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
Answer & explanation
Correct answer: option 1
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 $=\frac{5}{6}$