x is a four digits positive integer. If x is chosen randomly

Question

x is a four digits positive integer. If x is chosen randomly, what is the probability that number x has only two similar digits?

A) 213/500
B) 54/125
C) 151/375
D) 2/5
E) 101/250

Posted from my mobile device

goodboy01 · Accepted Answer

X = PQRS

Total 4 digit nos. = 10000-1000 = 9000 (sample space)...0). eqn.

X = PQRS (4digits)

Using counting principle - 
P - Can take 9 digits (1-9)
Q - Can take 1 digit (It has to be same as P)
R - Can take 9 digits (Because 1 & 2nd digit are same 3rd digit has to be different)
S - Can take 8 digits (Because 1 & 2nd digit are same 3rd & 4th digits have to be different)

P.Q.R.S = 9*1*9*8 = 648 ...i) eqn.

Suppose, we have number 1102 - it can be arranged in 6 WAYS -- 1102, 1120, 1021, 1012, 1210, 1201 - Note 1st no. is fixed ...ii) eqn.

Multiplying both i). and ii). eqn. = 648*6 = 3888...iii). eqn. (Total nos. with both digits same)

Dividing 0). eqn. and iii) eqn.

Probability = 3888/9000 = 54/125

Harshjha001 · Answer

shahwazkhan01  
Can you please explain how will "0" fit into your explanation . Does your solution deal with double presence of "0".
As you are assuming that P (thousands place) is the number that will occur twice , but P will never have value 0 .

goodboy01 · Answer

P cannot take '0', that was the reason we took value 9 (1-9) rather than 10 (0-10).
Q will take the value same as P, so we took value as '1'.
R will take all values except 'P' - that was the reason we took 9 (removing 1 no. from 0-10 nos.)
S will take all values except 'P' and 'R' -- if not there could be chance where both back digits can be same as front digits - that was the reason we took 8 (removing 2 no. from 0-10 nos.).

P.Q.R.S - We have fixed 'P' because we need total combinations because if we take 4! ways then pattern will repeat.

SaidNassar1991 · Answer

To calculate the probability we can count all the possible choices where we have only two similar digits, and we divide it over all numbers between 1,000 and 9,999 included which are 9,000 numbers.

First digit of X can be 1,2,...9 but not zero.
The other three digits can be 0,1,..9.

We should have only two similar digits.
So we have C2of4= 6 possible similarities:
Either the first and the second are similar, or the first and the third, or the first and the forth, or the second and the third, or the second and the forth, or the third and the forth.

AFirst method:

First and second similar: for first digit we have 9 choices, the second we have 1 choice, the third we have 9 choices, the forth we have 8 choices.

First and third similar: for first digit we have 9 choices, the second we have 9 choices, the third we have 1 choice, the forth we have 8 choices.

First and forth similar: for first digit we have 9 choices, the second we have 9 choices, the third we have 8 choices, the forth we have 1 choice.

Second and third similar: for first digit we have 9 choices, the second we have 9 choices, the third we have 1 choice, the forth we have 8 choices.

Second and forth similar: for first digit we have 9 choices, the second we have 9 choices, the third we have 8 choices, the forth we have 1 choice.

Third and forth similar: for first digit we have 9 choices, the second we have 9 choices, the third we have 8 choices, the forth we have 1 choice.

So numbers having only 2 similar digits are 6*9*9*8=3888

Second method:

We can do it another way, if want see clearly what happens with the 0 that cannot be in the first digit, and therefore differentiate the cases where the repeated number is 0 and where not.

1. If first digit is one of the two similar, we have 3 choices with each having 9*1*9*8 numbers, so we have 3*9*1*9*8= numbers. (Here 0 cannot be one of the similar)

2. If first digit is not one of the two similar, we have 3 choices but in each choice we should differentiate whether the two similar are 0 or not.

a. If the two similar digits are 0, we have 9*1*1*8. We have 3 places of the two 0 so we have 3*9*1*1*8 numbers.

b. If the two similar digits are not 0, we have 3 places to the two none 0. But we should differentiate for each case again if the third one of the the three right digits is 0 or not:
 I. If the third digit is 0 we have 9*1*1*8
 II. If the third digit is not 0: we have left 9*1*8*7

So we have in b. (3*9*1*1*8+3*9*8*7)=3*9*8*8

So the numbers with only 2 similar digits are 3*9*1*9*8+ 3*9*1*1*8+3*9*8*8=6*9*9*8=3888 we arrive at the same number going step by step.

