A box contains cards number 1-10. A card is drawn at random, without replacement, and a second card is drawn. What is the probability that the first number is a multiple of five and the second number is a multiple of three?

Total outcome is 10 and favorable outcome is 2 ⇒ the probability that the first number is a multiple of five is [tex]\frac{2}{10}[/tex] = [tex]\frac{1}{5}[/tex]

Total outcome is 9 and favorable outcome is 3 ⇒ the probability that the first number is a multiple of three is [tex]\frac{1}{3}[/tex] if first draw was successful, and [tex]\frac{2}{9}[/tex] if the first number was multiple of three.

joaobezerra joaobezerra · Answer 2 · 2021-12-03T11:45:07+01:00

Using conditional probability, it is found that there is a 0.0667 = 6.67% probability that the first number is a multiple of five and the second number is a multiple of three.

A probability is the number of desired outcomes divided by the number of total outcomes.

Conditional Probability

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: First number is a multiple of 5.
Event B: The second number is a multiple of 3.

There are 10 numbers, of which 2(5 and 10) are multiples of 5, hence [tex]P(A) = \frac{2}{10} = 0.2[/tex]

Then, supposing a multiple of 5 is taken, there will be 9 numbers, of which 3(3, 6 and 9) are multiples of 3, hence [tex]P(B|A) = \frac{3}{9} = \frac{1}{3} = 0.3333[/tex]

Then:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(A)P(B|A)[/tex]

[tex]P(A \cap B) = 0.2(0.3333)[/tex]

[tex]P(A \cap B) = 0.0667[/tex]

0.0667 = 6.67% probability that the first number is a multiple of five and the second number is a multiple of three.

A similar problem is given at https://brainly.com/question/14398287