the combinations ,thanks
  4 red,4 blue (asked exactly 3r or 3b)
  probability of taking red hat is=4(red hats)/8(total)// for all probabilites will be same in this problem
  exactly 3 red =4C3
  exactly 3 blue=4c3
  probability for drawing red hat is 4/8 note: he returning every hat to drawer
  firstly exactly 3 red is= 4C1{4/8*4/8*4/8*(4/8)}=1/4
  exactly 3 blue= 4C1{4/8*4/8*4/8*(4/8)}=1/4
  OR =1/4+1/4=1/2
• 12 years agoHelpfull: Yes(36) No(38)
• 4 red,4 blue (asked exactly 3r or 3b)
  probability for drawing red hat is 4/8 note: he returning every hat to drawer
  firstly exactly 3 red is= 4/8*4/8*4/8*(4/8)=1/16
  exactly 3 blue= 4/8*4/8*4/8*(4/8)=1/16
  OR =1/16+1/16=1/8
  hey this correct ans
  
• 12 years agoHelpfull: Yes(29) No(22)
• 4c3*4c1/8c4=1/8
• 11 years agoHelpfull: Yes(19) No(16)
• we need RRRB or BBBR for exact three red and blue hats. and in first case (RRRB)
  we get these 3 R in one case by probability
  (4c1/8c1)(R)*(4c1/8c1)(R)*(4c1/8c1)(R)*(4c1/8c1)(B)=(4c1/8c1)^4
  and we have 4 cases for exactly 3 R :(RRRB)(RRBR)(RBRRR)(BRRR)
  SO THE PROB FOR EXACTLY 3 R IS 4*((4c1/8c1)^4)
  similarly for 3 blue is 4*((4c1/8c1)^4)
  and for both cases 4*((4c1/8c1)^4)+4*((4c1/8c1)^4) =1/2
  ans. 1/2
  
  
  
• 12 years agoHelpfull: Yes(15) No(18)
• notice that here we have to take rrrb or bbbr
  case 1: rrrb
  in dis we can take first "r" den return it .hence for "rrr" we hav probability 1/2.aftr that "b" can take.all the hats are der.and probabilty to take "b" is also 1/2.
  so for take rrrb we have 1/2*1/2=1/4
  similarly case2:bbbr is 1/2*1/2=1/4
  
  hence 1/4+1/4=2/4=1/2
  
  
• 12 years agoHelpfull: Yes(9) No(4)
• Option (2) Correct
  pos s i bl e out comes=
  (RRRR)(RRRB)(RRBR)(RBRR)(BRRR)(BBBR)(BBRB)(BRBB)(RBBB)(RRBB)(BBRR)(RBBR)(BRRB)(RBRB)(BRBR)(BBBB)
  s o t ot al 16
  n we want exact l y 3 bl ue hat s
  s o 4/16=1/4 ans .
• 10 years agoHelpfull: Yes(9) No(5)
• @NAGARJUN..
  correct ansr is 1/2.can u plz elaborate ur ansr more..
• 12 years agoHelpfull: Yes(4) No(8)
• first of all find the n(S) which is taking 4 hats randomly out 0f 8.its mentioned that we should return the hat we took before taking next hat so n(s)=8c1*8c1*8c1*8c1=8*8*8*8.then find the n(a) which is getting exactly 3 red hats that means 3 red and one blue hats. n(a)=4c1*4c1*4c1*(4c1)=4*4*4*4.therefore p(A)=n(a)/n(s)=4*4*4*4/8*8*8*8=1/16.Similarly p(b)=n(b)/n(s)=(4)*(4)*(4)*4/8*8*8*8=1/16.OR=1/16+1/16=2/16=1/8.
• 10 years agoHelpfull: Yes(4) No(2)
• read it carefully you want either red or blue. answer seems to be 1/8 but "actually its 1/4" there are 4 red and 4 blue and u immediately return hat before taking out next so
  probability of getting red = probability of getting blue = 4/8 = 1/2
  now lets begin... I picked one hat. for the 1st pick color doesn't matter (i want either three red or three blue hats remember?) lets say i picked red now i returned it now i want two more red hats there are 4 red out of 8 probability of getting red is now 1/2
  in short,
  probability of getting any(red/blue) color * probability of getting same color * probability of getting same color again is (probability of getting any color is 1 always)
  1* 1/2 * 1/2 = 1/4 is the correct answer.
  
