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MHT CET202522 Apr 2025Morning ShiftMathematicsProbabilityActual

A coin is tossed until one head appears or a tail appears 4 times in succession. The probability distribution of the number of tosses is

Options

  1. AX 1 2 3 4 P ( X =x) 1 8 1 8 1 2 1 4
  2. BX 1 2 3 4 P ( X =x) 1 4 1 2 1 8 1 8
  3. CX 1 2 3 4 P ( X =x) 1 8 1 4 1 8 1 2
  4. DX 1 2 3 4 P ( X =x) 1 2 1 4 1 8 1 8

Correct answer

D. X 1 2 3 4 P ( X =x) 1 2 1 4 1 8 1 8

Step-by-step solution

The process terminates when a head appears or four consecutive tails occur. The number of tosses, X , has possible values 1 , 2 , 3 , and 4 . An outcome of a head on the first toss gives X = 1 , with probability P(X=1) = 1 2 . For X = 2 , the sequence is TH , so P(X=2) = 1 2 1 2 = 1 4 . For X = 3 , the sequence is TTH , giving P(X=3) = ( 1 2 )^3 = 1 8 . X = 4 occurs either with TTTH or TTTT , each with probability ( 1 2 )^4 = 1 16 , so P(X=4) = 1 16 + 1 16 = 1 8 . Summing the probabilities confirms 1 2 + 1 4 + 1 8

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