Three coins are tossed. If at least two coins show head, the probability of getting one tail is:

- a)3/4
- b)1/3
- c)1
- d)2/3

Correct answer is option 'A'. Can you explain this answer?

3 Answers

Amhe Nadi
answered
Aug 16, 2018

Sample space = HHH, HHT, HTH, THH, TTH, THT, HTT, TTTGetting atleast two heads(F) = HHH, HHT, HTH, THHProbability of getting atleast two heads(F) = 4/8 =1/2Getting one tail (E) = HHT, HTH, THHE intersection F = HHT, HTH, THHP(E intersection F) = 3/8Required probability = P(E intersection F) / P(F) = 3/8 ÷ 1/2 =3/4

Seblewongel Girma
answered
May 03, 2019

Subset={HHH , HHT,HTH,HTT,THH,THT,TTH,TTT}

P(at least two head and one tail)=6/8=3/4

P(at least two head and one tail)=6/8=3/4

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