
Find the chance of throwing head and tail alternatively in three successive tossings of the coin.
Answer
602.7k+ views
Hint: First of all, consider an experiment of throwing a coin thrice and then find out the total number of possible outcomes and the total number of favorable outcomes. Then find the probability of that considered experiment to get the required answer. So, use this concept to reach the solution to the given problem.
Complete step-by-step solution:
Consider an experiment of throwing a coin thrice.
The sample space of the experiment can be written as \[S = \left\{ {HHH,HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}\]
Hence, the total number of outcomes = 8
Let \[E\]: event of getting head and tail alternatively in three successive trails.
So, the only possible outcomes are \[\left\{ {HTH,THT} \right\}\].
Hence, the number of possible outcomes = 2
We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Therefore, \[P\left( E \right) = \dfrac{2}{8} = \dfrac{1}{4}\]
Thus, there is a chance of throwing head and tail alternatively in three successive tossings of the coin is \[\dfrac{1}{4}\].
Note: In probability theory, an experiment or trail is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as sample space. The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. The probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}\].
Complete step-by-step solution:
Consider an experiment of throwing a coin thrice.
The sample space of the experiment can be written as \[S = \left\{ {HHH,HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}\]
Hence, the total number of outcomes = 8
Let \[E\]: event of getting head and tail alternatively in three successive trails.
So, the only possible outcomes are \[\left\{ {HTH,THT} \right\}\].
Hence, the number of possible outcomes = 2
We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Therefore, \[P\left( E \right) = \dfrac{2}{8} = \dfrac{1}{4}\]
Thus, there is a chance of throwing head and tail alternatively in three successive tossings of the coin is \[\dfrac{1}{4}\].
Note: In probability theory, an experiment or trail is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as sample space. The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. The probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}\].
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