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Sagot :
To find the probability that exactly 4 students out of 12 have taken chemistry, given that 40% of students at the high school take chemistry, we will use the binomial probability formula:
[tex]\[ P(k \text{ successes }) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where:
- [tex]\( n \)[/tex] is the number of trials (students surveyed).
- [tex]\( k \)[/tex] is the number of successes (students taking chemistry).
- [tex]\( p \)[/tex] is the probability of success on a single trial (probability of a student taking chemistry).
Given:
- [tex]\( n = 12 \)[/tex]
- [tex]\( k = 4 \)[/tex]
- [tex]\( p = 0.40 \)[/tex]
First, let's compute [tex]\(\binom{n}{k} \)[/tex], which is the number of combinations of [tex]\( n \)[/tex] items taken [tex]\( k \)[/tex] at a time:
[tex]\[ \binom{n}{k} = \frac{n!}{(n-k)! \cdot k!} \][/tex]
Substituting in the values for [tex]\( n \)[/tex] and [tex]\( k \)[/tex]:
[tex]\[ \binom{12}{4} = \frac{12!}{(12-4)! \cdot 4!} = \frac{12!}{8! \cdot 4!} \][/tex]
Next, simplify the factorials:
[tex]\[ 12! = 12 \times 11 \times 10 \times 9 \times 8! \quad \text{and} \quad 8! \text{ cancels out, so we get} \][/tex]
[tex]\[ \frac{12 \times 11 \times 10 \times 9}{4!} \][/tex]
Calculating [tex]\( 4! \)[/tex]:
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]
So:
[tex]\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{24} = \frac{11880}{24} = 495 \][/tex]
Now, substitute [tex]\(\binom{12}{4}\)[/tex], [tex]\( p = 0.40 \)[/tex], and [tex]\( (1-p) = 0.60 \)[/tex] into the binomial probability formula:
[tex]\[ P(4 \text{ successes}) = 495 \times (0.40)^4 \times (0.60)^8 \][/tex]
Calculate [tex]\( (0.40)^4 \)[/tex]:
[tex]\[ (0.40)^4 = 0.40 \times 0.40 \times 0.40 \times 0.40 = 0.0256 \][/tex]
Calculate [tex]\( (0.60)^8 \)[/tex]:
[tex]\[ (0.60)^8 = 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 = 0.016797 \][/tex]
Finally, multiply these results together with the combination:
[tex]\[ 495 \times 0.0256 \times 0.016797 = 0.21284093952 \][/tex]
Rounding this result to the nearest thousandth:
[tex]\[ 0.21284093952 \approx 0.213 \][/tex]
Therefore, the probability that exactly 4 students out of 12 surveyed have taken chemistry is [tex]\( 0.213 \)[/tex]. The correct answer is 0.213.
[tex]\[ P(k \text{ successes }) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where:
- [tex]\( n \)[/tex] is the number of trials (students surveyed).
- [tex]\( k \)[/tex] is the number of successes (students taking chemistry).
- [tex]\( p \)[/tex] is the probability of success on a single trial (probability of a student taking chemistry).
Given:
- [tex]\( n = 12 \)[/tex]
- [tex]\( k = 4 \)[/tex]
- [tex]\( p = 0.40 \)[/tex]
First, let's compute [tex]\(\binom{n}{k} \)[/tex], which is the number of combinations of [tex]\( n \)[/tex] items taken [tex]\( k \)[/tex] at a time:
[tex]\[ \binom{n}{k} = \frac{n!}{(n-k)! \cdot k!} \][/tex]
Substituting in the values for [tex]\( n \)[/tex] and [tex]\( k \)[/tex]:
[tex]\[ \binom{12}{4} = \frac{12!}{(12-4)! \cdot 4!} = \frac{12!}{8! \cdot 4!} \][/tex]
Next, simplify the factorials:
[tex]\[ 12! = 12 \times 11 \times 10 \times 9 \times 8! \quad \text{and} \quad 8! \text{ cancels out, so we get} \][/tex]
[tex]\[ \frac{12 \times 11 \times 10 \times 9}{4!} \][/tex]
Calculating [tex]\( 4! \)[/tex]:
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]
So:
[tex]\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{24} = \frac{11880}{24} = 495 \][/tex]
Now, substitute [tex]\(\binom{12}{4}\)[/tex], [tex]\( p = 0.40 \)[/tex], and [tex]\( (1-p) = 0.60 \)[/tex] into the binomial probability formula:
[tex]\[ P(4 \text{ successes}) = 495 \times (0.40)^4 \times (0.60)^8 \][/tex]
Calculate [tex]\( (0.40)^4 \)[/tex]:
[tex]\[ (0.40)^4 = 0.40 \times 0.40 \times 0.40 \times 0.40 = 0.0256 \][/tex]
Calculate [tex]\( (0.60)^8 \)[/tex]:
[tex]\[ (0.60)^8 = 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 = 0.016797 \][/tex]
Finally, multiply these results together with the combination:
[tex]\[ 495 \times 0.0256 \times 0.016797 = 0.21284093952 \][/tex]
Rounding this result to the nearest thousandth:
[tex]\[ 0.21284093952 \approx 0.213 \][/tex]
Therefore, the probability that exactly 4 students out of 12 surveyed have taken chemistry is [tex]\( 0.213 \)[/tex]. The correct answer is 0.213.
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