A drawer has 4 red hat & 4 blue hat .find d prblty f getting xatly 3 blue hats when taking out 4 hats randomly out f d drawer and immediatly returning every hat to the drawer before taking out the next???

• (4C3*4C1)/8C4
• 9 years agoHelpfull: Yes(28) No(9)
• 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
  ans :1/4
• 9 years agoHelpfull: Yes(21) No(2)
• probability for taking first blue hat is 4/8
  second one also 4/8
  for third one also 4/8
  now the fourth hat must be red because they mentioned they require exactly 3 blue hats.
  probability for red is also 4/8
  so in total = (4/8)^4
• 9 years agoHelpfull: Yes(7) No(3)
• 4c3*4c1/(8c4)=8/35
• 9 years agoHelpfull: Yes(4) No(1)
• first of all we have to take one blue hat out of 8 hats which can be done in 4c1/8c1 ways.
  and then we return it to drawer and take another blue hat in 4c1/8c1 ways
  and then we return it to drawer and take another blue hat in 4c1/8c1 ways
  next 4th hat is taken in 8c1/8c1 ways
  so total probability is (4c1*4c1*4c1*8c1)/(8c1*8c1*8c1*8c1)
• 9 years agoHelpfull: Yes(2) No(2)
• p(r)=1/4 p(b)=1/4
  now our objective=bbbr*4!/3!
  as blue hats are returned back so p(b) remains 1/4
  our ans=1/4*1/4*1/4*1/4*4!/3!=1/64
• 9 years agoHelpfull: Yes(2) No(1)
• 1/5
  
  (1/2*1/2*1/2)/(1/2*1/2*1/2+1/2)
• 9 years agoHelpfull: Yes(1) No(0)
• (4c3*4c1)/8c4
• 9 years agoHelpfull: Yes(1) No(1)
• P(BBBR) = (1/2)^4
  P(BBRB) = (1/2)^4
  P(BRBB) = (1/2)^4
  P(RBBB) = (1/2)^4
  total = p(BBBR)+P(BBRB)+P(BRBB)+P(RBBB)=1/4
  ans :1/4
• 9 years agoHelpfull: Yes(1) No(0)
• sabse pehla wala galat hai soln
  
  
  
• 9 years agoHelpfull: Yes(1) No(2)
• ((1/4)^4)*4/8c1 multiply 4 becoz 4cases exits
• 9 years agoHelpfull: Yes(0) No(0)
• to take a blue hat - 4c1/8c1.Give it immediately.
  another blue hat - 3c1/7c1,give it.
  another - 2c1/6c1.
  Exactly three blue hats done and given.
  Now 4th hat to be red hat - 4c1/5c1 (5c1 bcoz,3blue hats given and 4 red n 1blue hat are the remaining ones)
  So, (4c1/8c1) * (3c1/7c1) * (2c1/6c1) * (4c1/5c1)
• 9 years agoHelpfull: Yes(0) No(3)
• 4c3*4c1/8c4
• 9 years agoHelpfull: Yes(0) No(0)
• (4c1/8c1)*4=1/16
  now 3 exactly blue hats can be obtained in following ways(bbbr),(bbrb),(brbb),(rbbb)so 4*(1/16)=1/4.
• 9 years agoHelpfull: Yes(0) No(0)
• RRRB HAT CAN BE ARRANGED IN 4 WAYS BECAUSE PERMUTATION OF RRRB IS 4!/3!=4
  PROBABILITY=RRRB+RRBR+RBRR+BRRR
  = 4*(4*4*4*4)/(8*8*8*8)=1/4
• 9 years agoHelpfull: Yes(0) No(0)
• plz any one say what is crct ans for this.
• 9 years agoHelpfull: Yes(0) No(0)
• (4c1*4c1*4c1*8c1)/(8c1*8c1*8c1*8c1)
• 9 years agoHelpfull: Yes(0) No(0)
• 4C1*4C1/8C1
• 9 years agoHelpfull: Yes(0) No(0)
• 4c1*4c1*4c1*4c1/8c4=128/35
• 9 years agoHelpfull: Yes(0) No(0)
• We can choose in (3,1)and(1,3) ways.
  so for 1st condition it is (4/8)*(4/8)*(4/8)*(4/8)*(4/8)=1/16
  similarly in second condition it is also 1/16
  so total probability is 1/16 + 1/16=2/8
• 8 years agoHelpfull: Yes(0) No(0)
• Two types solution exists here:
  1)
  if all balls are identical then
  8c3/10!
  
  2)
  if all balls are not identical :
  7!*8c3*3!/10!
• 8 years agoHelpfull: Yes(0) No(1)
• 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
• 8 years agoHelpfull: Yes(0) No(0)