Question

1. a bag contains 40 slips of paper numbered 1 - 40. if one is randomly chosen, what is the probability of not choosing a multiple of 3?

$\frac{3}{40}$
$\frac{3}{21}$
$\frac{9}{10}$
$\frac{27}{40}$
clear all

Explanation:

Step1: Find number of multiples of 37

The multiples of 37 between 1 - 40 is 37 itself. So there is 1 multiple of 37 in the set of numbers from 1 - 40.

Step2: Calculate probability of choosing a multiple of 37

The probability formula is $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Here, the total number of outcomes is 40 (since there are 40 slips of paper), and the number of favorable outcomes (choosing a multiple of 37) is 1. So $P(\text{multiple of 37})=\frac{1}{40}$.

Step3: Calculate probability of not choosing a multiple of 37

We know that $P(\text{not }A)=1 - P(A)$. So $P(\text{not multiple of 37})=1-\frac{1}{40}=\frac{40 - 1}{40}=\frac{39}{40}$. But it seems there is a mistake in the options. If we assume the question was about multiples of 4 (as the options seem to be related to such a problem), the multiples of 4 between 1 - 40 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, so there are 10 multiples of 4. Then $P(\text{multiple of 4})=\frac{10}{40}=\frac{1}{4}$, and $P(\text{not multiple of 4})=1-\frac{1}{4}=\frac{3}{4}=\frac{30}{40}$. If we assume the question was about multiples of 3, the multiples of 3 between 1 - 40 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, so there are 13 multiples of 3. Then $P(\text{multiple of 3})=\frac{13}{40}$, and $P(\text{not multiple of 3})=1 - \frac{13}{40}=\frac{27}{40}$.

Answer:

$\frac{27}{40}$ (assuming the intended number was 3 instead of 37 based on the options)