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

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