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Lottery tickets numbered consecutively from 100 through 999 are placed

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Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post Updated on: 12 Aug 2018, 23:01
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e-GMAT Question of the Week #5


Lottery tickets numbered consecutively from 100 through 999 are placed in a box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd?

    A. \(\frac{1}{150*50}\)

    B. \(\frac{2}{150*50}\)

    C. \(\frac{1}{54}\)

    D. \(\frac{1}{27}\)

    E. \(\frac{3}{8}\)


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Originally posted by EgmatQuantExpert on 29 Jun 2018, 04:29.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:01, edited 3 times in total.
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Re: Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 29 Jun 2018, 04:54
EgmatQuantExpert wrote:
e-GMAT Question of the Week #5


Lottery tickets numbered consecutively from 100 through 999 are placed in a box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd?

    A. \(\frac{1}{150*50}\)

    B. \(\frac{2}{150*50}\)

    C. \(\frac{1}{54}\)

    D. \(\frac{1}{27}\)

    E. \(\frac{3}{8}\)


To access all the questions: Question of the Week: Consolidated List


Actually if you understand the question, you would know that the denominator has to be a multiple of 4 and max of 9 with just 2s and two 3s that is 2^n*3^2
Only E left


Let's solve it and why denominator should be in form of \(2^n*3^2\)Number of prime digits - 2,3,4,7 so 4 of them
So each of the three digit can be filled with any of three..
Total ways one can select - 4*3*2 and similarly second in 4*3*2 ways..
Combined 4*4*3*3*2*2 and this will come in denominator and that is why the denominator should have just 2s or max 9

Only way the sum could be odd is when one ends with 2 and other with a odd prime number..
First chooses - 3*2*(1)......1 because only 2 can be placed
Second chooses - 3*2*(3)... 3 because 2 is not included
Total 3*2*1*3*2*3*(2).. .....2 because within themselves they can pick either

Probability = \(\frac{2*(3*2*3*2*3}{4*4*3*3*2*2)=}\frac{3}{8}\)

E


Also the entire thing is related to last digit....
For odd units digit.. one number has to be 2 and the other can be any of 3 odd number, so 1*3*2.... 2 because each can interchange
Total ways .. both numbers can have any of the last digits so 4*4


Probability = \(\frac{1*3*2}{4*4}=\frac{3}{8}\)

E
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Re: Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 29 Jun 2018, 08:26
chetan2u wrote:
EgmatQuantExpert wrote:
e-GMAT Question of the Week #5


Lottery tickets numbered consecutively from 100 through 999 are placed in a box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd?

    A. \(\frac{1}{150*50}\)

    B. \(\frac{2}{150*50}\)

    C. \(\frac{1}{54}\)

    D. \(\frac{1}{27}\)

    E. \(\frac{3}{8}\)


To access all the questions: Question of the Week: Consolidated List


Actually if you understand the question, you would know that the denominator has to be a multiple of 4 and with just 2s, that is 2^n
Only E has 8..

Let's solve it and why denominator should be in form of 2^n

Number of prime digits - 2,3,4,7 so 4 of them
So each of the three digit can be filled with any of three..
Total ways one can select - 4*4*4 and similarly second in 4*4*4 ways..
Combined 4*4*4*4*4*4 and this will come in denominator and that is why the denominator should have just 2s

Only way the sum could be odd is when one ends with 2 and other with a odd prime number..
First chooses - 4*4*1.......1 because only 2 can be placed
Second chooses - 4*4*3.... 3 because 2 is not included
Total 4*4*1*4*4*3 *2.... 2 because within themselves they can pick either

Probability = 2*(4*4*4*4*3/4*4*4*4*4*4)=3/8

E


The question says the numbers are distinct. So shouldn’t denominator be 4*3*2*4*3*2??

