a fair coin is tossed 3 times in succession.if the first toss produced a head then the probability of getting exactly two heads in 3 tosses (including frist toss) is

• Probability=1/2
  Let getting exactly two heads be P(A)
  And Total possibilities be P(B)
  
  Then P(A)=(H,H,T),(H,T,H)=>2
  P(B)=(H,T,T),(H,H,T),(H,T,H),(H,H,H)=>4
  
  Probability=P(A)/P(B)=2/4=1/2
  
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• HTT
  HHH
  HTH
  HHT
  
  2/4=1/2
• 10 years agoHelpfull: Yes(2) No(0)
• In second and third the prob of head
  3/4
  1*3/4=3/4
• 10 years agoHelpfull: Yes(0) No(0)
• we need a one head in two tosses.........
  hh
  ht
  th
  tt
  prob is=1/2
  n first is already head=1
  1/2 and 1=1/2*1=1/2
• 10 years agoHelpfull: Yes(0) No(0)
• 3/4
  (h,h,h),(h,h,t),(h,t,h),(t,h,h)=4
  3 tosses=>
  3/4
• 10 years agoHelpfull: Yes(0) No(0)
• P(H)=1/2 , P(T)=1/2
  ANS:- P(H) * P(T) * P(H)+ P(H) * P(H) *P(T)=(1/2 * 1/2 * 1/2)+(1/2 * 1/2 *1/2)
  =(1/6)+(1/6)=2/6=1/3
  
• 10 years agoHelpfull: Yes(0) No(0)