Question

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Answer 1

Choice number one:

`\displaystyle (5)/(10)\cdot (4)/(9)`.

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) = (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) = (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) = (5)/(10)\cdot (4)/(9)`.