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
