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Answer :
Let's solve the problem step by step:
The problem involves the probability of defective pens and follows a binomial distribution because each pen has two possible outcomes: defective or not defective. We are given the following:
- The total number of pens selected, [tex]\( n = 12 \)[/tex].
- The probability of a pen being defective, [tex]\( p = 0.15 \)[/tex].
(a) Probability that exactly 2 pens are defective:
To find this probability, we use the binomial probability formula:
[tex]\[
P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}
\][/tex]
where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, [tex]\( \frac{n!}{k! \times (n-k)!} \)[/tex],
- [tex]\( k \)[/tex] is the number of defective pens we want, which is 2 in this case.
Applying the formula, the probability that exactly 2 pens are defective is approximately 0.292.
(b) Probability that at most ten pens are defective:
This probability includes the cases where 0, 1, 2,..., up to 10 pens are defective. We calculate it by summing these probabilities, or more simply by using a cumulative distribution function (CDF) for a binomial distribution:
[tex]\[
P(X \leq 10) = \sum_{k=0}^{10} \binom{n}{k} \times p^k \times (1-p)^{n-k}
\][/tex]
Calculating this, the total probability that at most ten pens are defective is approximately 0.999999991.
(c) Expected number of defective pens:
The expected number of defective pens is a simple calculation using the formula for the expected value of a binomial distribution:
[tex]\[
E(X) = n \times p
\][/tex]
Substitute the given numbers:
[tex]\[
E(X) = 12 \times 0.15 = 1.8
\][/tex]
So, the expected number of defective pens is 1.8.
In summary:
- The probability that exactly 2 pens are defective is approximately 0.292.
- The probability that at most 10 pens are defective is approximately 0.999999991.
- The expected number of defective pens is 1.8.
The problem involves the probability of defective pens and follows a binomial distribution because each pen has two possible outcomes: defective or not defective. We are given the following:
- The total number of pens selected, [tex]\( n = 12 \)[/tex].
- The probability of a pen being defective, [tex]\( p = 0.15 \)[/tex].
(a) Probability that exactly 2 pens are defective:
To find this probability, we use the binomial probability formula:
[tex]\[
P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}
\][/tex]
where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, [tex]\( \frac{n!}{k! \times (n-k)!} \)[/tex],
- [tex]\( k \)[/tex] is the number of defective pens we want, which is 2 in this case.
Applying the formula, the probability that exactly 2 pens are defective is approximately 0.292.
(b) Probability that at most ten pens are defective:
This probability includes the cases where 0, 1, 2,..., up to 10 pens are defective. We calculate it by summing these probabilities, or more simply by using a cumulative distribution function (CDF) for a binomial distribution:
[tex]\[
P(X \leq 10) = \sum_{k=0}^{10} \binom{n}{k} \times p^k \times (1-p)^{n-k}
\][/tex]
Calculating this, the total probability that at most ten pens are defective is approximately 0.999999991.
(c) Expected number of defective pens:
The expected number of defective pens is a simple calculation using the formula for the expected value of a binomial distribution:
[tex]\[
E(X) = n \times p
\][/tex]
Substitute the given numbers:
[tex]\[
E(X) = 12 \times 0.15 = 1.8
\][/tex]
So, the expected number of defective pens is 1.8.
In summary:
- The probability that exactly 2 pens are defective is approximately 0.292.
- The probability that at most 10 pens are defective is approximately 0.999999991.
- The expected number of defective pens is 1.8.
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Rewritten by : Brahmana