Probability

There are 3 white marbles that could have been picked (2 from Bag A, 1 from Bag C). In 2 of those 3 cases, the remaining marble is white (Bag A). Thus, the probability is 2/3.

This is the classic two-coin probability puzzle. There are 3 heads faces that could be shown (HH coin has 2, HT coin has 1). If we see heads, 2 out of 3 cases are from the double-headed coin, so P(other side is heads) = 2/3. Many students mistakenly think it's 1/2.

This is the St. Petersburg paradox. Expected value = sum of (1/2^n × 2^n) = sum of 1 = infinity. Each term contributes $1 to the expectation. This is a famous result in probability theory showing infinite expectation from a finite-prize game.

P(at least one hit in 4 shots) = 1 - P(no hits) = 1 - (3/4)^4 = 1 - 81/256 = 175/256. Each shot is independent with P(hit) = 1/4.

Total outcomes = 6^4 = 1296. Outcomes with no 3s = 5^4 = 625. So outcomes with at least one 3 = 1296 - 625 = 671.

To solve this question, the user needs to know the concept of conditional probability. Conditional probability is the probability of an event occurring given that another event has already occurred. In this case, we want to find the probability of selecting a white marble given that we have already selected a white marble from the same bag.

Let's use Bayes' theorem to solve the problem. Bayes' theorem states that the probability of A given B is equal to the probability of B given A multiplied by the probability of A, divided by the probability of B. In this case, we want to find the probability of selecting a white marble given that we have already selected a white marble from the same bag.

Let A be the event that the remaining marble is white, and let B be the event that we have already selected a white marble from the same bag.

So, we want to find P(A|B), which is the probability that the remaining marble is white given that we have already selected a white marble from the same bag.

From the problem statement, we know that there are three bags, and we selected one of them randomly. Thus, the probability of selecting any one of the bags is 1/3.

Now, let's consider each bag separately:

• If we selected Bag A, then the probability of selecting a white marble is 1, since both marbles in Bag A are white. Thus, the probability that the remaining marble is white is also 1.

• If we selected Bag B, then the probability of selecting a white marble is 0, since both marbles in Bag B are black. Thus, the probability that the remaining marble is white is 0.

• If we selected Bag C, then the probability of selecting a white marble is 1/2, since one of the marbles in Bag C is white and the other is black. If we selected the white marble, then the remaining marble must be black. Alternatively, if we selected the black marble, then the remaining marble must be white. Thus, the probability that the remaining marble is white is 1/2.

Now, we can use the law of total probability to find the probability of selecting a white marble from any one of the bags:

P(B) = P(B|A) * P(A) + P(B|B) * P(B) + P(B|C) * P(C)

P(B) = 1 * (1/3) + 0 * (1/3) + 1/2 * (1/3)

P(B) = 1/2

Thus, the probability of selecting a white marble from any one of the bags is 1/2.

Now, we can apply Bayes' theorem:

P(A|B) = P(B|A) * P(A) / P(B)

P(A|B) = 1 * (1/3) / (1/2)

P(A|B) = 2/3

Therefore, the probability that the remaining marble from the same bag is also white is 2/3.

Option B is the correct answer.

The Answer is: B. 2/3

This is a fundamental property of standard six-sided dice (cubes). The opposite sides are arranged so that they always sum to seven: 1 opposite 6, 2 opposite 5, and 3 opposite 4. This design ensures balanced weight distribution.

This is total probability theorem. P(Red) = P(choose bag 1) × P(Red|bag 1) + P(choose bag 2) × P(Red|bag 2) = (1/2) × (3/5) + (1/2) × (4/10) = 3/10 + 2/10 = 5/10 = 1/2. Each bag is equally likely to be chosen, then we combine the weighted probabilities.

Total ways to choose any subset from 12 balls = 2^12 = 4096. Subtract selections with no green (2^7 = 128) and no blue (2^8 = 256). Add back selections with neither (2^3 = 8) since counted twice. 4096 - 128 - 256 + 8 = 3720.

On a standard six-sided die, opposite faces are always arranged to sum to seven. The pairs are: 1 opposite 6, 2 opposite 5, and 3 opposite 4. This design ensures balanced distribution of numbers across the cube.

This is the two-box paradox. The tempting reasoning that switching gives $12.50 expected value is flawed. When you see $10, the possible pairs are ($5,$10) or ($10,$20). Since you chose the $10 box, if the pair was ($5,$10) you definitely picked the larger one (probability 1), but if the pair was ($10,$20) you only had a 50% chance of picking the smaller one. This creates selection bias: seeing $10 is more likely from the ($5,$10) pair (2/3) than from ($10,$20) (1/3). So the other box contains $5 with 2/3 probability and $20 with 1/3 probability, giving expected value of $10 - no gain from switching.

Using Bayes' theorem or simple counting: there are 3 white marbles total, and 2 of them are in Bag A (where the remaining is white). If you drew a white marble, the probability it came from Bag A is 2/3. Only Bag A guarantees a white remaining; Bag C has a black remaining. Therefore the probability is 2/3.

When a standard die is thrown once, there are 6 equally likely outcomes (1,2,3,4,5,6). The numbers 'between 2 and 6' (excluding endpoints) are 3, 4, and 5 - that's 3 favorable outcomes. Probability = 3/6 = 1/2. Option C is correct. Option B (2/6) would be incorrect, and option A (1) is impossible for probability.

On a standard six-sided die, the opposite faces are designed to always sum to seven (1+6, 2+5, 3+4). Therefore, the statement that they add up to eight is false.

Each coin toss has a 1/2 probability of being heads. For 4 independent tosses, the probability of all 4 being heads is (1/2) * (1/2) * (1/2) * (1/2) = 1/16.