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Probability strikes me again

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Probability strikes me again

by cypherskull » Mon May 07, 2012 12:38 pm
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
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by aneesh.kg » Mon May 07, 2012 1:14 pm
And I'm here to help again!

From 100 to 999,there are 900 hundred 3-digit numbers.
Lets find how many three-digit numbers don't have even a single 7 in them.

Make three dashes.
_ _ _

Write down the number of possibilities at every dash.
First dash: 0 and 7 can't come so 8C1 = 8
Second dash: only 7 can't come so 9C1 = 9
Third dash: only 7 can't come so 9C1 = 9

_ _ _
8*9*9 = 648

The unfavourable probability is when any of these numbers is chosen. So, this probability has to be subtracted from 1.

Required probability = 1 - (8*9*9)/(9*10*10)
1 - (8*9)/(10*10)
1 - 18/25
[spoiler]7/25[/spoiler]

[spoiler](C)[/spoiler] is the answer
Last edited by aneesh.kg on Mon May 07, 2012 1:17 pm, edited 1 time in total.
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by Bill@VeritasPrep » Mon May 07, 2012 1:16 pm
We have a total of 900 3-digit numbers: 100-999, inclusive. The question asks us to figure out how many of these have 1, 2, or 3 7's as digits, which leaves only one other possibility: no 7's.

To create a 3-digit number without using a 7, we have 8 options for the hundreds place (10 digits 0-9 with the exception of 0 and 7), 9 options for the tens place (0-9 with the exception of 7), and 9 options for the units place (0-9 with the exception of 7).

8*9*9=648; since we used the complement method, we must remember to subtract that from the total of 900: 900-648 = 252.

Thus, the probability of a 3-digit number having at least one 7 is 252/900, which reduces to 7/25.
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