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Thread: New dice problem for Alan...

  1. #1
    "Given two fair dice, what is the probability that the sum of their numbers is 4 if exactly one die shows a 3?"

    I think that the experts might agree with you on this one.

    Answer(s) at https://gmatclub.com/forum/given-two...um-205280.html

    Gone to the next life.
    Last edited by Bill Yung; 12-28-2017 at 02:52 PM.
    78255585899=317*13723*17989=(310+7)*[(13730-7)*(100*100+7979+10)]-->LOVE avatar@137_371_179_791, or 137_371_17[3^2]_7[3^2]1, 1=V-->Ace, low. 78255585899-->99858555287=(99858555288-1)=[-1+(72*2227)*(722777-100000)]={-1+(72*2227)*[(2000+700777+20000)-100000]}-->1_722_227_277_772_1. 7×8×2×5×5×5×8×5×8×9×9=362880000=(1000000000-6√97020000-100000)-->169_721. (7/8×2/5×5/5×8/5×8/9×9)={[(-.1+.9)]^2×(6+1)}-->1961=√4*2.24; (1/7×8/2×5/5×5/8×5/8×9/9)={1/[7×(-.2+1)^2]}-->1721=[(10*10/4)/(√4+110)].

  2. #2
    Notice in the responses that someone there even knows the difference between "exactly one die" and "at least one die." Alan thinks they mean the same when in reality, they're NOT.

  3. #3
    Here's a new link to this problem. The old one seems to have died.

    https://www.quora.com/Given-two-fair...-die-shows-a-3
    78255585899=317*13723*17989=(310+7)*[(13730-7)*(100*100+7979+10)]-->LOVE avatar@137_371_179_791, or 137_371_17[3^2]_7[3^2]1, 1=V-->Ace, low. 78255585899-->99858555287=(99858555288-1)=[-1+(72*2227)*(722777-100000)]={-1+(72*2227)*[(2000+700777+20000)-100000]}-->1_722_227_277_772_1. 7×8×2×5×5×5×8×5×8×9×9=362880000=(1000000000-6√97020000-100000)-->169_721. (7/8×2/5×5/5×8/5×8/9×9)={[(-.1+.9)]^2×(6+1)}-->1961=√4*2.24; (1/7×8/2×5/5×5/8×5/8×9/9)={1/[7×(-.2+1)^2]}-->1721=[(10*10/4)/(√4+110)].

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