Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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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 $7$ | ||
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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 $7 = 8*9*9$ | + | |
| − | Total no. of ways including $7 = 9 * 10 * 10$ | + | So, total no. of ways excluding $7$ = $8*9*9$ |
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| + | Total no. of ways including $7$ = $9 * 10 * 10$ | ||
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So, ans = $(8*9*9)/(9*10*10) = 18/25$ | So, ans = $(8*9*9)/(9*10*10) = 18/25$ | ||
{{Template:FBD}} | {{Template:FBD}} | ||
| − | [[Category:Probability | + | [[Category: Non-GATE Questions from Probability]] |
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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 = $(8*9*9)/(9*10*10) = 18/25$
GATECSE