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Re: Let C be the collection of all integers between 100,000 and 999,999 in [#permalink]
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NoviceBoy wrote:
Let C be the collection of all integers between 100,000 and 999,999 inclusive. What is the probability that the number randomly selected from C has exactly 5 occurrences of the digit 3 in the decimal representation?

A) 1/150,000
B) 53/899,999
C) 54/899,999
D) 53/900,000
E) 3/50,000



Assuming no digit is zero, we have:
Number of ways of choosing the positions of 3 = 6C5 = 6 ways
Number of ways of choosing the 6th digit = 8 ways (ignoring 0 and 3)
Thus, total number of such numbers = 6 * 8 = 48

Assuming zero is present, the first digit must be 3. Zero can occur in the 2nd, 3rd, 4th, 5th and 6th positions i.e. 5 ways

Thus, total favorable cases = 48 + 5 = 53
Total cases = 999999 - 100000 + 1 = 900000
Probability = 53/900000

Answer D
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Let C be the collection of all integers between 100,000 and 999,999 in [#permalink]
Case 1: Dont starts with 3

In this case we have (considering that the first number cant be 0)

8/9 x 1/10 x 1/10 x 1/10 x 1/10 x 1/10 = 8/9 x 1/100.000

Case 2: Starts with 3

In this case we have (considering that the first number cant be 0)

1/9 x 9/10 x 1/10 x 1/10 x 1/10 x 1/10 Notice that after the first 3, the rest can be ordered un 5 possible ways, 5!/4!, so:

5 x 1/100.000

Finally, adding both cases:

8/9 x 1/100.000 + 5 x 1/100.000

(1/100.000)(5 + 8/9)

(1/100.000)(53/9)

53/900.000­
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Let C be the collection of all integers between 100,000 and 999,999 in [#permalink]
 
NoviceBoy wrote:
Let C be the collection of all integers between 100,000 and 999,999 inclusive. What is the probability that the number randomly selected from C has exactly 5 occurrences of the digit 3 in the decimal representation?

A) 1/150,000
B) 53/899,999
C) 54/899,999
D) 53/900,000
E) 3/50,000

­smart guess can be helpful to solve this question < 30sec
scan the options
B C D are very close so answer can be either of them 
the total number of outcomes is the total number of integers which in the case is 900,000
so only option d has the said value

answer - D­
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Re: Let C be the collection of all integers between 100,000 and 999,999 in [#permalink]
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Expert Reply
NoviceBoy wrote:
Let C be the collection of all integers between 100,000 and 999,999 inclusive. What is the probability that the number randomly selected from C has exactly 5 occurrences of the digit 3 in the decimal representation?

A) 1/150,000
B) 53/899,999
C) 54/899,999
D) 53/900,000
E) 3/50,000

­We have all 900,000 six digit numbers here.
If we have five 3s and we have one occurence of an additional  digit, then we need to consider 2 cases:

Any additional digit except 0 and 3: 
A digit can be chosen in 8 ways from 1/2/4/5/.../9. 
Five 3s and another digit can be arranged  in 6 distinct ways (333334, 333343, 333433... 433333)
This gives us 8*6 = 48 numbers

The additional digit is 0:
0 cannot be the first digit from left but it can take any other 5 positions such as 333330, 333303 etc.
Hence 5 more numbers. 

Total we have 53 such numbers. 

Required Probability = Favorable Cases/ Total number of cases = 53/900,000

Answer (D)

Video on Probability:
https://youtu.be/0BCqnD2r-kY
 
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Re: Let C be the collection of all integers between 100,000 and 999,999 in [#permalink]
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