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please can anybody help me out here I have an assigment but I don't know how to go about it

QUESTION ONE
in a survey of 100 students, those studying various languages were found to be: Spanish 28; German 30; French 42; Spanish and French 10; Spanish and German 8; German and French 5; all the three languages 3
(I) how many students studied no language?
(ii) how many students had French as their only language?

QUESTION TWO
in a manufacturing process 20% of items produced are defective. 6 items are chosen from the production line, determine the probability of getting the following
(I) exactly one?
(ii) none?
(iii) more than two?

I will be very grate full if any mathematician can help me out pleasssss tanks

Am sorry I can't help. I suck when it comes to mathematics. Would have loved to help.

chiibekee:
Am sorry I can't help. I suck when it comes to mathematics. Would have loved to help.

OK but tanks anyway

**Question One:**

**(I) How many students studied no language?**

To find the number of students who studied no language, we can use the principle of inclusion-exclusion. First, add the individual counts of students studying each language:

Spanish: 28
German: 30
French: 42

Now, account for those studying more than one language:

Spanish and French: 10
Spanish and German: 8
German and French: 5

Also, subtract the count of students studying all three languages:

All three languages: 3

Let's calculate:

Total students studying at least one language = (Spanish) + (German) + (French) - (Students studying two languages) + (Students studying all three languages)
Total students studying at least one language = 28 + 30 + 42 - (10 + 8 + 5) + 3
Total students studying at least one language = 100

Therefore, the number of students studying no language = Total students - Students studying at least one language
Number of students studying no language = 100 - 100
Number of students studying no language = 0

So, there are **0 students** who studied no language.

**(ii) How many students had French as their only language?**

To find the number of students who studied only French, subtract the students studying French and another language(s) from the total number studying French.

Students studying only French = French - (French and Spanish) - (French and German) + (All three languages)

Let's calculate:

Students studying only French = French - (French and Spanish) - (French and German) + (All three languages)
Students studying only French = 42 - 10 - 5 + 3
Students studying only French = 30

So, there are **30 students** who had French as their only language.

**Question Two:**

**(I) Probability of exactly one defective item out of 6 chosen?**

This can be calculated using the binomial probability formula:

Probability of getting exactly \(k\) successes in \(n\) trials = \(\binom{n}{k} \times p^k \times (1-p)^{n-k}\)

Where:
\(n\) = total trials (6 items chosen)
\(k\) = number of successes (exactly one defective item)
\(p\) = probability of success (probability of choosing a defective item = 20% or 0.2)

Let's calculate:

Probability of exactly one defective item = \(\binom{6}{1} \times 0.2^1 \times (1-0.2)^{6-1}\)
Probability of exactly one defective item = \(6 \times 0.2 \times 0.8^5\)
Probability of exactly one defective item ≈ \(0.329\)

Therefore, the probability of getting exactly one defective item out of 6 chosen is approximately **0.329**.

**(ii) Probability of none of the 6 items being defective?**

The probability of not choosing a defective item in a single trial is \(1 - 0.2 = 0.8\). To find the probability of none of the 6 items being defective, we raise the probability of not choosing a defective item to the power of the number of trials.

Probability of none defective in 6 items = \(0.8^6\)
Probability of none defective in 6 items ≈ \(0.262\)

Therefore, the probability of none of the 6 items being defective is approximately **0.262**.

**(iii) Probability of more than two defective items out of 6 chosen?**

To find the probability of more than two defective items, you can find the individual probabilities of getting 0, 1, and 2 defective items, and then subtract their sum from 1 to find the probability of getting more than two defective items.

Probability of more than two defective items = 1 - (Probability of 0 defective + Probability of 1 defective + Probability of 2 defective)

We've already calculated the probabilities of 0 and 1 defective items. To find the probability of exactly 2 defective items:

Probability of exactly two defective items = \(\binom{6}{2} \times 0.2^2 \times (1-0.2)^{6-2}\)
Probability of exactly two defective items = \(15 \times 0.04 \times 0.8^4\)
Probability of exactly two defective items ≈ \(0.331\)

Now, let's calculate the probability of more than two defective items:

Probability of more than two defective items = 1 - (Probability of 0 defective + Probability of 1 defective + Probability of 2 defective)
Probability of more than two defective items = 1 - (0.262 + 0.329 + 0.331)
Probability of more than two defective items ≈ \(0.078\)

Therefore, the probability of getting more than two defective items out of 6 chosen is approximately **0.078**.


I am an English student, but chatgpt did this.

Pls check well ooo, I don't know if it is totally correct