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3 years ago

7

When three pipes fill a pool, they can finish in 12 hours. Two of the pipes can finish in 18 hours if they are working together.

How long would it take a third pipe to fill the pool on its own?

1 answer:

Ymorist [56]3 years ago

6 0

The number of pipes * the time is a constant proportional to the pool size.

3*12 = 2*18 = 1*36. So one pipe would take 36 hours.

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Then, n(S) = 12 + 8 + 5 = 25.

Let, E, F and G represents the events of pulling a red marble, a blue marble and a green marble respectively.

The, n(E) = 12, n(F) = 8 and n(G) = 5.

Therefore, the probabilities of each of these three events E, F and G will be

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{12}{25}=\dfrac{12}{25}\times100\%=48\%,\\\\\\P(F)=\dfrac{n(F)}{n(S)}=\dfrac{8}{25}=\dfrac{8}{25}\times100\%=32\%,\\\\\\P(G)=\dfrac{n(G)}{n(S)}=\dfrac{5}{25}=\dfrac{5}{25}\times100\%=20\%.$

Now, the probability of pulling three green marbles out with replacement is given by

$P(G)\times P(G)\times P(G)=\dfrac{5}{25}\times\dfrac{5}{25}\times \dfrac{5}{25}=\dfrac{1}{125}=\dfrac{1}{125}\times100\%=0.8\%.$

Thus, the probability of pulling a red marble is 48%, probability of pulling a blue marble is 32%, probability of pulling a green marble is 20%

and

the probability of pulling three green marbles out with replacement is 0.8%.

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3 years ago