Question

type the correct answer in the box. use numerals instead of words. for this problem, a non - integer answer should be entered as a fraction in simplest form using / as the fraction bar. malik randomly picked two numbers from 1 to 9 (including 1 and 9). the same number could be picked more than once. the first of the two numbers he picks is odd and less than 5. what is the probability that the sum of the two numbers malik picks is less than 5, given that the first number is odd and less than 5?

Explanation:

Step1: Determine the possible first - numbers

The first number is odd and less than 5. So the possible first - numbers are 1 and 3.

Step2: Calculate the total number of cases for the second number

Since the second number can be from 1 to 9 (9 possibilities for each first - number), the total number of cases for the two - number pairs (given the first - number condition) is \(n = 2\times9=18\).

Step3: Find the pairs where the sum is less than 5

If the first number \(a = 1\):

• Let the second number be \(b\). We want \(a + b<5\), so \(1 + b<5\), which gives \(b < 4\). So \(b = 1,2,3\) (3 possibilities).

If the first number \(a = 3\):

• We want \(a + b<5\), so \(3 + b<5\), which gives \(b < 2\). So \(b = 1\) (1 possibility).

The number of favorable cases \(m=3 + 1=4\).

Step4: Calculate the probability

The probability \(P=\frac{m}{n}\). Substituting \(m = 4\) and \(n = 18\), we get \(P=\frac{4}{18}=\frac{2}{9}\).

Answer:

\(\frac{2}{9}\)