Question

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Answer 1

Answer:

Choice number one:

$\displaystyle \frac{5}{10}\cdot \frac{4}{9}$.

Step-by-step explanation:

• Let $A$ be the event that the number on the first card is even.
• Let $B$ be the event that the number on the second card is even.

The question is asking for the possibility that event $A$ and $B$ happen at the same time. However, whether $A$ occurs or not will influence the probability of $B$. In other words, $A$ and $B$ are not independent. The probability that both $A$ and $B$ occur needs to be found as the product of

• the probability that event $A$ occurs, and
• the probability that event $B$ occurs given that event $A$ occurs.

5 out of the ten numbers are even. The probability that event $A$ occurs is:

$\displaystyle P(A) = \frac{5}{10}$.

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:

$\displaystyle P(B|A) = \frac{4}{9}$.

The probability that both $A$ and $B$ occurs will be:

$\displaystyle P(A \cap B) = P(B\cap A) = P(A) \cdot P(B|A) = \frac{5}{10}\cdot \frac{4}{9}$.