Question

Amit tosses a fair coin twice, and let \[X\] be defined as the number of heads he observes. Find the Probability mass function \[{P_X}\].
A) \[\dfrac{1}{3}\]
B) \[\dfrac{1}{8}\]
C) \[\dfrac{1}{4}\]
D) None of these

Answer 1

Hint:
We are required to find the Probability mass function of the given event. The Probability mass function characterizes the distribution of the discrete random variables. Here, we will first find the sample space of the given event. Based on the sample space, we will calculate the probability mass function.
Formula Used: We will use the formula \[{\text{Probability = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}\] in solving the question.

Complete step by step solution:
Let \[Z\] be a discrete random variable of a function, then the probability mass function of the random variable \[Z\] is given as follows –
\[{P_z}\left[ Z \right] = P\left[ {Z = z} \right]\], for all \[z\] belongs to the range of \[Z\].
Here the Range \[\left[ Z \right]\] is a countable set which is written as \[\left[ {{z_1},{z_2},{z_3},....} \right]\]. Hence, the random variable \[Z\] takes the values as \[{z_1},{z_2},{z_3},.....\].
Also, the main conditions for the Probability mass function are –
\[\begin{array}{l}{P_z}\left[ Z \right] \ge 0\\\sum\nolimits_{z \in Range\left[ Z \right]} {{P_z}\left[ Z \right] = 1} \end{array}\]
Now, let us solve the question.
Let the event \[X\] be defined as the number of heads observed when two coins are tossed.
Sample space when two coins are tossed is as follows,
\[S = \left\{ {HH, HT, TH, TT} \right\}\]
Now we know that the formula for the Probability is –
\[{\text{Probability = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}\]
From the above sample space, we can see that –
There is only one favorable outcome \[\left\{ {TT} \right\}\], out of the total four outcomes. Now, we will find the probability of getting 0 head using the formula of probability.
\[\begin{array}{c}{\text{Probability = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}\\ = \dfrac{1}{4}\end{array}\]
 So, the probability of getting 0 head is \[\dfrac{1}{4}\].
There are only two favorable outcomes \[\left\{ {HT,TH} \right\}\], out of the total four outcomes. Now, we will find the probability of getting 1 head using the formula of probability.
\[\begin{array}{c}{\text{Probability = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}\\ = \dfrac{2}{4}\end{array}\]
So, the probability of getting 1 head is \[\dfrac{2}{4}\].
There is only one favorable outcome \[\left\{ {HH} \right\}\], out of the total four outcomes. . Now, we will find the probability of getting 2 heads using the formula of probability.
\[\begin{array}{c}{\text{Probability = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}\\ = \dfrac{1}{4}\end{array}\]
 So, the probability of getting 2 heads is \[\dfrac{1}{4}\].
Based upon the above information the Probability mass function can be made as follows –

\[X\]	0	1	2
\[P\left[ X \right]\]	\[\dfrac{1}{4}\]	\[\dfrac{2}{4}\]	\[\dfrac{1}{4}\]

Now we will verify our Probability mass function. We know that when the sum of all the probabilities is 1, then, it is a Probability mass function.
Hence, we add all the Probabilities,
\[\begin{array}{c}P\left[ {X = 0} \right] + P\left[ {X = 1} \right] + P\left[ {X = 2} \right] = \dfrac{1}{4} + \dfrac{2}{4} + \dfrac{1}{4}\\ = \dfrac{4}{4}\end{array}\]
Dividing 4 by 4, we get
\[P\left[ {X = 0} \right] + P\left[ {X = 1} \right] + P\left[ {X = 2} \right] = 1\]
As the sum of all the probabilities has come 1, hence, the Probability mass function is verified.
We can observe that \[\dfrac{1}{4}\] represents one of the probability from the probability mass function.
Now out of all the provided choices, only option [C] represents a probability from this Probability mass function.

\[\therefore\] Option (C)is the correct answer.

Note:
In probability theory, the sample space of an experiment or random trial can be defined as the set of all the possible outcomes or results of that experiment. The sample space is usually denoted with the help of the set notation. All the possible ordered outcomes are listed as the elements in the set.