Question

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 ?

Answer 1

Answer:

$P_{f} = \frac{7}{24}$

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, $[1,24]$:

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, $[1,24]$:

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 $4 + 3 = 7$. Since the total number of cases is $24$, the probability of favorable outcomes, $P_{f}$, is:

$P_{f} = \frac{number of favorable outcomes}{Total number of cases} = \frac{7}{24}$