Difference between revisions of "Probability qn 1"

Revision as of 21:18, 29 June 2014

The probability that a number selected at random between $100$ and $999$ (both inclusive) will not contain the digit $7$ is:

(a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> (d)<math>18/25</math>

Solution by Arjun Suresh

First digit can be chosen in $8$ ways from $1-9$ excluding $7$ Second digit can be chosen in $9$ ways from $0-9$ excluding $7$ and similarly the third digit in $9$ ways. So, total no. of ways excluding $7 = 8*9*9$ Total no. of ways including $7 = 9 * 10 * 10$ So, ans = $(8*9*9)/(9*10*10) = 18/25$

@@ Line 6: / Line 6: @@
 ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
- First digit can be chosen in $8$ ways from $1-9$ excluding $7$
+First digit can be chosen in $8$ ways from $1-9$ excluding $7$
- Second digit can be chosen in $9$ ways from $0-9$ excluding $7$ and similarly the third digit in $9$ ways.
+Second digit can be chosen in $9$ ways from $0-9$ excluding $7$ and similarly the third digit in $9$ ways.
- So, total no. of ways excluding $7 = 8*9*9$
+So, total no. of ways excluding $7 = 8*9*9$
- Total no. of ways including $7 = 9 * 10 * 10$
+Total no. of ways including $7 = 9 * 10 * 10$
- So, ans = $(8*9*9)/(9*10*10) = 18/25$
+So, ans = $(8*9*9)/(9*10*10) = 18/25$
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