Answer:
Choice number one:
[tex]\displaystyle \frac{5}{10}\cdot \frac{4}{9}[/tex].
Step-by-step explanation:
- Let [tex]A[/tex] be the event that the number on the first card is even.
- Let [tex]B[/tex] be the event that the number on the second card is even.
The question is asking for the possibility that event [tex]A[/tex] and [tex]B[/tex] happen at the same time. However, whether [tex]A[/tex] occurs or not will influence the probability of [tex]B[/tex]. In other words, [tex]A[/tex] and [tex]B[/tex] are not independent. The probability that both [tex]A[/tex] and [tex]B[/tex] occur needs to be found as the product of
- the probability that event [tex]A[/tex] occurs, and
- the probability that event [tex]B[/tex] occurs given that event [tex]A[/tex] occurs.
5 out of the ten numbers are even. The probability that event [tex]A[/tex] occurs is:
[tex]\displaystyle P(A) = \frac{5}{10}[/tex].
In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:
[tex]\displaystyle P(B|A) = \frac{4}{9}[/tex].
The probability that both [tex]A[/tex] and [tex]B[/tex] occurs will be:
[tex]\displaystyle P(A \cap B) = P(B\cap A) = P(A) \cdot P(B|A) = \frac{5}{10}\cdot \frac{4}{9}[/tex].
