VJesus12 wrote:What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?
A. 271/900
B. 27/100
C. 7/25
D. 1/9
E. 1/10
The OA is the option C.
Could someone give me a good explanation of how can I get the correct answer here? Please. I need some help. Thanks in advance.
Hello Vjesus12.
This is the way I'd solve it.
We need to build a 3-digit positive number. Now,
* The first digit has 9 different options (1,2,3,...,9).
* The second digit has 10 different options (0,1,2,3,...,9).
* The third digit has 10 different options (0,1,2,3,...,9).
So, the total amount of numbers that we can build is 9*10*10=
900.
On the other hand, the favorable cases are:
* One 7:
i) In the first place: there are 1*9*9=
81 options.
ii) In the second place: there are 8*1*9=
72 options.
ii) In the third place: there are 8*9*1=
72 options.
* Two 7's:
i) One in the first place and other in the second place: there are 1*1*9=
9 options.
ii) One in the first place and other in the third place: there are 1*9*1=
9 options.
ii) One in the second and one in the third place: there are 8*1*1=
8 options.
* Three 7's:
The only option is 777. So,
1 option.
Therefore, the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits is given by $$P=\frac{total\ favorable\ cases}{total\ options}$$ $$P=\frac{81+72+72+9+9+8+1}{900}$$ $$P=\frac{252}{900}$$ $$P=\frac{63}{225}$$ $$P=\frac{7}{25}.$$
And this is why the correct answer is the option
C.
I hope it helps you.
Regards. <i class="em em-smiley"></i>
