A bag contains 19 cards numbered 1 through 19. A card is randomly chosen from the bag. What is the probability that the card has an even number on it? Write your answer as a fraction in simplest form. willllll get brainest

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Answer:

[tex]\dfrac{9}{19}[/tex]

Step-by-step explanation:

It is given that there are a total of 19 cards ({1,2,3,4, ...... ,19}).

Let S be the set of total cards.

S ={1,2,3,4, ...... ,19}

Total number of observations, n(S) = 19

Even number of cards is the set {2,4,6,8,10,12,14,16,18}

Total number of cards with even number on it = 9

Let E be the event of selecting a card with even number on it.

n(E) = 9

Probability of an event A can be formulated as:

[tex]P(A) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]

In this case, formula of P(E) is:

[tex]P(E) = \dfrac{n(E)}{n(S)}[/tex]

[tex]\Rightarrow \dfrac{9}{19}[/tex]

The probability that the card chosen at random is an even number is:

[tex]\dfrac{9}{19}[/tex]

The probability that the card will have an even number will be 0.4737.

  • From the information given, we are informed that a bag contains 19 cards numbered 1 through 19.

  • Based on the information, the even numbers will be 2, 4, 6, 8, 10, 12, 14, 16, and 18. Therefore, the probability of having when numbers will be:

     = 9/19 = 0.4737

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