Question

Write the probability distribution when three coins are tossed.
(a)

\[X\]	0	1	2	3
\[P\left( X \right)\]	\[\dfrac{1}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{1}{8}\]

 (b)

\[X\]	0	1	2	3
\[P\left( X \right)\]	\[\dfrac{1}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{5}{8}\]	\[\dfrac{7}{8}\]

(c)

\[X\]	0	1	2	3
\[P\left( X \right)\]	\[\dfrac{7}{8}\]	\[\dfrac{5}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{1}{8}\]

(d)

\[X\]	0	1	2	3
\[P\left( X \right)\]	\[\dfrac{1}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{5}{8}\]	\[\dfrac{1}{8}\]

Answer 1

Hint: We solve this problem by using the Bernoulli trials or binomial distribution. Here, if \['n'\] is number of times the event repeated and \['p,q'\] are probabilities of getting the particular result and not getting the particular result respectively then the probability distribution is given as
\[P\left( X=x \right)={}^{n}{{C}_{x}}.{{p}^{x}}.{{q}^{n-x}}\]
By using the above formula we find the probability distribution for \[x=0,1,2,3\].

Complete step by step answer:
We are given that the coin is tossed 3 times, so, let us assume
\[\Rightarrow n=3\]
Let us find the probability distribution of getting the head.
We know that is a coin is tossed then the probability of getting a head is given as
\[p=\dfrac{1}{2}\]
Similarly, we know that the probability of not getting a head as
\[q=\dfrac{1}{2}\]
Here, we can say that this distribution is Bernoulli trails.
We know that if \['n'\] is number of times the event repeated and \['p,q'\] are probabilities of getting the particular result and not getting the particular result respectively then the probability distribution is given as
\[P\left( X=x \right)={}^{n}{{C}_{x}}.{{p}^{x}}.{{q}^{n-x}}\]
Now, by substituting the required values in above formula we get
\[\Rightarrow P\left( X=x \right)={}^{3}{{C}_{x}}.{{\left( \dfrac{1}{2} \right)}^{x}}.{{\left( \dfrac{1}{2} \right)}^{3-x}}........equation(i)\]
Now, let us find the probability distribution for\[x=0,1,2,3\]
By substituting \[x=0\] in equation (i) we get
\[\begin{align}
  & \Rightarrow P\left( X=0 \right)={}^{3}{{C}_{0}}.{{\left( \dfrac{1}{2} \right)}^{0}}.{{\left( \dfrac{1}{2} \right)}^{3-0}} \\
 & \Rightarrow P\left( X=0 \right)=1\times 1\times \dfrac{1}{8}=\dfrac{1}{8} \\
\end{align}\]
Similarly, by substituting \[x=1\] in equation (i) we get
\[\begin{align}
  & \Rightarrow P\left( X=1 \right)={}^{3}{{C}_{1}}.{{\left( \dfrac{1}{2} \right)}^{1}}.{{\left( \dfrac{1}{2} \right)}^{3-1}} \\
 & \Rightarrow P\left( X=1 \right)=3\times \dfrac{1}{2}\times \dfrac{1}{4}=\dfrac{3}{8} \\
\end{align}\]
Similarly, by substituting \[x=2\] in equation (i) we get
\[\begin{align}
  & \Rightarrow P\left( X=2 \right)={}^{3}{{C}_{2}}.{{\left( \dfrac{1}{2} \right)}^{2}}.{{\left( \dfrac{1}{2} \right)}^{3-2}} \\
 & \Rightarrow P\left( X=2 \right)=3\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{3}{8} \\
\end{align}\]

Similarly, by substituting \[x=3\] in equation (i) we get
\[\begin{align}
  & \Rightarrow P\left( X=3 \right)={}^{3}{{C}_{3}}.{{\left( \dfrac{1}{2} \right)}^{3}}.{{\left( \dfrac{1}{2} \right)}^{3-3}} \\
 & \Rightarrow P\left( X=3 \right)=1\times \dfrac{1}{8}\times 1=\dfrac{1}{8} \\
\end{align}\]
Now, let us create a table containing the values of \[x,P\left( X \right)\] to get the probability distribution.

\[X\]	0	1	2	3
\[P\left( X \right)\]	\[\dfrac{1}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{3}{8}\]	\[\dfrac{1}{8}\]

So, the correct answer is “Option a”.

Note: Students will make mistakes in calculating the probability distribution for \[x=1,2,3\]. We have the formula for binomial distribution as
\[\Rightarrow P\left( X=x \right)={}^{3}{{C}_{x}}.{{\left( \dfrac{1}{2} \right)}^{x}}.{{\left( \dfrac{1}{2} \right)}^{3-x}}\]
While calculating the probability distribution for \[x=1,2,3\], they may calculate as
\[\Rightarrow P\left( X\le 1 \right)=P\left( X=0 \right)+P\left( X=1 \right)\]
\[\Rightarrow P\left( X\le 2 \right)=P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right)\]
\[\Rightarrow P\left( X\le 3 \right)=P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right)+P\left( X=3 \right)\]
This will give the wrong answer because the above mentioned formulas are for getting the head at least 1, 2, 3 respectively. But as we need only distribution we no need to go for at least. This part needs to be taken care of.