2. What is the probability that a 3digit 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
probability haunts

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 shashank.ism
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Solve a book of probability by K.C.Sinha , or a problem book of IITJEE by TMH, or problem plus by aschit Das gupta.gmatnmein2010 wrote:2. What is the probability that a 3digit 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 any good book of IITJEE . you will surely be able to solve any problem.
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 shashank.ism
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Total no. of 3 digit = 99999 =100gmatnmein2010 wrote:2. What is the probability that a 3digit 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 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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imo c
Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1P(no 7)=1648/900=252/900=7/25
Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1P(no 7)=1648/900=252/900=7/25

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i got that too, but counting each hundreds, tens and units digit.... how do u come up with 8*9*9?thephoenix wrote:imo c
Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1P(no 7)=1648/900=252/900=7/25
There are only 10 numbers : {0,1,2,3,4,5,6,7,8,9}gmatnmein2010 wrote:2. What is the probability that a 3digit 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
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?
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
 Alfred A. Montapert, Philosopher.
 Alfred A. Montapert, Philosopher.
This is how i solved.
probability : 1  ( nonfavorable outcomes /total no: of outcomes )
Total NO: we have three digits and 10 digits (09)
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?
probability : 1  ( nonfavorable outcomes /total no: of outcomes )
Total NO: we have three digits and 10 digits (09)
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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i would solve it simply like thisgmatnmein2010 wrote:2. What is the probability that a 3digit 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
supposed ABC is a 3digit 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
Why can't the first place have 7???ramana wrote:This is how i solved.
probability : 1  ( nonfavorable outcomes /total no: of outcomes )
Total NO: we have three digits and 10 digits (09)
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?
What do you mean?
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
 Alfred A. Montapert, Philosopher.
 Alfred A. Montapert, Philosopher.
Gurpinder wrote:Why can't the first place have 7???ramana wrote:This is how i solved.
probability : 1  ( nonfavorable outcomes /total no: of outcomes )
Total NO: we have three digits and 10 digits (09)
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?
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 nonfavorable outcome which should be ' Number selected at random has no '7' in it'
clears?
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,
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,
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
 Alfred A. Montapert, Philosopher.
 Alfred A. Montapert, Philosopher.
hmm, that condition is applied only for counting nonfavorable outcomes.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,
of course the first digit can be 7 according to the question stem.
have i answered your question now?
Yes,ramana wrote:hmm, that condition is applied only for counting nonfavorable outcomes.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,
of course the first digit can be 7 according to the question stem.
have i answered your question now?
Thanks
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
 Alfred A. Montapert, Philosopher.
 Alfred A. Montapert, Philosopher.
 Maciek
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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
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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