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: Here we find the possible outcomes of fair coin which is tossed two times. Write the possibilities of the number of heads amit observes. Taking X as the number of heads observed we write the probability of the number of heads in each case. Calculate the probability mass function as the sum of all the probabilities obtained.
* If X is a discrete random variable of a function. Then the probability mass function of a random variable X is given by \[{P_X}(x) = P(X = x)\]\[\forall x \in \] Range of x
* Probability of an event is given by dividing the number of favorable outcomes by total number of outcomes.

Complete step-by-step answer:
We are given Amit tossed a fair coin twice.
So, we take the possibilities in terms of ordered pairs where the first element represents the first toss and second element represents the second toss.
Let H denote that heads appears on the toss of a fair coin.
Let T denote that tails appears on the toss of a fair coin.
Then for two tosses, possible outcomes are\[(HH,HT,TH,TT)\]
Total number of outcomes is 4
Let X denotes the number of times Amit observes heads.
We see from the total outcomes that heads appears 0 times, 1 time and 2 times.
Therefore, \[X = \left\{ {0,1,2} \right\}\]
Now we find probabilities for each value of x belonging to X.
Zero heads appear on toss:
Number of times zero heads appear is 1
Total number of outcomes is 4
Probability of heads observed zero times by Amit \[ = \dfrac{1}{4}\]
So, \[P(0) = \dfrac{1}{4}\]
One head appear on toss:
Number of times one head appears in two tosses is 2
Total number of outcomes is 4
Probability of one heads observed by Amit \[ = \dfrac{2}{4}\]
So, \[P(1) = \dfrac{2}{4}\]
Cancel same terms from numerator and denominator
\[P(1) = \dfrac{1}{2}\]
Two heads appear on toss:
Number of times two heads appears in two tosses is 1
Total number of outcomes is 4
Probability of two heads observed by Amit \[ = \dfrac{1}{4}\]
So, \[P(2) = \dfrac{1}{4}\]
We draw a Frequency table for the given data

Number of heads observed (x)	0	1	2
Probability mass function (P(x))	\[\dfrac{1}{4}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{4}\]

Then using the formula we can write probability mass function of X as
\[{P_X}(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{1}{4}}&{;x = 0,2} \\
  {\dfrac{1}{2}}&{;x = 1}
\end{array}} \right\}\]
Therefore, Probability mass function is \[{P_X}(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{1}{4}}&{;x = 0,2} \\
  {\dfrac{1}{2}}&{;x = 1}
\end{array}} \right\}\]
So, option D is the correct answer.

Note: Students might make mistakes in writing the probabilities as they might ignore the case with zero heads as there are no heads to observe. But we have to take the case where Amit doesn’t observe a head also, which will give probability \[\dfrac{1}{4}\]
Also, keep in mind that we take the observations in ordered pair form because the coin was tossed two times. Always write the probability to its simplest form i.e. without any common factor between the numerator and the denominator.