Acc to me the solution is as follows.. Please check

For denominator
Number of ways in which person 1 picks a number acc to te conditions-4*3*2
Similarly for person 2 - 4*3*2
So total ways -4*3*2*4*3*2

For numerator
For the sum to be odd one of the number should end in 2 - 3*2*1 -6
And the second number should end with an odd prime number -
1) a number with no 2 at all - 3*2*1- 6
2) a number with 2 at first and second place - 1*3*2 + 3*1*2-12
total - 6+12=18

Total numerator 6*18*2 (as both of them can have either)
Probability -( 6*18*2)/(4*3*2*4*3*2)
=3/8
E

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Re: Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 29 Jun 2018, 09:22
Antreev wrote:
chetan2u wrote:
EgmatQuantExpert wrote:
e-GMAT Question of the Week #5


Lottery tickets numbered consecutively from 100 through 999 are placed in a box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd?

    A. \(\frac{1}{150*50}\)

    B. \(\frac{2}{150*50}\)

    C. \(\frac{1}{54}\)

    D. \(\frac{1}{27}\)

    E. \(\frac{3}{8}\)


To access all the questions: Question of the Week: Consolidated List


Actually if you understand the question, you would know that the denominator has to be a multiple of 4 and with just 2s, that is 2^n
Only E has 8..

Let's solve it and why denominator should be in form of 2^n

Number of prime digits - 2,3,4,7 so 4 of them
So each of the three digit can be filled with any of three..
Total ways one can select - 4*4*4 and similarly second in 4*4*4 ways..
Combined 4*4*4*4*4*4 and this will come in denominator and that is why the denominator should have just 2s

Only way the sum could be odd is when one ends with 2 and other with a odd prime number..
First chooses - 4*4*1.......1 because only 2 can be placed
Second chooses - 4*4*3.... 3 because 2 is not included
Total 4*4*1*4*4*3 *2.... 2 because within themselves they can pick either

Probability = 2*(4*4*4*4*3/4*4*4*4*4*4)=3/8

E


The question says the numbers are distinct. So shouldn’t denominator be 4*3*2*4*3*2??

Acc to me the solution is as follows.. Please check

For denominator
Number of ways in which person 1 picks a number acc to te conditions-4*3*2
Similarly for person 2 - 4*3*2
So total ways -4*3*2*4*3*2

For numerator
For the sum to be odd one of the number should end in 2 - 3*2*1 -6
And the second number should end with an odd prime number -
1) a number with no 2 at all - 3*2*1- 6
2) a number with 2 at first and second place - 1*3*2 + 3*1*2-12
total - 6+12=18

Total numerator 6*18*2 (as both of them can have either)
Probability -( 6*18*2)/(4*3*2*4*3*2)
=3/8
E

Posted from my mobile device


Yes you are correct. I missed out on reading the distinct part.
We still got the answer because the question has to do more with the last digit
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 29 Jun 2018, 09:27
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EgmatQuantExpert wrote:
e-GMAT Question of the Week #5


Lottery tickets numbered consecutively from 100 through 999 are placed in a box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd?

    A. \(\frac{1}{150*50}\)

    B. \(\frac{2}{150*50}\)

    C. \(\frac{1}{54}\)

    D. \(\frac{1}{27}\)

    E. \(\frac{3}{8}\)


To access all the questions: Question of the Week: Consolidated List



Given the consecutive #'s from 100 to 999, as the #'s on a ticket.

Jack & Rose each pick a # which has distinct primes as its digits. Hence the digits are {2,3,5,7}

Total 3 digit #'s with distinct primes as its digits are = 4*3*2 = 24

Now for the sum of the #'s picked to be Odd, we need one # as even & other as odd, hence the numbers can be _ _ 2 & _ _ O

One # has units digit as even & other is an # 3 digit #. There are (3*2*1) = 6 such even #'s & (2*3*3) = 18

Now Case 1 consider Jack picks the even # & Rose picks the Odd #, hence the probability = (6/24)*(18/24) = 3/16

Now Case 2 consider Rose picks the even # & Jack picks the Odd #, hence the probability = (6/24)*(18/24) = 3/16

Therefore the required probability = 3/16 + 3/16 = 3/8

Answer E.


