AAPL wrote: ↑

Fri Feb 05, 2021 10:29 am

Manhattan Prep

Raffle tickets ere consecutively from 201 through 450 are placed in a certain box. What is the probability of a ticket picked from the box at random having a number with at least one of its digits as 3?

A. 52/ 125
B. 58/125
C. 133/250
D. 71/125
E. 77/ 125

OA C

Solution:

We see that there are a total of 450 - 201 + 1 = 250 raffle tickets or integers. So 250 will be the denominator of the desired probability, and the numerator will be the number of integers that have at least one of its digits as 3. Let’s determine that now.

All the integers in the 300s have at least one digit of 3, so there are 100 such integers. For the integers in the 200s, 19 of them have at least one digit of 3 (9 have a units digit of 3 only, another 9 have the tens digit of 3 only, and 1 has both the tens and units digits of 3). For the integers in the 400s (up to 450), 14 of them have at least one digit of 3 (4 have the units digit of 3 only, another 9 have the tens digit of 3 only, and 1 has both the tens and units digits of 3). Therefore, the total number of integers that has at least one digit of 3 is 100 + 19 + 14 = 133, and therefore, the probability is 133/250.

Answer: C