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10. From the numbered cards from 1 to 30, a card is drawn at random. Find the probabilities of the following events:

a) The number is either divisible by 4 or by 5.
b) The number is either divisible by 5 or by 6.
c) The number is either a multiple of 3 or a multiple of 4.
d) The number is either a prime or an even number.



Answer :

GinnyAnswer
Certainly! Let's go through each part of the question step-by-step, calculating the probabilities for each specified event.

### Part (a)

#### (i) The number is either divisible by 4 or by 5

To determine the probability of drawing a number that is either divisible by 4 or by 5:

1. Total numbers: 30 (numbered 1 to 30).
2. Divisibility by 4: Numbers divisible by 4 between 1 and 30 are: 4, 8, 12, 16, 20, 24, 28.
- Count: 7.
3. Divisibility by 5: Numbers divisible by 5 between 1 and 30 are: 5, 10, 15, 20, 25, 30.
- Count: 6.
4. Divisibility by both 4 and 5 (LCM is 20): Numbers divisible by 20 between 1 and 30 are: 20.
- Count: 1.
5. Using the principle of inclusion and exclusion:
[tex]\[ \text{P(4 or 5)} = \frac{\text{Count(4)} + \text{Count(5)} - \text{Count(4 and 5)}}{\text{Total numbers}} = \frac{7 + 6 - 1}{30} = \frac{12}{30} = 0.4 \][/tex]

So, the probability is [tex]\(0.4\)[/tex].

#### (ii) The number is either divisible by 5 or by 6

To determine the probability of drawing a number that is either divisible by 5 or by 6:

1. Total numbers: 30.
2. Divisibility by 5: Count: 6 (as above).
3. Divisibility by 6: Numbers divisible by 6 between 1 and 30 are: 6, 12, 18, 24, 30.
- Count: 5.
4. Divisibility by both 5 and 6 (LCM is 30): Numbers divisible by 30 between 1 and 30 are: 30.
- Count: 1.
5. Using the principle of inclusion and exclusion:
[tex]\[ \text{P(5 or 6)} = \frac{\text{Count(5)} + \text{Count(6)} - \text{Count(5 and 6)}}{\text{Total numbers}} = \frac{6 + 5 - 1}{30} = \frac{10}{30} \approx 0.333 \][/tex]

So, the probability is approximately [tex]\(0.333\)[/tex].

#### (iii) The number is either a multiple of 3 or a multiple of 4

To determine the probability of drawing a number that is either a multiple of 3 or a multiple of 4:

1. Total numbers: 30.
2. Multiple of 3: Numbers multiple of 3 between 1 and 30 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
- Count: 10.
3. Multiple of 4: Count: 7 (as above).
4. Multiple of both 3 and 4 (LCM is 12): Numbers multiple of 12 between 1 and 30 are: 12, 24.
- Count: 2.
5. Using the principle of inclusion and exclusion:
[tex]\[ \text{P(3 or 4)} = \frac{\text{Count(3)} + \text{Count(4)} - \text{Count(3 and 4)}}{\text{Total numbers}} = \frac{10 + 7 - 2}{30} = \frac{15}{30} = 0.5 \][/tex]

So, the probability is [tex]\(0.5\)[/tex].

#### (iv) The number is either a prime or an even number

To determine the probability of drawing a number that is either a prime or an even number:

1. Total numbers: 30.
2. Even numbers: Numbers divisible by 2 between 1 and 30 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.
- Count: 15.
3. Prime numbers: Numbers between 1 and 30 that are prime: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
- Count: 10.
4. Prime and even numbers: The only even prime number is 2.
- Count: 1.
5. Using the principle of inclusion and exclusion:
[tex]\[ \text{P(Prime or Even)} = \frac{\text{Count(Prime)} + \text{Count(Even)} - \text{Count(Prime and Even)}}{\text{Total numbers}} = \frac{10 + 15 - 1}{30} = \frac{24}{30} = 0.8 \][/tex]

So, the probability is [tex]\(0.8\)[/tex].

### Part (b)

Unfortunately, part (b) seems cut off or incomplete.

If you could clarify or complete the question, I would be happy to assist further!

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