The probability that we will get blue 1, 2, or 3 times can also be expressed as the probability that we will NOT get blue 0 or 4 times. In other words:

P(blue = 1 + blue = 2 + blue= 3) = 1 - P(blue = 0 + blue = 4)

The probability that we will get blue from a bag is 25/100 = 1/4. The probability that we will get something other than blue from a bag is 75/100 = 3/4.

So the probability that we will get something other than blue for all four bags (blue = 0) is 3/4 *3/4 * 3/4 * 3/4, and the probability that we will get only blue for all four bags (blue = 4) is 1/4 * 1/4 * 1/4 * 1/4. Putting that in our equation gives:

1 - (3/4 *3/4 * 3/4 * 3/4 + 1/4 * 1/4 * 1/4 * 1/4) = 1 - (81/256 + 1/256) = 1 - (82/256) = 1 - (41/128) = 87/128

BTGmoderatorRO wrote:Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256

OA is c.

I got A as my answer but option B looks more correct to me. Can some Expert help me? Thanks alot

We should first note that the condition "the number of blue disks chosen is no less than 1 and no greater than 3" is equivalent to the condition "the number of blue disks is neither 0 nor 4."

We can use the following equation:

P(the number of blue disks chosen will be no less than 1 and no greater than 3) = 1 - P(selecting 0 blue disks) - P(selecting 4 blue disks)

Let's first determine P(selecting 0 blue disks):

75/100 x 75/100 x 75/100 x 75/100 = (3/4)^4 = 81/256

Next let's determine P(selecting 4 blue disks):

25/100 x 25/100 x 25/100 x 25/100 = (1/4)^4 = 1/256

Thus:

P(the number of blue disks chosen will be no less than 1 and no greater than 3) is:

1 - 81/256 - 1/256 = 1 - 82/256 = 174/256 = 87/128

Answer: C