Question

A student has to match three historical events Dandi March, Quit India movement and Mahatma Gandhi Assassination with years 1948, 1930 and 1942. The student has no knowledge of the correct answers and decides to match the events and years randomly. If X denote the number of correct answers he gets, the mean of x is:
(a) 0.5
(b) 1
(c) 1.5
(d) 2

Answer 1

Hint: First, before proceeding for this, we must know the following question is used to get the probability of the cases in different conditions. Then, we have the total value of the outcome as 6. Then, we can draw a table with random variable X representing the number of correct matches as 0, 1, 2 and 3 and P(X) being their corresponding probability. Then, by using the formula of the mean E(X) for the discrete data is given by $ E\left( X \right)=\sum\limits_{i=0}^{n}{{{p}_{i}}{{x}_{i}}} $ .

Complete step-by-step answer:
In this question, we are supposed to find the value of the mean of the match problem in which a student has the randomness to match any event with any year.
So, before proceeding for this, we must know the following question is used to get the probability of the cases in different conditions.
Then, we require the total number of cases for the above question is given by:
 $ 3\times 2\times 1=6 $
So, we have the total value of the outcome as 6.
Now, we need to find the probability of each case in which student gets zero correct, 1 correct, 2 correct and all 3 correct.
Here, we can see that there are three questions in which exactly 2 correct can't be a case as if two are correct then automatically third will also be correct.
So, the probability of two being correct is zero.
Also, from the above discussion we get the probability of three being correct as $ \dfrac{1}{6} $ .
Now, the probability of none is correct will be in 2 cases only, so the probability of none correct is $ \dfrac{2}{6}=\dfrac{1}{3} $ .
Now, the condition for only one be correct is having 3 cases in which either 1, 2, 3 is correct, so the probability of 1 being correct is $ \dfrac{3}{6}=\dfrac{1}{2} $ .
Then, we can draw a table with random variable X representing the number of correct matches as 0, 1, 2 and 3 and P(X) being their corresponding probability as:

X	P(X)
0	$ \dfrac{1}{3} $
1	$ \dfrac{1}{2} $
2	0
3	$ \dfrac{1}{6} $

So, now by using the formula of the mean E(X) for the discrete data is given by:
 $ E\left( X \right)=\sum\limits_{i=0}^{n}{{{p}_{i}}{{x}_{i}}} $
Then, by substituting the values of the random variable and their probabilities from the table, we get:
 $ \begin{align}
  & E\left( X \right)=0\times \dfrac{1}{3}+1\times \dfrac{1}{2}+2\times 0+3\times \dfrac{1}{6} \\
 & \Rightarrow E\left( X \right)=0+\dfrac{1}{2}+0+\dfrac{1}{2} \\
 & \Rightarrow E\left( X \right)=1 \\
\end{align} $
So, the mean of the above question is 1
So, the correct answer is “Option B”.

Note: In this type of the question, we must be careful before finding the probability of different cases as required in the question as if we get one of the probabilities wrong, then the entire solution will go in vain. Moreover, we should be aware of the terms like mean, variance and standard deviation for the data as:
Mean is given by $ E\left( X \right)=\sum\limits_{i=0}^{n}{{{p}_{i}}{{x}_{i}}} $
Variance is given by $ V\left( X \right)=E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}} $
Standard deviation is given by $ \sigma =\sqrt{E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}}} $ .