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− | 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: |
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− | + | (a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> '''(d)<math>18/25</math>''' | |
− | + | ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}=== | |
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− | + | First digit can be chosen in $8$ ways from $1-9$ excluding $7$ | |
− | [[Category:Probability]] | + | Second digit can be chosen in $9$ ways from $0-9$ excluding $7$ and similarly the third digit in $9$ ways. |
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+ | So, total no. of ways excluding $7$ = $8*9*9$ | ||
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+ | Total no. of ways including $7$ = $9 * 10 * 10$ | ||
+ | |||
+ | So, ans = $(8*9*9)/(9*10*10) = 18/25$ | ||
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+ | {{Template:FBD}} | ||
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+ | [[Category: Non-GATE Questions from Probability]] |
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)16/25 (b)(9/10)^3 (c)27/75 (d)18/25
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