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2 years ago

9

There are 5 different coloured cups in a row.

1 answer:

Marizza181 [45]2 years ago

6 0

There are 12 possible arrangements.

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Considering that the blue cup has to be on one end, we consider the number of arrangements possible with the other 4 cups, then multiply by 2(considering the blue on each end).

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Considering green and yellow before red and purple, there are $2 \times 2 = 4$ outcomes(G-Y-R-P, G-Y-P-R, Y-G-R-P and Y-G-P-R).
Another two possible outcomes are yellow-purple and green-red, or vice-versa, thus 4 + 2 = 6 outcomes.
To account for the blue on the end, multiplying by 2. $2 \times 6 = 12$
There are 12 possible arrangements.

A similar problem is given at brainly.com/question/24617788

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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

exis [7]

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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Lapatulllka [165]

Answer:

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