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probability haunts

by gmatnmein2010 » Mon Feb 08, 2010 5:34 am
2. 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

pls suggest books to improve my probability
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by shashank.ism » Mon Feb 08, 2010 5:39 am
gmatnmein2010 wrote:2. 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

pls suggest books to improve my probability
Solve a book of probability by K.C.Sinha , or a problem book of IIT-JEE by TMH, or problem plus by aschit Das gupta.


Solve any good book of IIT-JEE . you will surely be able to solve any problem.
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by shashank.ism » Mon Feb 08, 2010 5:49 am
gmatnmein2010 wrote:2. 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

pls suggest books to improve my probability
Total no. of 3 digit = 999-99 =100
total no. of required 3- digits which can be formed = either 1 7's , 2 7's , all 3 7's
= (8.9.1+9.1.9+1.9.9)+(1.1.9+1.9.1+8.1.1)+1.1.1=72+81+81+9+9+8+1 = 271


Ans A
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by thephoenix » Mon Feb 08, 2010 6:48 am
imo c

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25
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by roh00kan » Sat Aug 28, 2010 1:40 am
What's the Official Answer?
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by shibal » Sat Aug 28, 2010 6:22 am
thephoenix wrote:imo c

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25
i got that too, but counting each hundreds, tens and units digit.... how do u come up with 8*9*9?
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by Gurpinder » Sat Aug 28, 2010 8:47 am
gmatnmein2010 wrote:2. 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
There are only 10 numbers : {0,1,2,3,4,5,6,7,8,9}

You want 1 or more = 7.

So the probability of the first one being 7 is 1/10
second one being 7 = 1/10 and third one = 1/10

1/10+1/10+1/10=3/30=1/10

So (E).

Isnt this the way to do it guys?
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by ramana » Sat Aug 28, 2010 8:59 am
This is how i solved.

probability :- 1 - ( non-favorable outcomes /total no: of outcomes )

Total NO: we have three digits and 10 digits (0-9)

last digit can be anyone in the range[0 9] : 10
middle digit can be anyone in the range[0 9] : 10
first Digit cannot be '0' -> [1 9] : 9

total 9*10*10 =900

NON- Favorable outcomes : No digit should contain 7 and the first place should not have 0 and 7

thus we have 8*9*9 = 648

apply the formula for Prob stated above,

= 252/900 reduced to 7/25

IMO C

what is the OA?
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by diebeatsthegmat » Sat Aug 28, 2010 9:55 am
gmatnmein2010 wrote:2. 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

pls suggest books to improve my probability
i would solve it simply like this
supposed ABC is a 3-digit positive integer
there are 10 numbers from 0 to 9 and A cant be 0 thus we will have 9*10*10=900
if only A = 7, we will have 9*9=81 ways to select B and C
if only B or C = 7 we will have only 9*8+9*8=72*2=144 ways to select a and C/B
if only AB or AC=7 we will have only 2*9=18 ways to select C or B, respectively
if only BC=7 we will have only 8 ways to select A because A cant be 0 or 7 anymore
if ABC all = 7 we will have 1 way to select em
thus totally we will have 144
+81+18+8+1=252 ways
so the probability will be 252/900 or 7/25
C is the answer
hope it helps
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by Gurpinder » Sat Aug 28, 2010 10:06 am
ramana wrote:This is how i solved.

probability :- 1 - ( non-favorable outcomes /total no: of outcomes )

Total NO: we have three digits and 10 digits (0-9)

last digit can be anyone in the range[0 9] : 10
middle digit can be anyone in the range[0 9] : 10
first Digit cannot be '0' -> [1 9] : 9

total 9*10*10 =900

NON- Favorable outcomes : No digit should contain 7 and the first place should not have 0 and 7

thus we have 8*9*9 = 648

apply the formula for Prob stated above,

= 252/900 reduced to 7/25

IMO C

what is the OA?
Why can't the first place have 7???

What do you mean?
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by ramana » Sat Aug 28, 2010 10:12 am
Gurpinder wrote:
ramana wrote:This is how i solved.

probability :- 1 - ( non-favorable outcomes /total no: of outcomes )

Total NO: we have three digits and 10 digits (0-9)

last digit can be anyone in the range[0 9] : 10
middle digit can be anyone in the range[0 9] : 10
first Digit cannot be '0' -> [1 9] : 9

total 9*10*10 =900

NON- Favorable outcomes : No digit should contain 7 and the first place should not have 0 and 7

thus we have 8*9*9 = 648

apply the formula for Prob stated above,

= 252/900 reduced to 7/25

IMO C

what is the OA?
Why can't the first place have 7???

What do you mean?

favorable outcomes according to question stem - one or more 7 in a three digit number selected.

I am going for the non-favorable outcome which should be ' Number selected at random has no '7' in it'

clears?
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by Gurpinder » Sat Aug 28, 2010 10:16 am
Hey,

No, I get what you are doing. You took all the non favorable outcomes. But I am just confused with your wording "the first place should not have 0 and 7". Why can't the first number be 7? The question said 1 or more 7's. So all 3 could be 7.

Do you get what i am saying?

thanks,
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by ramana » Sat Aug 28, 2010 10:22 am
Gurpinder wrote:Hey,

No, I get what you are doing. You took all the non favorable outcomes. But I am just confused with your wording "the first place should not have 0 and 7". Why can't the first number be 7? The question said 1 or more 7's. So all 3 could be 7.

Do you get what i am saying?

thanks,
hmm, that condition is applied only for counting non-favorable outcomes.

of course the first digit can be 7 according to the question stem.

have i answered your question now?
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by Gurpinder » Sat Aug 28, 2010 10:23 am
ramana wrote:
Gurpinder wrote:Hey,

No, I get what you are doing. You took all the non favorable outcomes. But I am just confused with your wording "the first place should not have 0 and 7". Why can't the first number be 7? The question said 1 or more 7's. So all 3 could be 7.

Do you get what i am saying?

thanks,
hmm, that condition is applied only for counting non-favorable outcomes.

of course the first digit can be 7 according to the question stem.

have i answered your question now?
Yes,

Thanks
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by Maciek » Sat Aug 28, 2010 10:32 am
Hi all!

this is wrong approach

number of possible outcomes
9 * 10 * 10 = 900
how many units are 7?
9 * 10 = 90
how many tens are 7?
9 * 9 = 81
how many hundreds are 7?
1*10*10 = 100
number of wanted outcomes
271
the probability is 271/900
it is choice A but it is wrong

hope it helps!
Best,
Maciek
Last edited by Maciek on Sat Aug 28, 2010 10:52 am, edited 1 time in total.
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