Question

Three fair coins are tossed. If both heads and tails appear, then the probability that exactly one head appears, is
a) $\dfrac{3}{8}$
b) $\dfrac{1}{6}$
c) $\dfrac{1}{2}$
d) $\dfrac{1}{3}$

Answer 1

Hint: Here we are asked to find the probability of getting exactly one head on tossing three coins and getting both heads and tails. First, we need to separate the events, that is the event of getting both heads and tails and the event of getting exactly one head. Then by using the formula we will find the required answer. The formula that we will be using for this problem is given in the formula section.

Formula: Formula that we need to know:
$P\left( {\dfrac{a}{b}} \right) = \dfrac{{P\left( {b \cap a} \right)}}{{P\left( a \right)}}$

Complete step-by-step answer:
It is given that three coins are tossed and if both heads and tails appear we aim to find the probability of getting exactly one head.
First let us find the size of the universal set, it is given that three coins are tossed so the size of the universal set will be ${2^3} = 8$ . Now let us define the events.
Let $A$ be the event of getting both heads and tails then its sample space will be $\left\{ {\left( {HHT} \right),\left( {HTH} \right),\left( {THH} \right),\left( {TTH} \right),\left( {THT} \right),\left( {HTT} \right)} \right\}$ . Thus, the size of the event $A$ is $6$ .
Let $B$ be the event of getting exactly one head then its sample space will be $\left\{ {\left( {TTH} \right),\left( {THT} \right),\left( {HTT} \right)} \right\}$ . Thus, the size of the event $B$ is $3$ .
Since we aim to find the probability of getting exactly one head when we get both tails and heads, we will use the formula $P\left( {\dfrac{a}{b}} \right) = \dfrac{{P\left( {b \cap a} \right)}}{{P\left( a \right)}}$ to find it.
Before finding it, we should know the probability of $P\left( {B \cap A} \right)$ let us find it first.
Here the event is an intersection between the events $B\& A$ . From the sample spaces of the events $B\& A$ , we get the size of the event $B \cap A$ to be $3$ .
Now let us move on to the formula. We need to have the probability of the events $B \cap A$ and $A$ .
Since we know that the size of the universal set is eight, we get $P\left( {B \cap A} \right) = \dfrac{3}{8}$ and $P\left( A \right) = \dfrac{6}{8}$ .
The probability of getting exactly one head when both are heads and tails is
$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {B \cap A} \right)}}{{P\left( A \right)}} = \dfrac{{\dfrac{3}{8}}}{{\dfrac{6}{8}}}$
On simplifying the above, we get
$P\left( {\dfrac{A}{B}} \right) = \dfrac{3}{8} \times \dfrac{8}{6}$
$P\left( {\dfrac{A}{B}} \right) = \dfrac{3}{6} = \dfrac{1}{2}$
Thus, we got the required probability and now let us find the correct option.
As we can see that options a, b, and d are not equivalent to $\dfrac{1}{2}$ thus, option c) $\dfrac{1}{2}$ is the correct option.

So, the correct answer is “Option c”.

Note: Probability is a branch of mathematics that deals with the occurring of random events. In simple words, probability is nothing but possibility. The probability of an event can be found by dividing the number of possible outcomes of an event by the size of its universal set (That is the number of all possible outcomes). Also, the probability of the sum of all the events in a sample space equals one.