In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

Respuesta :

Answer:

[tex]P_{f} = \frac{7}{24}[/tex]

Step-by-step explanation:

Considering that the set has 24 cards on it, numbered from 1 to 24 inclusive, it means that the total number of cases is 24. Now, we have to find the number of favorable outcomes.

  • Cases where the number is divisible by both 2 and 3:

In the interval 1 to 24 inclusive, [tex][1,24][/tex]:

Numbers divisible by 2 => 2 4 6 8 10 12 14 16 18 20 22 24

Numbers divisible by both 2 and 3 => 6 12 18 24

Which means that there are 4-favorable outcomes for this scenario.

  • Cases where the number is divisible by 7:

In the interval 1 to 24 inclusive, [tex][1,24][/tex]:

Numbers divisible by 7 => 7 14 21

Which means that there are 3-favorable outcomes for this scenario.

  • The total number of favorable outcomes is [tex]4 + 3 = 7[/tex]. Since the total number of cases is [tex]24[/tex], the probability of favorable outcomes, [tex]P_{f}[/tex], is:

[tex]P_{f} = \frac{number of favorable outcomes}{Total number of cases} = \frac{7}{24}[/tex]