Difference between revisions of "Probability qn 1"

Latest revision as of 12:04, 15 July 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>

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 1: / Line 1: @@
-The probability that a number selected at random between 100 and 999 (both
+The probability that a number selected at random between $100$ and $999$ (both
-inclusive) will not contain the digit 7 is:
+inclusive) will not contain the digit $7$ is:
-(a)16/25 (b)(9/10)^3 (c)27/75 '''(d)18/25'''
+(a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> '''(d)<math>18/25</math>'''
-===Solution===
+==={{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 in 9 ways from 0-9 excluding 7 and third digit also 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
-{{Template:FB}}
+Second digit can be chosen in $9$ ways from $0-9$ excluding $7$ and similarly the third digit in $9$ ways.
-[[Category:Probability]]
+So, total no. of ways excluding $7$ = $8*9*9$
+Total no. of ways including $7$ = $9 * 10 * 10$
+So, ans = $(8*9*9)/(9*10*10) = 18/25$
+{{Template:FBD}}
+[[Category: Non-GATE Questions from Probability]]