11 coins draw sequentially.what is the probability of 6th and 8th no. coin given a head

• as the events are independent
  answer=probability of getting head at 6th pos * probability of getting head at 8th pos
  =1/2 * 1/2
  =1/4
• 4 years agoHelpfull: Yes(6) No(0)
• option please?
• 5 years agoHelpfull: Yes(5) No(3)
• There is 2^11 outcomes
  Out of it 2^9 gives head at 6&8 so probability is 1/4
• 5 years agoHelpfull: Yes(3) No(0)
• answer is 1/2 as each case will have the same probabilty of one head and one tail similarly 6th and 8th will also have the probabilty of that only
• 4 years agoHelpfull: Yes(2) No(2)
• Head= H
  Tail= T
  Probability- T T T T T H T H _ _ _
  Till 6th no it will be constant and after that the probability will change so it will be 3/2^3 = 3/8
• 5 years agoHelpfull: Yes(1) No(16)
• (a)1/2^8 (b)1/2^9 (b)1/4 (c)1/4*2^9
• 5 years agoHelpfull: Yes(0) No(6)
• always solve to possibilty...one is wrong and one is right..all coins have 1/2 possibility..then 1/2*1/2..11term
  then answer is 1/4*2^9
  bt other solution
  first coin always given head or tail so 1,2,3,4,5,7,9,10,11 and head or tail comes possibilty in these coins always (1+1)/2..then answer always given 1/4....so i m confuse choosen one answer..
• 5 years agoHelpfull: Yes(0) No(6)
• you gave exam in september?
• 5 years agoHelpfull: Yes(0) No(21)
• for 6th coin probability of getting head= 1/2
  for 8th coin probability of getting head= 1/2
  for Both the cases P(E)= 1/2.
• 3 years agoHelpfull: Yes(0) No(0)