Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
||
Line 10: | Line 10: | ||
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 = $(8*9*9)/(9*10*10) = 18/25$ |
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 = $(8*9*9)/(9*10*10) = 18/25$
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 = $(8*9*9)/(9*10*10) = 18/25$