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Kazeer [188]
3 years ago
15

Suppose you have a bag containing 2 black marbles and 3 red marbles. You reach into the bag, select a marble, see what color it

is and replace it in the bag. Then you repeat this process a second time. What is the probability of picking a red marble both times?
Mathematics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

\dfrac{9}{25}

Step-by-step explanation:

Given that the bag contains black and red marbles.

Number of black marbles in the bag = 2

Number of red marbles in the bag = 3

Total number of marbles in the bag = Number of black marbles + Number of red marbles = 2 + 3 = 5

Let us have a look at the formula for probability of an event E, which can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(\text{First red marble}) = \dfrac{\text{Number of red marbles}}{\text{Total number of marbles}} = \dfrac{3}{5 }

Now, the marble chosen at first is replaced.

Therefore, the count remains the same.

P(\text{Second red marble}) = \dfrac{\text{Number of red marbles}}{\text{Total number of marbles}} = \dfrac{3}{5}

Now, the <em>required probability</em> can be found as:

P(\text{First red marble})\times P(\text{Second red marble}) = \dfrac{3}{5}\times \dfrac{3}{5} = \bold{\dfrac{9}{25} }

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The Fundamental Theorem of Calculus regarding geometry states that 
\int\limits^b_a g{(x)} \, dx = F(b)-F(a)

Where F is the indefinite integral of g(x)

The first step is to integrate g(x)
\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}

Then substitute the value of b and a=1 into 2x^{2}

[2 (b)^{2}]-[2 (1)^{2}] = 240
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b^{2}= \frac{242}{2}
b^{2}=121
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Hence the limit of the area under g(x) is between a=1 and b=11

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3 years ago
P/2 = 3/4 + p/3 solve for p<br> Show your work<br> It is URGENT
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Step-by-step explanation:

p/2 = 3/4 + p/3

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p/6 = 3/4

multiply both sides by 6

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3 years ago
Conditional Distribution, Marginal Distribution, Joint Distribution. <br> What’s the difference?
lutik1710 [3]

Explanation:

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Conditional distribution: This distribution contrasts from the previous one in the sense that we are restricting the universe of events to specific condition for other variable, making a modification of our marginal results. If we know that throwing a dice will give us a result higher than 2, then to in order to calculate the probability of the dice being a multiple of 3 using that condition, we have two favourable cases (3 and 6) from 4 total possible results (3,4,5 and 6) discarding the impossible values (1 and 2) from this universe since they dont match the condition given (note that the restrictions given can also reduce the total of favourable cases).

The joint distribution calculates the probabilities for two different events (related to two different random variables) occuring simultaneously. If we want to calculate the joint probability of a dice being multiple of 3 and greater than 2 at the same time, our possible cases in this case are 3 and 6 from 6 possible results. We are not discarding 1 or 2 as possible results because we are not assuming, that the dice is greater than 2, that is another condition that we should met in the combination of events.

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Serhud [2]
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5 0
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I dont understand the circled question ....please help!
irina1246 [14]

Parallel lines NEVER intersect/meet, this is because they have the same slope and they go in the same direction.

These lines are parallel because they have the same slopes of 3/4 (because they go 3 units up and 4 units to the right. slope = rise/run)

7 0
3 years ago
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