- Home
- Standard 11
- Mathematics
14.Probability
hard
If $m$ rupee coins and $n$ ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins is
A
$^{m + n}{C_m}/{n^m}$
B
$\frac{{n\,(n - 1)}}{{(m + n)\,(m + n - 1)}}$
C
$^{m + n}{P_m}/{m^n}$
D
$^{m + n}{P_n}/{n^m}$
Solution
(b) $m$ rupee coins and $n$ ten paise coins can be placed in a line in $\frac{{(m + n)\,\,!}}{{m\,\,!\,\,n\,\,!}}$ ways.
If the extreme coins are ten paise coins, then the remaining $n – 2$ ten paise coins and $m$ one rupee coins can be arragned in a line in $\frac{{(m + n – 2)\,\,!}}{{m\,\,!(n – 2)\,\,!}}$ ways.
Hence the required probability
$ = \frac{{\frac{{(m + n – 2)\,\,!}}{{m\,\,!(n – 2)\,\,!}}}}{{\frac{{(m + n)\,\,!}}{{m\,\,!\,\,n\,\,!}}}} = \frac{{n(n – 1)}}{{(m + n)(m + n – 1)}}$.
Standard 11
Mathematics
