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

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

& \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{aligned}$$ Similarly, by substituting $$x=1$$ in equation (i) we get $$\begin{aligned} & \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{aligned}$$ Similarly, by substituting $$x=2$$ in equation (i) we get $$\begin{aligned} & \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{aligned}$$ Similarly, by substituting $$x=3$$ in equation (i) we get $$\begin{aligned} & \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{aligned}$$ Now, let us create a table containing the values of $$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.