A coin is tossed 3 times. The probability of getting a head and a tail alternately is:

  1. \(\dfrac 1 4\)
  2. \(\dfrac1 8\)
  3. \(\dfrac 1 2\)
  4. \(\dfrac 3 8\)

Answer (Detailed Solution Below)

Option 1 : \(\dfrac 1 4\)
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Detailed Solution

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Concept:

The probability of the occurrence of an event A, out of total possible outcomes N, is given by P(A) = \(\rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.

 

Calculation:

The total number of different possible outcomes (N) in tossing a coin 3 times is 23 = 8.

For getting a head and a tail alternately, the possibilities are HTH, THT → 2 possibilities n(A).

∴ Required probability = \(\rm \dfrac{n(A)}{N}=\dfrac{2}{8}=\dfrac{1}{4}\)

Alternate Method

The possible set of A coin is tossed 3 times is {HHH}{HHT}{HTH}{HTT}{TTT}{TTH}{THT}{THH} = 8

The probability of getting a head and a tail alternately is {HTH}{THT} = 2

So, required probability = \(\rm \dfrac{n(A)}{N}=\dfrac{2}{8}=\dfrac{1}{4}\)

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