Question

Draw a card from a pack of n cards marked 1, 2,....., n. The card is replaced in the pack and B draws a card. Find the probability that A draws (i) the same card as B, (ii) a higher card than B.
(a) $\left( i \right)\dfrac{1}{n}\left( ii \right)\dfrac{n-1}{n}$
(b) $\left( i \right)\dfrac{1}{{{n}^{2}}}\left( ii \right)\dfrac{n-1}{n}$
(c) $\left( i \right)\dfrac{1}{n}\left( ii \right)\dfrac{n-1}{2n}$
(d) $\left( i \right)\dfrac{2}{n}\left( ii \right)\dfrac{n-1}{n}$

Answer 1

Hint: First, we should know that probability is the ratio of the favourable outcomes to the total number of outcomes. Then, we need to calculate the probability that B chooses the same card as A chooses. Then, we go for the second case and calculate the probability of A drawing a card with number greater than B. For solving that we use the general expression of the sum of series as $\dfrac{n-1}{n}+\dfrac{n-2}{n}+........+\dfrac{1}{n}=\dfrac{\left( n-1 \right)n}{2n}$.

Complete step-by-step answer:
In this question, we are supposed to find the two different cases probability by using the given conditions.
First of all we should know that probability is the ratio of the favourable outcomes to the total number of outcomes.
So, for each case we want to get the probability we need to find these two things very carefully.
Then, we will firstly go with the total number of cases for drawing a card from a pack of n cards is given as n.
So, total outcomes=n
Now, for the first case, we want A draws the same card as B draws which means they both are drawing the same numbered card after replacement.
So, it has only one case that A and B choose the same numbered card.
Now, to get the probability that B chooses the same card as A chooses is given by:
P(B chooses the same card as A)=P(A chooses a card)*P(B chooses the same card)
So, from the above conclusion the probability that B chooses the same card as A is $\dfrac{1}{n}$.
Now, we go to the second case in which we are asked to find the probability that A draws a higher card than B which forms a large number of cases.
Suppose A draws a card numbered 8 and B draws a card numbered 5, in this similar way we have a lot of cases.
So to solve it we make a general expression as:
P(card of A greater than B)=P(B draws card 1)*P(A draws any card>1)+ P(B draws card 2)*P(A draws any card>2)+...........+ P(B draws card n-1)*P(A draws any card>n-1).
Now, to get this as probability we has it as an expression as:
$\dfrac{1}{n}\times \dfrac{n-1}{n}+\dfrac{1}{n}\times \dfrac{n-2}{n}+........+\dfrac{1}{n}\times \dfrac{1}{n}$
Now, to solve this expression, we will take $\dfrac{1}{n}$ common and use the formula of the series as:
$\dfrac{n-1}{n}+\dfrac{n-2}{n}+........+\dfrac{1}{n}=\dfrac{\left( n-1 \right)n}{2n}$
Then , the above expression becomes as:
$\dfrac{1}{n}\times \dfrac{\left( n-1 \right)n}{2n}=\dfrac{n-1}{2n}$
So, the probability of the card drawn by A being larger than the card drawn by B is given by $\dfrac{n-1}{2n}$.
Hence, option (c) is correct.

Note: Here, the special thing about this question is that in the second case calculation we need some general expression of the sum of series. So, we should know some of the general sum of series as:
$\begin{align}
  & \dfrac{n-1}{n}+\dfrac{n-2}{n}+........+\dfrac{1}{n}=\dfrac{\left( n-1 \right)n}{2n} \\
 & 1+2+3+.........+n=\dfrac{n\left( n-1 \right)}{2} \\
\end{align}$