Question

A two-sided coin (not the ‘Sholay’ coin) is tossed three times. What is the probability that ‘head’ will be the result exactly twice?
A.$\dfrac{3}{4}$
B.$\dfrac{2}{3}$
C.$\dfrac{1}{2}$
D.$\dfrac{3}{5}$
E.$\dfrac{3}{8}$

Answer 1

Hint: We are given in the question that a two-sided coin is not the ‘Sholay’ coin; it means the coin doesn’t have a head on both sides. A two-sided coin is a fair coin i.e., a coin has both tail and head. We will collect all the possible combinations for sample space. From that, we get the total number of possible outcomes. Then we get the probability by using the formula.

Complete step-by-step answer:
The coin is tossed three times, the possible number of outcomes is $2 \times 2 \times 2 = 8$.
Now each individual coin is likely to come up heads or tails then possible combinations we have are:
$HHH,HHT,HTT,HTH,TTT,TTH,THH,THT$
These are the sample space, sample space is nothing but a set of all possible outcomes of a random experiment. We denoted sample space in set notation, it is represented by $S$.
So we can write sample space in this format,
$S = \left\{ {HHH,HHT,HTT,HTH,TTT,TTH,THH,THT} \right\}$
To find the probability we need to divide the favorable outcomes by the number of outcomes. A favorable outcome means that the number of events we want from the experiment.
Here we need to find the probability that heads will be the result exactly twice. To determine the results we have to check from the sample space,
Here the favorable outcomes are $HHT,HTH,THH$
Therefore, the number of favorable outcomes is $3$.
So here we have the probability as,
$ \Rightarrow P\left( A \right) = \dfrac{3}{8}$
The probability that heads will be the result exactly twice is $\dfrac{3}{8}$.
So, the correct answer is “Option E”.

Note: This can also be solved by combinations method.
Probability that head comes twice = $^3{C_2} \times {\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{1}{2}$
 ( here $^3{C_2}$ means selecting two tosses out of three because we need exactly two heads, ${\left( {\dfrac{1}{2}} \right)^2}$ means probability of head occurring twice and $\dfrac{1}{2}$ means probability of tails occurring once ).
$ \Rightarrow \dfrac{{3 \times 2 \times 1}}{2} \times \dfrac{1}{4} \times \dfrac{1}{2}$
$ \Rightarrow \dfrac{3}{8}$