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Papessa [141]
3 years ago
10

A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five

percent of all type A goggles are returned within 10 days after the sale, whereas only two percent of type B are returned. If a pair of goggles is returned within the first 10 days after the sale, the probability that the goggles returned are of type B is
Mathematics
1 answer:
exis [7]3 years ago
7 0

Answer:

\dfrac{14}{29}

Step-by-step explanation:

Let <em>P(A) </em>be the probability that goggle of type A is manufactured

<em>P(B) </em>be the probability that goggle of type B is manufactured

<em>P(E)</em> be the probability that a goggle is returned within 10 days of its purchase.

According to the question,

<em>P(A)</em> = 30%

<em>P(B)</em> = 70%

<em>P(E/A)</em> is the probability that a goggle is returned within 10 days of its purchase given that it was of type A.

P(E/B) is the probability that a goggle is returned within 10 days of its purchase given that it was of type B.

P(A \cap E) will be the probability that a goggle is of type A and is returned within 10 days of its purchase.

P(B \cap E) will be the probability that a goggle is of type B and is returned within 10 days of its purchase.

P(E \cap A) = P(A) \times P(E/A)

P(E \cap A) = \dfrac{30}{100} \times \dfrac{5}{100}\\\Rightarrow P(E \cap A) = 1.5 \%

P(E \cap B) = P(B) \times P(E/B)

P(E \cap B) = \dfrac{70}{100} \times \dfrac{2}{100}\\\Rightarrow P(E \cap B) = 1.4 \%

P(E) = 1.5 \% + 1.4 \% \\P(E) = 2.9\%

If a goggle is returned within 10 days of its purchase, probability that it was of type B:

P(B/E) = \dfrac{P(E \cap B)}{P(E)}

\Rightarrow \dfrac{1.4 \%}{2.9\%}\\\Rightarrow \dfrac{14}{29}

So, the required probability is \dfrac{14}{29}.

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Step-by-step explanation:

See attached for visual

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<h3>How many liters of tea does Ben drink every 6 days?</h3>

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