nahid078 wrote:A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1

P(2nd ball is yellow or 3rd ball is not yellow) = P(2nd ball is yellow) + P(3rd ball is not yellow) - P(2nd ball is yellow AND 3rd ball is not yellow).

Rule:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.

P(Y on the 2nd pick) = P(Y on the first pick) = 1/4. (Of the 4 balls, 1 is Y.)
P(R or B on the 3rd pick) = P(R or B on the first pick) = 3/4. (Of the 4 balls, 3 are R or B.)

There is an OVERLAP between the two probabilities above.
If the 2nd ball is yellow, then the 3rd ball is automatically NOT yellow.
Since the 3rd ball will be not yellow in every case in which the 2nd ball is yellow, we get:
P(2nd ball is yellow and 3rd ball is not yellow) = P(2nd ball is yellow) = 1/4.

Plugging these probabilities into the blue equation above, we get:
P(2nd ball is yellow or 3rd ball is not yellow) = 1/4 + 3/4 - 1/4 = 3/4.

The correct answer is B.

nahid078 wrote:A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1

We are given that a box contains 1 blue ball, 1 yellow ball, and 2 red balls. We need to determine, when selecting 3 balls from the box, the probability that the second ball is yellow OR the third ball is not yellow.

Let's look at the requirements in two separate scenarios:

Scenario 1: The second ball is yellow

The probability of not choosing the yellow ball in the first draw is 3/4, and the probability of choosing the yellow ball from the remaining 3 balls in the second draw is 1/3 (we do not care which ball is selected in the third draw in this scenario). Therefore, the probability of choosing the yellow ball in the second draw is:

3/4 x 1/3 = 1/4

Scenario 2: The third ball is not yellow. There are several ways this can happen, as follows:

a. The first ball is yellow. If this happens, we know that the third ball cannot be yellow because there is only one yellow ball in the box.

Since there are four balls, the probability of choosing the yellow ball in the first draw is 1/4 (we do not care which balls are selected in the second or third draws in this scenario).

b. The second ball is yellow. If this happens, we know that the third ball cannot be yellow because there is only one yellow ball in the box. We already considered this outcome in Scenario 1, so we WILL NOT include it again here.

c. The yellow ball is not selected at all. This outcome also satisfies the requirement that the third ball is not yellow.

The probability of choosing a non-yellow ball in the first, second, and third draws is 3/4, 2/3, and 1/2, respectively. So the probability of not choosing a yellow ball in three draws is:

3/4 x 2/3 x 1/2 = 1/4

Since any of these outcomes would satisfy the given condition and they are mutually exclusive, we can sum the individual probabilities of each outcome:

1/4 + 1/4 + 1/4 = 3/4

Answer: B

We could also just set up our probability equation:

P(#2 yellow) + P(#3 not yellow) =

P(#1 not yellow)*P(#2 yellow) + P(#3 not yellow) =

(3/4)*(1/3) + P(first two both not yellow)*P(#3 not yellow) + P(#1 yellow) =

(3/4)*(1/3) + (3/4)*(2/3)*(1/2) + (1/4) =

1/4 + 1/4 + 1/4