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Nataly_w [17]
3 years ago
7

Three bottles of different sizes contain different compositions of red and blue candy. The largest bottle contains eight red and

two blue pieces, the mid-size bottle has five red and seven blue, the small bottle holds four red and two blue. A monkey will pick one of these three bottles, and then pick one piece of candy from it. Because of the size differences, there is a probability of 0.5 that the large bottle will be picked, and a probability of 0.4 that the mid-size bottle is chosen. Once a bottle is picked, it is equally likely that the monkey will select any of the candy inside, regardless of color.
a. What is the probability that a blue candy is picked?
b. If a blue candy is picked, what is the probability that the large bottle was selected?
Mathematics
1 answer:
choli [55]3 years ago
3 0

Answer:

0.3667

0.2727

Step-by-step explanation:

a)

P (Blue) = P (large and Blue) + P (mid and Blue) + P (small and Blue)

P (Blue) = (0.5 * 0.2) + (0.4*7 / 12) + (0.1* 1 / 3)

P (Blue) = 0.3667

b)

Conditional Probability:

P (Large / Blue ) = P (large and Blue) / P ( Blue)

P (Large / Blue ) = 0.5*0.2 / 0.36667

P (Large / Blue ) = 0.2727

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Let \vec r(t),\vec v(t),\vec a(t) denote the rocket's position, velocity, and acceleration vectors at time t.

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\vec r(0)=\langle0,0,10\rangle\,\mathrm m

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Immediately after launch, the rocket is subject to gravity, so its acceleration is

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where g=9.8\frac{\rm m}{\mathrm s^2}.

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\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}

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(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}

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