Lottery tickets numbered consecutively from 100 through 999 are placed

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

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.


Thanks,
GyM

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
E

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

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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This one is a very intellectual solution. Short and effective. Thank You!­

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Prime digits from 0 to 9 are 2,3,5 and 7. We have to keep a few things in mind.
1. Even prime is 2, odd primes are 3,5,7.
2. The required sum is odd and we know even+odd=odd, so either Jack or Rose has to select a lottery which ends with 2 and the other person has to select a lottery where unit digit is odd. Then only the sum will be odd.
Total number of ways in which Jack can select 4*3*2=24, Rose can select 4*3*2=24. So 24*24=576.
Let, Jack select a lottery that ends with 2. Then Rose selects a lottery that ends with an odd digit.
Jack can select in 3*2*1=6 ways and Rose in 3*2*3=18 ways.
The reverse of the above can also happen where Rose selects a lottery ending with 2 and Jack selects a lottery that ends with an odd digit. So total ways are 2*(6*18)=216.
The probability is 216/576=3/8. Option (E) is correct.