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Question: Numbers \[1,2,3.......,2n\,\,\,(n\in N)\] are written down on \[2n\,\,\] cards. One number is select...

Numbers 1,2,3.......,2n   (n∈N)1,2,3.......,2n\,\,\,(n\in N) are written down on 2n  2n\,\, cards. One number is selected at random from these cards. If the probability of drawing a number 'n' is proportional to 'r', then the probability of drawing an even number in one draw is
1. 12n+1\dfrac{1}{2n+1}
2. n+2n+4\dfrac{n+2}{n+4}
3. n2n+1\dfrac{n}{2n+1}
4. None of these

Explanation

Solution

In this particular problem there are total 2n cards and it is mentioned that if P(r)P(r) is the probability that the number r is drawn in one draw, it is given that means P(r)  ∝  rP(r)\,\,\propto \,\,r that means we can also write p(r)  =  krp(r)\,\,=\,\,kr after applying this simplify further you will get the probability of even number.

Complete step-by-step solution:
In the above problem series is given up to 2n2n that is
1,2,3.......,2n  1,2,3.......,2n\,\,Where, (n∈N)(n\in N)
In the question it has mentioned another condition that
If P(r)P(r) is the probability that the number r is drawn in one draw, it is given that means P(r)  ∝  rP(r)\,\,\propto \,\,r
We can also write P(r)  =  krP(r)\,\,=\,\,kr
By remembering the concept of probability that ∑Pi(x)=1\sum{{{P}_{i}}(x)}=1
It can also be written as P(1)+P(2)+P(3)+.....+P(2n)=1P(1)+P(2)+P(3)+.....+P(2n)=1
As above condition is mentioned that P(r)  =  krP(r)\,\,=\,\,kr
That means
k(1+2+3+.....+2n)=1−−(1)k(1+2+3+.....+2n)=1--(1)
As we know, the sum of the number of series up to n is n(n+1)2\dfrac{n(n+1)}{2}
Similarly the sum of the number of series up to 2n so we can replace n as 2n in the above formula we get:
=2n(2n+1)2=\dfrac{2n(2n+1)}{2} Substitute in equation (1)
nk(2n+1)=1nk(2n+1)=1
From here we get the value of k
k=1n(2n+1)k=\dfrac{1}{n(2n+1)}
Now, we have to find the probability of even numbers that is
The required probability =P(2)+P(4)+P(6)+.......+P(2n)−−(2)=P(2)+P(4)+P(6)+.......+P(2n)--(2)
As above condition is mentioned that p(r)  =  krp(r)\,\,=\,\,kr by taking the 2 common and also As we know that sum of number of series up to n is n(n+1)2\dfrac{n(n+1)}{2} and substitute in equation (2)
2k(1+2+3+.....+n)=2kn(n+1)22k(1+2+3+.....+n)=\dfrac{2kn(n+1)}{2}
Substitute the value of k is:
2k(1+2+3+.....+n)=2n(2n+1)×n(n+1)22k(1+2+3+.....+n)=\dfrac{2}{n(2n+1)}\times \dfrac{n(n+1)}{2}
By simplifying further we get:
2k(1+2+3+.....+n)=(n+1)(2n+1)2k(1+2+3+.....+n)=\dfrac{(n+1)}{(2n+1)}
So, the correct option is “option 4”.

Note: In this particular problem we should always remember that the formula for the sum of the number of series is n. If the sum of series goes up to 2n then in the formula we have to replace it as 2n. We have to keep this in mind because most students make silly mistakes while writing formulas. So, the above solution can be preferred for such types of problems.