A probablility question about 4 checkers. 2 black and 2 red.?

Question

Questions a,b,c,d,e,f
Lets say 4 checkers(2 red and 2 black) were placed into a paper sack..Shakin up...and then you reached in and grabbed 2 of them. (a)What is the probability that you will get 2 of the same color?

Put the checkers back in the paper sack. Shake it up. Reach in the sack and pull out 1 checker. Without looking at its color place it in a box. Reach back in the paper sack and pull out a 2nd checker. (b)What is the probability the 2nd checker is the same color at the 1st checker(the one in the box)? (c)Does putting the 1st checker in a box change the probability that the 2nd checker will match? (d)What is the probability of both checkers being the same color? (e) What is the probability that the 1st checker(the one in the box) will be a different color than the 2nd checker?

Were these questions easy or hard?

Answer 1

a. 1/3
b. 1/3
c. no
d. 1/3
e. 2/3

Answer 2

A) the probablilty of getting two of the same color is 1/6-- on the first draw, you have a 2:4 (or 1/2) probabilty of getting either color- on the next draw you have a 1/3 probability of getting the remaining checker of the color withdrawn before (3 checkers left, only one with the color already drawn) these two are not mutually exclusive (the first MUST happen for the second to happen) so you multiply the probabilities together 1/2 * 1/3 = 1/6

b) is the same answer
c) no
d) 1/6 same question
e) 5/6- the inverse of them being the same

relatively easy

Answer 3

a. 50%
b. 33.33%
c. yes, drawing them separately does.
d. this is the same as b.
e. 66.67%

f. It is easy enough to enumerate the possibilities with only 4 checkers.

(BTW, B and E have to add up to 1, either they're the same color or not.)

I'll enumerate them here:
a)
Red Red
Red Black
Black Red
Black Black

So there are 4 possibilities, 2 are of the same color, thus this is 50%.

b) Case 1 Red is chosen first.
Black
Black
Red (same)

Case 2 Black is chosen first.
Red
Red
Black (same)

So there are 6 possibilities here, 2 give you the same color, thus it is 1/3.

In other words: The probability they're both black is 1/6, and the probability that they're both red is also 1/6.

c) since a and b are different, then the answer is yes.

d) I really don't see how this is any different than b.

e) Look at my enumeration in B, count the number of both color choices (4/6) or 2/3.

Answer 4

i believe:
a) a 33% chance exists of pulling the matching checker given the first one is removed and not rinserted into the bag. given four, one has to be red or black. given it is black and not replaced this leaves one black and 2 reds or a 33% percent of pulling a black one. if the black checker is reinserted into the bag the ratio is 2:2 or you have a fifty percent chance for a match since the first draw becomes inconsequential due to checker replacement. this actually answers all situations from a-d.
e) however has a 66% percent chance of success. not knowing the color of the checker pulled is of no consequence. given you pull a checker (lets say it is black for the sake of argument) this leaves on black and 2 reds, leaving a 2 out of 3 scenario or 66% chance of having two different colored checkers drawn.
they weren't too difficult. just have to read the ?'s carefully

Answer 5

finding out on first black checker is 10/20 or 50% finding out on second black checker with change is likewise 10/20 or 50% Multiply mutually to discover the prospect of finding out on 2 black checkers with change a million/2 x a million/2=a million/4 each and each individual determination is self sustaining, yet combining to %. to relies upon

Answer 6

a. 1/6 that you will get 2 of the same color.
b. 1/6
c. no
d. 1/6
e. 1/2

Answer 7

a: 1/2
b: 1/12
c: yes
d: same as b
e: 2/3

Hell yes. your doing gr 10 or11 probibility. Good luck!

@pfmonkey, how can it be 1/6? there are only 4 sets of combination.

Answer 8

50%

Answer 9

1,6 4,6