Question

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? ( _______ )

Answer 1

Answer:

The required probability is $\frac{2}{9}$

Step-by-step explanation:

Let us take the events,

A = The sum of the two numbers is less than 5

B = The first number is odd and is less than 5.

Now, each of the two numbers can take value from 1 to 9.

So, the total number of outcomes = 9 × 9 = 81.

Also, the first number is odd and less than 5.

Then, the possibility for the first number = 1 and 3

While the possibility for the second number = 1 to 9

So, P(B) = $\frac{2\times 9}{81}=\frac{18}{81}$

Now, we see that,

$A\bigcap B$ = The event having sum of the two numbers less than 5 with first number odd and less than 5.

The possible numbers are, (1,1), (1,2), (1,3) and (3,1).

So, $P(A\bigcap B)$ = $\frac{4}{81}$

Hence,

$P(A|B) = \frac{P(A\bigcap B)}{P(B)}$

i.e. $P(A|B) = \frac{\frac{4}{81}}{\frac{18}{81}}$

i.e. $P(A|B) = \frac{4}{18}$

i.e. $P(A|B) = \frac{2}{9}$

i.e. P(A|B) = $\frac{2}{9}$

Thus, the required probability is $\frac{2}{9}$.

Answer 2

Answer:

2/9

Step-by-step explanation: