Probability Question?

Question

What is the probabily of drawing 3 red marbles from a bag that contains 8 red, 6 blue, and 4 yellow? What is the probability of drawing a read and then a blue? A blue and red on the first and second try?

Please explain

Just take them out and dont put them back in.

Answer 1

Are you replacing the marbles after you draw them?

Answer 2

The first question is a little ambiguous. Do you mean drawing 3 red marbles in a row? If so, then the answer is 0.069. There are 18 marbles total, so you have (8/18 x 7/17 x 6/16). Each time you remove a red marble from the bag you also decrease the total number of marbles available.

As for drawing a red and then a blue, or vice versa, the answer is the same 0.157. Red then blue: (8/18 x 6/17). Blue then red: (6/18 x 8/17).

Answer 3

The bag contains 16 marbles. Eight are red, eight are blue, and four are yellow.

The probability of drawing a red marble the first time is 1 in 2 because there are eight red marbles. Eight out of sixteen chances. With each red marble you draw (and your question seems to imply consecutively), your chance of drawing a red marble goes down (e.g., your probability of drawing the second is 7 in 15, the third is 6 in 14 which equals 3 in 8).

To draw a red first, again, is 1 in 2. To draw a blue second is 6 in 15 because there are fifteen marbles left in the bag, and only six of them are blue. Reduce that to 2 in 5.

A blue on the first try (there are six of them in the sixteen-marble bag) would be, aptly, 6 in 16. That would be reduced to lowest terms of 3 in 8. Then to draw a red marble second would be 8 in 15.

Probabilities would be written with the two numbers on opposite sides of a colon (e.g., 8:15). Hope this helps.

Answer 4

On your first pick you have a 8 out of 18 chance of picking a red marble (8 red marbles, 18 total marbles). Assuming that you pick a red marble and remove it from the bag, you have a 7 out of 17 chance to pick a red marble. Again, if you pick red and remove it from the bag, you have a 6 out of 16 chance to pick red. The probability is calculated by multiplying each of these probabilities together, that is: 8/18 x 7/17 x 6/16 or 336 out of 4896 or 1 out of 14.57142857 or roughly 7%.

The probability of drawing a red then a blue is 8/18 x 6 (number of blue balls)/17 or 48 out of 306 chance or 1 out of 6.375

The probability of drawing a blue then a red would be 6/18 x 8/17 or 48 out of 306 chance or 1 out of 6.375, the same odds as drawing a red then a blue.

Answer 5

A fairly easy way to look at what you need to do to figure this out is to look at what your question is. When you're looking at the (total) probability of events that aren't dependent from one another, you need to determine whether or not you're looking at AND statements or OR statements.

An example of an OR statement: "Probability of drawing a red OR a blue marble on the first try?" On "ORs," you can add those individual probabilities together. In this case, it'd be the probability of drawing a red out of the bag (8/18) + probability of drawing a blue out of the bag (6/18), or 14/18.

Your question has a couple of ANDs, even if you don't realize it. What's the probability of drawing a red marble from the bag on the first try, AND a red marble on the 2nd try, AND a red marble on the third try. In the case of AND statements, you multiply the probabilities together instead of adding.

So!

Red marble on first try: (total red marbles/total marbles) 8/18
Red marble on 2nd try (total red marbles - what you pulled out/total marbles - what you pulled out) (8-1)/(18-1) = 7/17
Red marble on 3rd try (total red marbles - what you pulled out/total marbles - what you pulled out (8-2)/(18-2) = 6/16

And your total probability becomes 8/18 * 7/17 * 6/16 ---

Which you can reduce into the following: 1/3 * 7/17 * 1/2

For a total of 7/102.

Answer 6

well 3 reds in a row:

the first one you have 8 of 18 chance of getting a red so its 8/18 then for the 2nd its 7 /17 and for the 3rd its 6 /16 ..for a total of 8/18*7/17*6/16 or about a 6.8% chance of doing that. in that order

dwg a red then a blue...ok...8/18 times 6/17 (the reason its 17 instead of 18--you only have 17 left to chose from) or a total about 15.6%

of dwg a blue first and then a red in that order? 6/18 (6 of the 18 are blue) times 8/17 (8 of the 17 left are red) also 15.6%

if not in that order:
of dwg a blue on the first try: = 6/18 or 33.33%
of dwg a red on the first try = 8/18 or 44%
on the 2nd try--well it depends on what was drawn on the first try...

Answer 7

p(drawing 3 red marbles)
=(number of favorable events)/(total events)
=(8C3)/(18C3)
=56/816
=7/102
=0.0686


p(drawing 1 red and 1 blue marbles)
=(number of favorable events)/(total events)
=(8C1*6C1)/(18C2)
=(8*6)/(18*17)
=8/51
=0.15686


and abt the last part...........drawing a blue and red on the first and second try?............is it like i draw a marble then without replacing it i draw the second????

i hope taht this helps u....

:)

Answer 8

(8/18) * (7/17) * (6/16) = (336/4896) = (7/102) = 0.0686

So probability is just under 7%