Question

A dice is tossed twice, A success in getting an odd number on a toss, find the mean and variance of the probability distribution of the number of success.

Answer 1

Hint:
We are given that, dice are tossed two times, to find the mean and variance of the given event, we will first find the probabilities of a different number of success.
We let X = number of successes. Then using $ p\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}} $ we find probabilities. Then we use $ \sum \left( X \right)=\sum Xp\left( X \right) $ to find the mean and we use $ Var\left( X \right)=\sum {{\left( X \right)}^{2}}-{{\left[ \sum \left( X \right) \right]}^{2}} $ to find the variance of given data.

Complete step by step answer:
We are given that, dice tossed twice, success getting an odd number. We are asked to find the mean and variance of the number of successes.
Now we have that, success in getting an odd number.
Let X be the number of success since we are given that dice is tossed twice so we can have that, success in both tosses or in just either one of the two tosses or none of the toss results in success.
So it means we have that X can take values 0, 1 and 2.
Now we will find the probability of each X.
As we know that, when we throw a dice we have 1, 2, 3, 4, 5, 6 as outcomes in which odd ones are 3, 1, 5 that is out of 6 there terms are odd. So, the probability of success is half. So, $ p\left( S \right)=\dfrac{1}{2} $ So, $ p=\dfrac{1}{2} $ .
And probability of failure = 1 - probability of success. So, $ p\left( F \right)=1-p\left( S \right)\Rightarrow 1-\dfrac{1}{2}=\dfrac{1}{2} $ So $ q=\dfrac{1}{2} $ .
Now we find the probability of X=0, 1 and 2.
We know p(X=r) is given as $ p\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}} $ . As $ p=\dfrac{1}{2},q=\dfrac{1}{2}\text{ and }n=2 $ and n = 2 (two toss are done) so in X=0 we have,
 $ p\left( X=0 \right)={}^{2}{{C}_{0}}{{p}^{0}}{{q}^{2-0}} $ .
Using above values we get, \[\Rightarrow {}^{2}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{0}}{{\left( \dfrac{1}{2} \right)}^{2}}\].
Simplifying we get,
  $ \begin{align}
  & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4} \\
 & \therefore p\left( X=0 \right)=\dfrac{1}{4} \\
\end{align} $ .
Now P(X=1) be the probability of getting odd number once $ p\left( X=1 \right)={}^{2}{{C}_{1}}{{p}^{1}}{{q}^{2-1}} $ .
Using above values \[\Rightarrow {}^{2}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{1}}{{\left( \dfrac{1}{2} \right)}^{1}}\].
Simplifying we get,
 $ \begin{align}
  & \Rightarrow 2\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{2} \\
 & \therefore p\left( X=1 \right)=\dfrac{1}{2} \\
\end{align} $ .
And now P(X=2) be probability of getting odd number twice p(X=2) we get,
$ p\left( X=2 \right)={}^{2}{{C}_{2}}{{p}^{2}}{{q}^{2-2}}\Rightarrow 1{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{0}} $
Simplifying we get $ \therefore p\left( X=2 \right)=\dfrac{1}{4} $ .
So we have probability distribution as,

X	0	1	2
p(X)	$ \dfrac{1}{4} $	$ \dfrac{1}{2} $	$ \dfrac{1}{4} $

Now we are asked to find mean and we know mean is defined as $ \sum{{{x}_{i}}}p\left( {{x}_{i}} \right) $ and denoted by $ \sum \left( X \right) $ .
We have X=0, 1, 2 so, we have mean as $ \sum \left( X \right)=\sum\limits_{1\cdot 1}^{3}{{{x}_{i}}}p\left( {{x}_{i}} \right) $ .
Using probability distribution we get, $ \sum \left( X \right)=0\times \dfrac{1}{4}+1\times \dfrac{1}{2}+2\times \dfrac{1}{4} $ .
Simplifying we get $ \sum \left( X \right)=0+\dfrac{1}{2}+\dfrac{1}{2}=1 $ .
So mean is $ \sum \left( X \right)=1 $ .
Now we are also asked to find the variable we know variance is given as, $ Var\left( X \right)=\sum {{\left( X \right)}^{2}}-{{\left[ \sum \left( X \right) \right]}^{2}} $ .
So first we calculate $ \sum {{\left( X \right)}^{2}} $ .
Now $ \sum {{\left( X \right)}^{2}}=\sum\limits_{1x}^{n}{{{x}_{i}}^{2}}p\left( {{x}_{i}} \right) $ .
Using probability distribution table we get, $ \sum {{\left( X \right)}^{2}}=0\times \dfrac{1}{4}+{{1}^{2}}\times \dfrac{1}{2}+{{2}^{2}}\times \dfrac{1}{4} $ .
Solving we get, $ \sum {{\left( X \right)}^{2}}=0+\dfrac{1}{2}+4\times \dfrac{1}{4}\Rightarrow \sum {{\left( X \right)}^{2}}=\dfrac{3}{2}\cdots \cdots \left( 1 \right) $ .
Now we have $ \sum \left( X \right)=1\text{ and }\sum {{\left( X \right)}^{2}}=\dfrac{3}{2} $ so,
 $ Var\left( X \right)=\sum {{\left( X \right)}^{2}}-{{\left[ \sum \left( X \right) \right]}^{2}}\Rightarrow \dfrac{3}{2}-1 $ .
Simplifying we get $ Var\left( X \right)=\dfrac{1}{2} $ .
Note:
While finding the number of successes, we have to look for all possibilities. X = 0 is also there, having all the failure is also a case in which we can redefine it as a case of zero success. Also while summing the product to find the mean and variance we need to be focused, a minute mistake happens at a sudden $ {{3}^{2}}=6 $ which is wrong.
Always remember we need the value of H or P or B, not their square so root is not needed to find always.