PLEASE HELP ITS MULTIPLE CHOICE i added extra points and i’ll mark BRAINLIEST !!!

Mathematics

2 answers:

Sphinxa [80]3 years ago

4 0

Answer:

Step-by-step explanation:

Artemon [7]3 years ago

3 0

Answer:

Step-by-step explanation:

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Ratling [72]

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:

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