A five digit number (having all different digits) is formed using the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. The probability that the formed number either begins or ends with an odd digit, is equal to 

$\frac{5}{6}$

Total formed numbers that begin with a odd digit

$={ }^5 C_1 .{ }^8 P_4=5(8)(7)(6)(5)$

Total formed numbers that end with a odd digit

$={ }^5 C_1 .{ }^8 P_4=(8)(7)(6)(5)$

Total formed number that begin with an odd digit and also end with an odd digit

$={ }^5 C_2 . 2 ! .{ }^7 P_3=5 .(4)(7)(6)(5)$

Thus total formed numbers that begin with an odd digit or end with an odd digits is equal to 5 . 7 . 6 . 60

Total formed numbers $={ }^9 P_5=9 . 8 . 7 . 6 . 5$

Thus, required probability $=\frac{5}{6}$