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Re: Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 29 Jun 2018, 13:59
EgmatQuantExpert wrote:
e-GMAT Question of the Week #5


Lottery tickets numbered consecutively from 100 through 999 are placed in a box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd?

    A. \(\frac{1}{150*50}\)

    B. \(\frac{2}{150*50}\)

    C. \(\frac{1}{54}\)

    D. \(\frac{1}{27}\)

    E. \(\frac{3}{8}\)


To access all the questions: Question of the Week: Consolidated List


all of the ticket numbers will have a units digit of 2,3,5,or 7
thus, in adding any of the two numbers, there are 4*4=16 possible combinations of units digits
of these 16, only 6 units digits combinations--2+3,2+5,2+7,3+2, 5+2, and 7+2--give an odd sum
6/16=3/8
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Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 02 Jul 2018, 09:23

Solution



Given:
    • Lottery tickets numbered consecutively from 100 through 999 are placed in a box
    • Jack picked up one ticket from the box, noted the number, and put the ticket back in the box
    • Then Rose came, picked up one ticket from the box, noted the number, and put the ticket back in the box
    • In both picked tickets, all the digits within the picked number are distinct and prime

To find:
    • The probability that the sum of the two ticket numbers is odd

Approach and Working:
When we say that sum of two numbers is odd, there can be two possible cases:
    • 1st number even and 2nd number odd [even + odd = odd]
    • 1st number odd and 2nd number even [odd + even = odd]

Therefore, for the favourable event, we can say:
    • If Jack picks an odd-numbered ticket, then Rose must pick an even-numbered ticket
    • Similarly, if Jack picks an even-numbered ticket, then Rose must pick an odd-numbered ticket

Now, it is also given that the digits within the numbers that they picked are distinct and prime
    • Single digit prime numbers = {2, 3, 5, 7}

The total possible numbers:
    • 3-digit even numbers with all distinct and prime digits = 3 * 2 * 1 = 6 [as it is an even number, the last digit must be 2. Hence, the possibilities remain for the other two places are 3 and 2 respectively]
    • 3-digit odd numbers with all distinct and prime digits = 3 * 3 * 2 = 18 [as it is an odd number, the last digit can be one of (3, 5, 7). Hence, the possibilities remain for the other two places are 3 and 2 respectively]
    • Total number of feasible cases = 6 + 18 = 24

Thus, if we calculate the event-wise probabilities,
    • P (Jack picks even AND Rose picks odd) = \(\frac{6}{24} * \frac{18}{24}\)
    • P (Jack picks odd AND Rose picks even) = \(\frac{18}{24} * \frac{6}{24}\)

Therefore, the final probability:
    • P (sum of the picked numbers is odd) = \(\frac{6}{24} * \frac{18}{24} + \frac{18}{24} * \frac{6}{24} = \frac{3}{8}\)

Hence, the correct answer is option E.

Answer: E

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Re: Lottery tickets numbered consecutively from 100 through 999 are placed  [#permalink]

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New post 07 Nov 2018, 19:19
Answer is E.

Ticket ranges from 100-999 -> there are 3 digits
Both Jack and Rose picked tickets with prime digits -> 4 possible digits [2, 3, 5, 7]
How many possible combinations consist of prime digits? = 4x3x2 = 24
How many possible combinations for even ticket that consist of prime digits? = 3 x 2 x 1 = 6 (since last digits has to be "2")
How many possible combinations for odd ticket that consist of prime digits? = 24 - 6 = 18

P(odd) = 18/24 = 3/4
P(even) = 6/24 = 1/4

If Jack pick odd and Rose pick even, then P(odd, even) = 3/4 x 1/4 = 3/16
If Jack pick even and Rose pick odd, then P(even, odd) = 1/4 x 3/4 = 3/16

So total P = 3/16 + 3/16 = 3/8
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Re: Lottery tickets numbered consecutively from 100 through 999 are placed &nbs [#permalink] 07 Nov 2018, 19:19
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Lottery tickets numbered consecutively from 100 through 999 are placed

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