Question

$A$ draws 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 $\left(i\right)$the same card as $B$ $\left(ii\right)$a higher card than $B$.

• A. $\left(i\right)\frac{1}{n}$ $\left(ii\right)\frac{(n-1)}{n}$
• B. $\left(i\right)\frac{1}{n^2}$ $\left(ii\right)\frac{(n-1)}{n}$
• C. $\left(i\right)\frac{1}{n}$ $\left(ii\right)\frac{(n-1)}{2n}$
• D. $\left(i\right)\frac{2}{n}$ $\left(ii\right)\frac{(n-1)}{n}$

Answer 1

Answer: (C)

$\left(i\right)\frac{1}{n}$ $\left(ii\right)\frac{(n-1)}{2n}$

P(same card as $A$) = P($A$ chooses a card) $\times$ P($B$ chooses the same card)
$=\frac{1}{n}$
(ii)P(card of $A$ greater than $B$)
$=$ P($B$ draws card $1$)$\times$P($A$ draws any card $>1$) $+$ P($B$ draws card $2$)$\times$P($A$ draws any card $>2$) $+$ ...... $+$P($B$ draws card $n-1$)$\times$P($A$ draws any card $>n-1$)
= $\frac{1}{n}\times \frac{n-1}{n}+\frac{1}{n}\times \frac{n-2}{n}+\frac{1}{n}\times \frac{n-3}{n}+....+\frac{1}{n}\times \frac{1}{n}$
= $\frac{1}{n}\times \frac{(n-1)n}{2n}=\frac{n-1}{2n}$