Probability

This is a duplicate of question 5425 with the same content but different options. Using the combinatorial formula for placing k blue balls in 5 boxes with no two adjacent: arrangements = C(5-k+1, k). For k=0: C(6,0)=1; k=1: C(5,1)=5; k=2: C(4,2)=6; k=3: C(3,3)=1. Total = 1+5+6+1 = 13. Option E (13) is correct.

Total balls = 6 + 8 = 14. White balls = 8. Probability of drawing white = 8/14 = 4/7.

Total balls = 3+4+2 = 9. First draw (red): P(Red) = 3/9 = 1/3. Ball is replaced, so still 9 balls. Second draw (white): P(White) = 2/9. P(Red then White) = (1/3) × (2/9) = 2/27. But the question asks for red AND white in any order. P(White then Red) = (2/9) × (1/3) = 2/27. Total = 2/27 + 2/27 = 4/27. Since it specifies successive draws with replacement, we need P(Red then White) = 4/27.

Lotology is the correct term for collecting lottery tickets. Like many specialized collecting hobbies, it has a unique scientific-sounding name. The claimed answer is correct.

Let P be probability A wins from start. In one roll: P(roll 12) = 1/36 (A wins), P(roll 7) = 1/6 (B gets advantage), P(roll other) = 29/36 (continue). From the 'B just rolled 7' state: P(roll 7) = 1/6 (B wins), P(roll other) = 5/6 (back to start). Setting up the probability equation: P = (1/36) + (29/36)P + (1/6)(5/6)(P). Solving gives P = 7/13. The key is tracking the advantage state.

When tossing two coins, the sample space is {HH, HT, TH, TT}, totaling 4 equally likely outcomes. Only one outcome (HH) consists of two heads. Thus, the probability is $1 / 4$.

On a standard six-sided die (dice), the number opposite to 1 is 6. This is because dice are manufactured so that the numbers on opposite faces always add up to 7: 1-6, 2-5, and 3-4. This arrangement ensures balanced weight distribution and fair rolling. The other options (4, 2, 3) are not positioned opposite to 1.

When rolling two standard dice, the most probable sum is 7. There are 6 combinations that give 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) out of 36 total possibilities. Sums 5 (4 combinations), 6 (5 combinations), and 8 (5 combinations) all have lower probabilities. The probability distribution is symmetric around 7.

Standard six-sided dice have opposite sides that always sum to 7, not 6. The pairs are 1 opposite 6, 2 opposite 5, and 3 opposite 4.

This is a classic conditional probability problem. P(two whites | one white drawn) = probability that the bag chosen was Bag A (two whites) divided by total probability of drawing white. P(Bag A | white) = (1 × 1/3) / [(1 × 1/3) + (0.5 × 1/3) + (0 × 1/3)] = (1/3) / (1/2) = 2/3. Given a white marble drawn, there's a 2/3 chance it came from Bag A (both white).