A man has 4 childrens,atleast 1 of whom is a girl. Find the probability that he has 3 girls and 1 boy.

• In the question it s given that he has atleast one girl, so accordingly we can say that the possibilities may be
  {(1girl,3boys),(2girl,2boys),(3girls,1boy)(4girls,0boys)}
  And the favourable event is 3girls 1boy
  so the probability is 1/4
• 10 years agoHelpfull: Yes(43) No(7)
• (4C3)/(4c1+4C2+4C3+4C4)
  =4/15
• 10 years agoHelpfull: Yes(9) No(4)
• 1/4..total chances(g,b)=(1,3),(2,2),(3,1)(4,0) out of which possible outcomes only 1 i.e.,(3,1)
• 10 years agoHelpfull: Yes(6) No(2)
• One is atleast girl so keeping aside that girl there are 3 children.
  Now 2 must be girls from these three(one we kept aside)
  So Favourable cases= 3 C 2
  Overall Cases= 2^3
  So, Probability= 3 C 2 / 2^3
  = 3/8
  
  Correct me if i am wrong....
• 10 years agoHelpfull: Yes(2) No(3)
• at least one is girl remaining children is 3.
  so two must be girls 3c2=3 and 1 is boy
  3*1=3
  
• 10 years agoHelpfull: Yes(0) No(7)
• the sample space is 1 girl or 2 girls or 3 girls or 4 girls.
  there the denominator will be 4c1+4c2+4c3+4c4=4+10+4+1=19
  therefore the prob. that the man has 3 girls and 1 boy is
  (4c3*1c1)/19=4/19.
• 10 years agoHelpfull: Yes(0) No(0)
• 1/8...as there is already one girl..so among rest of 3,having 2 girls and 1 boy's probability is 1/2 *1/2 *1/2=1/8
  
• 10 years agoHelpfull: Yes(0) No(0)
• 4/15 is correct answer :)
• 8 years agoHelpfull: Yes(0) No(1)