Elitmus
Exam
Numerical Ability
Probability
11 coins draw sequentially.what is the probability of 6th and 8th no. coin given a head
Read Solution (Total 9)
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- 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 - 5 years agoHelpfull: Yes(6) No(0)
- option please?
- 6 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
- 5 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 - 6 years agoHelpfull: Yes(1) No(16)
- (a)1/2^8 (b)1/2^9 (b)1/4 (c)1/4*2^9
- 6 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.. - 6 years agoHelpfull: Yes(0) No(6)
- you gave exam in september?
- 6 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)
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