Out of 800 families with 4 children each, how many families would you expect to have 2 boys and 2 girls in each family?

500
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100
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300

• for this type of problem,always remembr d follwng patterns:The sample space for a family having four children is:
  
  {(BBBB), (GGGG), (BGGG), (GBGG), (GGBG), (GGGB), (BBGG), (BGBG), (BGGB), (GBGB), (GBBG), (GGBB), (BBBG), (BBGB), (BGBB), (GBBB)}
  
  B - stands for Boy and G - stands for girl
  
  a) For a family to have two boys and two girls it is: {(BBGG), (BGBG), (BGGB), (GBGB), (GBBG), (GGBB)} = 6 elements
  Its probability = 6/16 = 3/8 = 0.375
  
  Hence, the probability for 800 families = (0.375) x(800) = 300
  
  b) A family to have atleast one boy = 1 - P(All gilrs) = 1 - 1/16 = 15/16 = 0.9375
  
  Hence, the probability for 800 families = (0.9375) x (800) = 750
  
  c) Probability of children of both sexes = 1 - {P(all boys) + P(all girls)}
  
  = 1 - 1/8 = 7/8 = 0.875
  
  Hence, the probability for 800 families = (0.875) x(800) = 700
• 10 years agoHelpfull: Yes(39) No(0)
• ans: 300
  
  number of ways in which there are 2girls and 2 boys is 4c2
  total cases is 2^4 = 16
  
  probability of 2boys2girls is (4c2)/16
  
  number of families is 800 * (6/16) = 300
• 10 years agoHelpfull: Yes(23) No(3)
• There are 16 cases in this one boy and three girls can happen so we get 4 cases in that BGGG,GBGG,GGBG,GGGB.Two boys and two girls we have 6 cases they are GGBB,BBGG,BGBG,GBGB,GBBG,BGGB.Three boys one girl 4 cases GBBB,BGBB,BBGB,BBBG.4 boys we have one case and 4 girls we have 1 case .So total 16 cases but in this we want two boys and two girls so 6 cases so probability is 6/16*800=300
• 10 years agoHelpfull: Yes(12) No(2)
• @ chandra sekhar reddy : good answer! but can you please explain how total cases are 2^4.
• 10 years agoHelpfull: Yes(6) No(0)
• Number of Boys: 0 1 2 3 4
  Number of girls: 4 3 2 1 0
  Total number of families = N = 800
  Probability of finding a boy = p = 1/2 = 0.5
  q = 1 - p = 1 - 1/2 = 1/2 = 0.5
  Number of children in each family = n = 4
  Use Binomial distribution
  P(r) = nCr*q^(n-r)*p^r
  a) Number of families expected to have 2 boys and 2 girls = 800*4C2*0.5^2*0.5^2
  = 800*6*0.25*0.25 = 300
  b) Number of families expected to have atleast 1 boy = 800*(1-P(r=0))
  = 800*(1-(4C0*0.5^4*0.5^0))
  = 800*(1-1*0.0625*1)
  = 800*0.9375 = 750
  c) Number of families expected to have children of both sexes = 800*(4C1*0.5^3*0.5^1 + 4C2*0.5^2*0.5^2 + 4C3*0.5^1*0.5^3)
  = 800*(4*0.125*0.5 + 6*0.25*0.25 + 4*0.5*0.125)
  = 800*(0.25 + 0.375 + 0.25)
  = 800* 0.875 = 700
• 10 years agoHelpfull: Yes(4) No(3)
• 300.
  
  No. of ways a family can have 2 girls & 2boys =4C2
  
  Total cases=4^2( as first child can b girl or boy..same as with other childs)
  
  Probabilty= 6/16
  
  Total families= 800*(6/16)
  =300
• 10 years agoHelpfull: Yes(3) No(1)
• Actually its not 4C2!!
  we can take it as arranging 2 boys & 2 girls in 4 places.
  so it will be= 4!/ (2! * 2!) { n!/p!*q!}
  = 6
• 10 years agoHelpfull: Yes(2) No(2)