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12 Jul 2026, 03:01
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If the question said sum should be more than 12 and above,the the probability would nothave been half?
12 Jul 2026, 03:14
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nikitathegreat
It would not be half.
The probability is half only because “higher than 10” means scores 11 through 18, and those scores pair perfectly with 3 through 10 around the middle value 10.5.
But if Joe needed more than 12, then only scores 13 through 18 would count. That is only the upper part of the distribution, so the probability would be less than 1/2.
Check this for more: https://gmatclub.com/forum/mary-and-joe ... ml#p788093
12 Jul 2026, 18:26
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it's a symmetry problem. Computing P(Joe > 10) directly means summing the counts for totals 11 through 18 — eight separate tallies. Symmetry collapses all of it into one line.
The key insight: the sum of three dice is symmetric about 10.5 (each die averages 3.5, so three average 10.5). The distribution mirrors around that axis:
P(sum = k) = P(sum = 21 − k)
So P(3) = P(18), P(4) = P(17), and so on.
Now map the question onto that mirror:
- Sums above 10: {11, 12, ..., 18}
- Sums 10 or below: {10, 9, ..., 3}
Under the reflection k ↔ 21 − k, the first set maps exactly onto the second. They're mirror images — and because 10.5 isn't an achievable sum, nothing sits on the axis. No leftover middle case.
The outcomes therefore split perfectly in half:
P(Joe > 10) = 1/2 = 108/216
The trap: if Mary had scored 11, the split would not be even — 11 lies above the axis, so you'd have to subtract the ties. The problem gives 10 precisely because the mirror lands cleanly on it.
The transferable move: when a distribution is symmetric, check whether the question is really just asking which side of the axis you're on. If it is, the answer falls out with no counting.
The key insight: the sum of three dice is symmetric about 10.5 (each die averages 3.5, so three average 10.5). The distribution mirrors around that axis:
P(sum = k) = P(sum = 21 − k)
So P(3) = P(18), P(4) = P(17), and so on.
Now map the question onto that mirror:
- Sums above 10: {11, 12, ..., 18}
- Sums 10 or below: {10, 9, ..., 3}
Under the reflection k ↔ 21 − k, the first set maps exactly onto the second. They're mirror images — and because 10.5 isn't an achievable sum, nothing sits on the axis. No leftover middle case.
The outcomes therefore split perfectly in half:
P(Joe > 10) = 1/2 = 108/216
The trap: if Mary had scored 11, the split would not be even — 11 lies above the axis, so you'd have to subtract the ties. The problem gives 10 precisely because the mirror lands cleanly on it.
The transferable move: when a distribution is symmetric, check whether the question is really just asking which side of the axis you're on. If it is, the answer falls out with no counting.
