Algebra

Probability

Medium

applied

It is known that 3% of plastic bags manufactured in a factory are defective. Using the Poisson distribution on a sample of 100 bags, the probability of at most one defective bag is:

$\frac{4}{e^2}$

$\frac{4}{e^3}$

$e^{-3}$

$\frac{1}{3}$

✅ Correct Option: 2

The factory produces plastic bags where 3% are defective. For a sample of 100 bags, the probability of at most one defective bag needs to be determined using the Poisson distribution.

The Poisson distribution applies when there is a large sample size with a small probability of occurrence.

Lambda (λ) represents the expected number of defective bags:

λ = n × p

Where n = 100 bags and p = 0.03

λ = 100 × 0.03

λ = 3

"At most 1 defective bag" means either 0 or 1 defective bag:

P(at most 1) = P(X = 0) + P(X = 1)

The Poisson formula is:

P(X = k) = $\frac{e^{-λ} \times λ^k}{k!}$

For P(X = 0):

P(X = 0) = $\frac{e^{-3} \times 3^0}{0!}$

P(X = 0) = $\frac{e^{-3} \times 1}{1}$

P(X = 0) = $e^{-3}$

For P(X = 1):

P(X = 1) = $\frac{e^{-3} \times 3^1}{1!}$

P(X = 1) = $\frac{e^{-3} \times 3}{1}$

P(X = 1) = $3e^{-3}$

P(at most 1) = P(X = 0) + P(X = 1)

P(at most 1) = $e^{-3} + 3e^{-3}$

P(at most 1) = $e^{-3}(1 + 3)$

P(at most 1) = $4e^{-3}$

P(at most 1) = $\frac{4}{e^3}$

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A die is thrown once then,

Match List-I with List-II

List-I	List-II
(A) Probability of getting odd number	(I) $\dfrac{1}{3}$
(B) Probability of getting number more than 4	(II) $\dfrac{2}{3}$
(C) Probability of getting number less than or equal to 4	(III) $\dfrac{1}{2}$
(D) Probability of getting number 2	(IV) $\dfrac{1}{6}$

Choose the correct answer from the options given below:

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Probability that a man speaks truth is $\frac{3}{4}$ . He throws a die and reports that it is a six. The probability that it is actually a six is

It is known that 3% of plastic bags manufactured in a factory are defective. Using the Poisson distribution on a sample of 100 bags, the probability of at most one defective bag is:

Solution

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