• 11 years agoHelpfull: Yes(3) No(8)
• Ans) 1/8
  Probability that exactly 3 red hats out of four : 1/2*1/2*1/2*1/2 (Return every hat)
  Probability that exactly 3 blue hats out of four : 1/2*1/2*1/2*1/2
  1/16+1/16=2/16 =>1/8
• 9 years agoHelpfull: Yes(2) No(1)
• DIVYA how can probability be greater than 1?
  
• 9 years agoHelpfull: Yes(2) No(0)
• they never asked for order to be RRRB OR BBBR..So ans should be 1/8
  
• 9 years agoHelpfull: Yes(2) No(0)
• Please Explain properly......!!!!
  Answer is right.... but how we get is still a mystery for me !!!!
  
  @supali: how we get p=1/16 ??
• 12 years agoHelpfull: Yes(1) No(4)
• i am not able to understand..plz elaborate it..
• 12 years agoHelpfull: Yes(1) No(5)
• 1/2 is the correct answer
• 10 years agoHelpfull: Yes(1) No(3)
• the answer is a) 1/2
• 9 years agoHelpfull: Yes(1) No(5)
• we have to take out exactly 3 red hat or three blue hat so our combination are 4c3*4c1 for red color hats and 1 blue color hat and again 4c1*4c3 for red and blue color hats..... and probability is not more than 1 ...so our solution is..
  
  (1/2*2*4c3*4c1)/8c4= 1/8 is the right answer.
• 9 years agoHelpfull: Yes(1) No(0)
• 8C4=70
  
  4C3=4 since two types of hats are there 4+4=8
  
  probability = 8/70 approx 1/8.7 but not confrim :(
  
  so 1/8
• 8 years agoHelpfull: Yes(1) No(1)
• ans : 1/2
  total possible outcomes are = RRRR RRRB RRBR RRBB RBRR RBRB RBBR RBBB BRRR BRRB BRBR BRBB BBRR BBRB BBBR BBBB
  probablitiy of getting exactly 3 red hats = 4/16
  3 blue hats = 4/16
  both= 4/16+4/16 = 1/2
• 7 years agoHelpfull: Yes(1) No(0)
• Total number of hats = 8
  We can select one hat from 8 hats in 8c1 ways.
  
  Total red hats = 4
  We can select one red hat from 4 red hats in 4c1 ways.
  
  Total blue hats = 4
  We can select one blue hat from 4 blue hats in 4c1 ways.
  
  Probability of selecting 3 red hats
  4c1/8c1 x 4c1/8c1 x 4c1/8c1 = 1/8
  
  Probability of selecting 3 blue hats
  4c1/8c1 x 4c1/8c1 x 4c1/8c1 = 1/8
  
  Add both of them (as OR is mentioned in question) -
  1/8 + 1/8 = 1/4
  
  Answer : 1/4
• 5 years agoHelpfull: Yes(1) No(0)
• ans is 1/4
  
• 9 years agoHelpfull: Yes(0) No(1)
• As the objects are replaced, the probability of drawing red or blue is equal.
  Probability to draw exactly 3 red hats and 1 blue hat = 12×12×12×12=116
  Similarly probability to draw exactly 3 blue hats and 1 red hat = 12×12×12×12=116
  Total probability = 116+116=18
• 9 years agoHelpfull: Yes(0) No(4)
• with replacement
  3 red hats can be possible in 4/8*4/8*4/8*4/8=1/16
  similarly 3 blue hats is 1/ 16
  therefore 1/16+1/16
  =2/16
  =1/8
• 5 years agoHelpfull: Yes(0) No(0)
• with replacement
  3 red hats can be possible in 4/8*4/8*4/8*4/8=1/16
  similarly 3 blue hats is 1/ 16
  therefore 1/16+1/16
  =2/16
  =1/8
• 5 years agoHelpfull: Yes(0) No(0)