Question

A bag contains n cards numbered as 1 , 2 , 3 ,....... upto n, well mixed up. A person 'P' draws a card from the bag and puts it back in the bag. Then a person 'Q' draws a card from the bag. The odds in favour of P getting a card with higher number than Q is :

• A. n -1 : n + 1
• B. n : n -1
• C. n + 1 : n
• D. (n + 1) : (n -1)

Answer 1

Answer: (A)

$S=\{n,(n-1),(n-2), \ldots \ldots . .3,2,1\}$
suppose ' $P$ ' draws the card number ' $n$ ' the probability of which is $\frac{1}{n}$
' $P ^{\prime}$ draws a higher number than ' $Q$ ' $=\frac{1}{ n ^2}[( n -1)+( n -2)+\ldots .+1]$
$=\frac{( n -1) n }{2 n ^2}=\frac{ n -1}{2 n }=\frac{ n -1}{( n -1)+ n +1}$
$\Rightarrow \quad$ odds in favour $=( n -1):( n +1) \Rightarrow( A )$