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- anonymous

A bag has 1 red marble, 4 blue marbles, and 3 green marbles. Peter draws a marble randomly from the bag, replaces it, and then draws another marble randomly. What is the probability of drawing 2 blue marbles in a row? Explain your answer.

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- anonymous

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- Valpey

Multiply the probability of drawing just one blue marble times itself.

- anonymous

wait. So 1*4?

- Valpey

The probability of drawing one blue marble is equal to the fraction: number of blue marbles over the total number of marbles.

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- Valpey

Which is more likely, drawing a blue marble or flipping a coin and getting tails?

- anonymous

drawing a blue marble i am assuming because there is only one side that has tails when there are four different blue marbles.

- anonymous

what am multiplying exactly though besides the one marble? The mean times the one marble?
the fraction would be 4/8 or simplified to 1/2

- Valpey

Right 1/2 or 50% of the time you draw a blue marble.

- Valpey

What percent of the time do you draw a blue marble twice in a row? (same as getting tails in two coin tosses)

- anonymous

thats the part I am confused about.

- Valpey

If I flip a coin twice, what is the chance it comes tails, tails?

- anonymous

it could be HH TT or TH. so I would say 1/3 or a chance...

- anonymous

of*

- anonymous

I am so confused
@Valpey

- Valpey

You left out HT which is just as likely as TH. So the four equally likely outcomes are:
HH, TH, HT, TT
Similarly let's say B for Blue Marble and N for Not Blue Marble. The four equally likely outcomes are:
BB, BN, NB, NN

- anonymous

ohhhhhh okay

- anonymous

But how does that go along with the problem? hehe sorry im slow

- anonymous

- Valpey

The answers are the same. Blue marbles represent 50% of the marbles so drawing one is equivalent to any other 50:50 outcome, like tossing a coin.

- Valpey

The chance of drawing two Blue Marbles in a row (with replacement) is the same as the chance of flipping a coin twice and getting tails, tails. This happens 1 in 4 times or 25%.

- Valpey

But we can change the problem slightly and see better how to extend this. Suppose instead there are 3 Blue Marbles and 5 Not Blue Marbles. Now we still have four possible outcomes:
BB, BN, NB, NN
but now they are not all equally likely. Each outcome is the product of the chance of the two independent events separately.
BB = \(\frac{3}{8}*\frac{3}{8} =\frac{9}{64}\)
BN = \(\frac{3}{8}*\frac{5}{8} =\frac{15}{64}\)
NB = \(\frac{5}{8}*\frac{3}{8} =\frac{15}{64}\)
NN = \(\frac{5}{8}*\frac{5}{8} =\frac{25}{64}\)

- anonymous

OHHHHHhhhhhhhhhhhHHHHHHHHhhhhhhhHHHHHHHhhhH

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