Probability of having only 2 similar digits is 3888/9000=54/125

Posted from my mobile device

GMATGuruNY · Answer

ALL POSSIBLE CASES:
A good 4-digit integer will be composed of EXACTLY 3 DISTINCT DIGITS.
The repeated digit must occupy TWO POSITIONS in the 4-digit integer.
From 4 positions, the number of ways to choose a pair of positions for the repeated digit = 4C2  = \frac{4*3}{2*1} = 6 
Number of options for the repeated digit = 10 (Any digit 0-9)
Number of options for the 3rd digit = 9 (Of the 10 digits 0-9, any but the one already used)
Number of options for the 4th digit = 8 (Of the 10 digits 0-9, any but the two already used)
To combine these options, we multiply:
6*10*9*8

BAD CASES:
A fraction of the options above must be discarded because they put 0 in the thousands place.
Each of the 10 digits has the same probability of appearing in the thousands place.
Thus,  \frac{1}{10}  of the options above will put 0 in the thousands place, with the result that the remaining  \frac{9}{10}  are viable:
 \frac{9}{10} * 6*10*9*8 = 9*6*9*8

Since there are 9000 integers between 1000 and 9999, inclusive, we get:
P(exactly 3 distinct digits)  = \frac{9*6*9*8}{9000} = \frac{6*9*8}{1000} = \frac{6*9}{125} = \frac{54}{125}

B

rsrighosh · Answer

i took a stupid method and gambled but it worked...

i saw that in any case the no. of 4 digit numbers will be 9x1x9x8 x (some number of how many ways we can arrange the digits = k)
the total number of 4 digit numbers is 9000

Probability is

\frac{(9*1*9*8 * k)}{9000}  =  \frac{(1*9*8 * k)}{1000}

We cannot simplify denominator further with 9

This means that the numerator has to be a number divisible by 9

Only B is such option.

Nikhil2369 · Answer

if the number was 1123 then we could've arranged it in 12 ways

Fdambro294 · Answer

Probability = Fav. Outcomes / Total Possible Outcomes

(1st)Total Possible Outcomes

____ ____ ____ ____

the 1,000s Digit: we can choose any digit except 0 --- 9 available options
100s Digit: any digit ---- 10 options
10s Digit: any digit ---- 10 options
Units Digit: any digit ----- 10 options

Total 4 Digit Numbers = 9 * (10)^3

(2nd)Favorable Outcomes

Scenario 1:
Start with the Scenario where 1 is in the 1,000s Digit Place (i.e., all the Digits from 10,000 thru 19,999)

we will have 2 Sub-Cases within Scenario 1:

Case 1: the 1 Digit is NOT repeated:
1,000s Digit: the 1 Digit is Fixed ------ 1 option

then for the remaining 3 Digits, we need to pick 2 Digit places where a Digit Other Than 1 will be repeated: we can do that in "3 choose 2" ways

then for the Digit that is repeated in those 2 places, we can pick any Digit OTHER THAN 1 ---- 9 options

and we can choose the last Digit in 8 ways ------ 8 options

Scenario 1, Case 1: 1 * "3 choose 2" * 9 * 8

OR

Scenario 1, Case 2: the Digit 1 IS repeated:

1,000s Digit: the 1 Digit is Fixed ----- 1 option

then we have 3 Digit Places, of which we need to choose 1 to place ANOTHER 1 DIGIT ------ "3 choose 1"

the remaining next Digit can not be a 1, because it has already been repeated twice ------ 9 options

the last digit can not be a 1 or the Digit chose before ------- 8 options

Scenario 1, Case 2: 1 * "3 choose 1" * 9 * 8

Scenario 1:

(1 * "3 choose 2" * 9 * 8) + (1 * "3 choose 1" * 9 * 8)

(3 * 9 * 8) + (3 * 9 * 8)

(2) * (3 * 9 * 8)

this same exact logic can be repeated for Each Digit from 1 thru 9 ------ so a total of 9 times -----which means the NUM = No. of Favorable Outcomes becomes:

(9) * (2) * (3 * 9 * 8)

Divide that NUM by the DEN of Total Possible Outcomes: 9 * (10)^3

(9 * 2 * 3 * 9 * 8) / (9 * 10 * 10 * 10)

----start reducing the fraction by canceling the 9s and grouping together the Prime Numbers----

(2'4th * 3'3rd) / (2'3rd * 5'3rd)

----the 2 to the 3rd Power cancels in the DEN leaving ---

(2 * 3'3rd) / (5'3rd) =

(2 * 27) / (125) =

54/125

Answer B

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x is a four digits positive integer. If x is chosen randomly

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