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Sep 13, 2006 5:22 am |
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re: re: re: re: re: re: The Man in the Bar.... a riddle! |
Sumanth Cidambi
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I am solving as follows:
12 houses, 9 planets, Houses 6, 8 and 12 to have no planets.
P(Houses 6, 8 and 12 having Planet 1) = 1/12 +1/12 +1/12 =3/12 or 1/4, because the events are mutually exclusive and P1 can be in H6 or H8 or H12 but not in all three houses at the same time. Therefore, the aliter probability is (1 - 1/4) = 3/4 or 0.75
Similarly, we get probability of 3/4 for P2 through P9 in Houses H1 through H12.
Now the fact that P1 is not in House 6 or in House 8 or in House 12 no way affects the behaviour of P2 through P9, i.e. the location of the planets in any house are mutually independent of each other
Hence the probability of their occurence is as follows:
3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 or
3/4 ^ 9 or 0.07508
If wrong, do educate me.
Cheers S
Private Reply to Sumanth Cidambi (new win) |
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