Question

Let \[p,q\] be chosen one by one from the set \[\{ 1,\sqrt 2 ,\sqrt 3 ,2,e,\pi \} \] with replacement. Now a circle is drawn taking \[\left( {p,q} \right)\] as its centre. Then the probability that at the most two rational points exist on the circle is (rational points are those points whose both the coordinates are rational)
Choose the correct option:-
A. $\dfrac{2}{3} $
B. $\dfrac{7}{8} $
C.$ \dfrac{8}{9} $
D.None of these.

Answer 1

Hint: For solving this particular question we have to consider the probability of the number chosen from the given set with replacement be $ 20 $ . Then \[P\](at most two rational number ) is equal to $ \dfrac{{4 \times {}^5{C_1}}}{{^5{C_1}{ + ^5}{C_2}}} $ , then we have to simplify the given combinations as the number of combinations of $ n $ different things taken $ r $ at a time is given be $ {}^n{C_r} $ .
Then, $ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $ .
Formula Used: The number of combinations of $ n $ different things taken $ r $ at a time is given be $ {}^n{C_r} $ .
Then, $ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $ .

Complete step by step solution:
A set is given that is , \[\{ 1,\sqrt 2 ,\sqrt 3 ,2,e,\pi \} \] .
The probability of the number chosen from the given set with replacement is $ 4 \times {}^5{C_1} = 20 $ ,
\[ \Rightarrow P\](at most two rational number ) is equal to $ \dfrac{{4 \times {}^5{C_1}}}{{^5{C_1}{ + ^5}{C_2}}} $ ,
 $ \Rightarrow \dfrac{{20}}{{5 + 10}} = \dfrac{4}{3} \times \dfrac{{{}^5{C_2}}}{{{}^5{C_3}}} $
 $ = \dfrac{8}{9} $
Hence, we get the required result, that is $ \dfrac{8}{9} $ .
Therefore, option ‘C’ is the correct option.
So, the correct answer is “Option C”.

Note: We have to simplify the given combinations as the number of combinations of $ n $ different things taken $ r $ at a time is given be $ {}^n{C_r} $ .
Then we used , $ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $ .
If we have questions similar in nature as that of above can be approached in a similar manner and we can solve it easily and can find the corresponding result.