An unbiased coin is tossed 3 times, if the third toss gets head what is the probability of getting at least one more head?

  1. \(3\over 4\)
  2. \(1\over 4\)
  3. \(1\over 2\)
  4. \(1\over 3\)

Answer (Detailed Solution Below)

Option 1 : \(3\over 4\)
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Detailed Solution

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Concept: 

  • The number of ways for selecting r from a group of n (n > r) = nCr 
  • The probability of particular case \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

Calculation

If it is known that third toss gets head, the possible cases:

(H, H, H), (H, T, H), (T, H, H), (T, T, H)

∴ Total cases possible = 4

Total favourable cases = 3 [(H, H, H), (H, T, H), (T, H, H)]

So, required probability P = \(\rm \text{Total favorable cases}\over\text{Total possible cases}\)

P = \(3\over4\)

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