Question

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.

Answer 1

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

Answer 2

wait. So 1*4?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

@Valpey @Hero

Answer 8

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

Answer 9

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

Answer 10

thats the part I am confused about.

Answer 11

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

Answer 12

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

Answer 13

of*

Answer 14

I am so confused


@Valpey

Answer 15

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

Answer 16

ohhhhhh okay

Answer 17

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

Answer 18

@Valpey

Answer 19

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.

Answer 20

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%.

Answer 21

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 = \(\frac{5}{8}*\frac{5}{8} =\frac{25}{64}\)

Answer 22

OHHHHHhhhhhhhhhhhHHHHHHHHhhhhhhhHHHHHHHhhhH