Official Solution:Mary and Joe each roll three dice. The total score for each person is the sum of the numbers on their three dice. If Mary scores 10, what is the probability that Joe will have a higher score in his attempt? A. \(\frac{24}{64}\)
B. \(\frac{32}{64}\)
C. \(\frac{36}{64}\)
D. \(\frac{40}{64}\)
E. \(\frac{42}{64}\)
To outscore Mary, Joe must achieve a score in the range of 11-18. The probability of scoring 3 is the same as the probability of scoring 18 (the 1-1-1 combination is opposite the 6-6-6 combination when considering the tops and bottoms of the dice). Similarly, the probability of scoring \(x\) is the same as the probability of scoring \(21 - x\). Consequently, the probability of scoring within the 11-18 range is equal to the probability of scoring within the 3-10 range. Since the range of 3-18 encompasses all possible outcomes, the probability of scoring within the 11-18 range is \(\frac{1}{2}\).
Alternative explanation The expected value when rolling a single die is \(\frac{1}{6}*(1+2+3+4+5+6)=3.5\).
Therefore, the expected value when rolling three dice is \(3*3.5=10.5\).
Mary scored 10, so the probability of getting a sum greater than 10 (11, 12, 13, ..., 18), or greater than the average, is the same as getting a sum less than the average (10, 9, 8, ..., 3). Since these two events are complementary (adding to 1) and symmetrical, the probability of each is 1/2:
The probability of getting a sum of 3 (the minimum possible sum) = the probability of getting a sum of 18 (the maximum possible sum);
The probability of getting a sum of 4 = the probability of getting a sum of 17;
The probability of getting a sum of 5 = the probability of getting a sum of 16;
...
The probability of getting a sum of 10 = the probability of getting a sum of 11.
Therefore, the probability distribution is symmetrical, and thus the probability of getting a sum from 3 to 10 = the probability of getting a sum from 11 to 18 = 1/2.
Answer: B
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