Question

There is a group of 50 people who are patriotic, out of which 20 believe in non-violence. Two persons are selected at random. What is the probability that either both are patriotic or both are non-violent? Explain the importance on non-violence in patriotism.

Answer 1

For finding probability as we know that probability of outcome $=\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcome}}$, so for this we need to find out the total number favourable outcome of both are patriotic or both are non-violent, which we can easily find out with the help of binomial distribution i.e. ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!},n\ge r$.

Complete step-by-step solution:
Moving ahead with the question in a stepwise manner, we want to find out the probability that from the selection of two people either both should be violent, or both should be non-violent.
So according to question;
Total number of people $=50$
Number of non-violent from total people $=20$
So remaining will be the violent people $50-20=30$
As we know that total number of favourable outcomes is the total possible outcomes of two people from the total people i.e.${}^{50}{{C}_{2}}$ so total outcomes is;
${}^{50}{{C}_{2}}=\dfrac{50!}{2!\left( 50-2 \right)!}$
On simplifying it we will get;
$\begin{aligned} & {}^{50}{{C}_{2}}=\dfrac{50!}{2!48!} \\\ & {}^{50}{{C}_{2}}=1225 \\\ \end{aligned}$
Total outcomes is equal to $1225$
Now Favourable outcome for two non-violent out of total non-violent people. So that will be ${}^{20}{{C}_{2}}$, so we will get;
${}^{20}{{C}_{2}}=\dfrac{20!}{2!\left( 20-2 \right)!}$
On simplifying it we will get;
$\begin{aligned} & {}^{20}{{C}_{2}}=\dfrac{20!}{2!18!} \\\ & {}^{20}{{C}_{2}}=190 \\\ \end{aligned}$
So favourable number of outcomes for two person being non-violent is $190$
In the same way to calculate the total number of favourable outcomes for coming out two people violent out of total violent people. So that will be ${}^{30}{{C}_{2}}$, so we will get;
${}^{30}{{C}_{2}}=\dfrac{30!}{2!\left( 30-2 \right)!}$
On simplifying it we will get;
$\begin{aligned} & {}^{30}{{C}_{2}}=\dfrac{30!}{2!28!} \\\ & {}^{30}{{C}_{2}}=435 \\\ \end{aligned}$
So favourable number of outcomes for two person being violent is $435$
Now to calculate the probability that either both are patriotic or both are non-violent we know that probability of outcome $=\dfrac{\text{Number of favourable outcome}}{\text{Total}~\text{number}~\text{of outcome}~}$, so number of favourable outcome of both are non-violent we know that probability of outcome will be possible outcome of non-violent in addition with possible outcome of violent. In other words, we want the probability of either one, which will increase the possible number of favourable outcomes which will be equal to the addition of both violent and non-violent outcomes.
So number of favourable outcome is $190+435=625$
So probability of either both are patriotic or both are non-violent we know that probability of outcome$=\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcome}}$, now put the value in the formula, so we will get;
$\dfrac{625}{1225}= \dfrac{25}{49}$
So probability of that either both are patriotic or both are non-violent $\dfrac{25}{49}$
Hence the answer is $\dfrac{25}{49}$.

Note: Binomial theorem is used to find out the can be thought of as simply the probability of a success or failure outcome in an experiment or survey that is repeated multiple times whose formula is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!},n\ge r$.