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0 . 7 5 Clue
THE PROBABILITY OF THE ANTS COLLIDING IS 0.75. EACH ANT CAN MOVE IN 2 DIFFERENT DIRECTIONS. BECAUSE THERE ARE 3 ANTS, THIS MEANS THAT THERE ARE 23 (8) POSSIBLE WAYS THAT THE ANTS CAN MOVE. NOW, THERE WILL NEVER BE A COLLISION BETWEEN ANY OF THE ANTS IF THEY ARE ALL WALKING IN THE SAME DIRECTION. AND, THE ONLY TIME THEY WILL BE WALKING IN THE SAME DIRECTION IS IF THEY ARE ALL WALKING EITHER CLOCKWISE OR COUNTER-CLOCKWISE AROUND THE TRIANGLE. SO, THERE ARE ONLY TWO SCENARIOS IN WHICH A COLLISION WILL NOT HAPPEN BETWEEN THE ANTS. THIS MEANS THAT THERE ARE 6 SCENARIOS WHERE THE ANTS WILL COLLIDE. AND 6 OUT OF 8 POSSIBLE SCENARIOS, MEANS THAT THE PROBABILITY OF COLLISION IS 6/8, WHICH EQUALS 3/4 OR 0.75. THUS, THE PROBABILITY OF THE ANTS COLLIDING IS 0.75 IT WILL TAKE 2 HOURS TO MEET. METHOD 1: IGNORE THE SPEED OF THE STREAM, AS THE BOBBER WILL BE CARRIED ALONG AT THREE MILES PER HOUR AS WILL YOU. IT TAKES TWO HOURS TO TRAVEL FOURTEEN MILES, AT A RATE OF SEVEN MILES PER HOUR. METHOD 2: AS THE BOBBER TRAVELS AT 3 MPH, IT WILL BE SIX MILES CLOSER TO YOU IN TWO HOURS. THE DISTANCE BETWEEN YOU AND THE BOBBER BECOMES 8 MILES (14 - 6). IN TWO HOURS YOU WOULD HAVE TRAVELLED 8 ( (7-3) X 2 ) MILES THE OPTION TO PULL THE TRIGGER AGAIN WILL GIVE A HIGHER PROBABILITY OF NOT BEING SHOT. THE KEY HINT HERE IS THAT THE BULLETS WERE LOADED ADJACENT TO EACH OTHER. THERE ARE 4 WAYS TO ARRANGE THE REVOLVER WITH CONSECUTIVE BULLETS SO THAT THE FIRST SHOT IS BLANK. THESE ARE THE POSSIBLE SCENARIOS: (XBBXXX) (XXBBXX) (XXXBBX) (XXXXBB) THE OTHER TWO SCENARIOS WOULD HAVE MEANT YOU GOT SHOT ON THE FIRST ATTEMPT. (BBXXXX) OR (BXXXXB). NOW LOOK AT THE SECOND SLOT IN THE 4 POSSIBLE SCENARIOS ABOVE. THE ODDS OF GETTING SHOT ARE 1⁄4 OR 25% (ONLY THE FIRST SCENARIO WOULD GET YOU SHOT). BUT IF YOU RESPIN, THERE ARE 2 BULLETS REMAINING AND A TOTAL OF 6 SLOTS, WHICH GIVES A PROBABILITY OF 2⁄6 OR 33% OF BEING SHOT ONE TRAIN WAS RUNNING TWICE AS FAST AS THE OTHER. LET: SPEED OF THE FAST TRAIN = F SPEED OF THE SLOW TRAIN = S TIME IT TAKES FOR THE TRAINS TO MEET (PASS EACH OTHER) = T SINCE BOTH TRAINS TRAVEL THE SAME TOTAL DISTANCE AND DISTANCE = TIME X SPEED: F(T+1) = S(T+4) WE'RE TRYING TO FIGURE OUT F/S WHICH IS EQUAL TO (T+4) / (T+1) FROM THE EQUATION ABOVE. SO WE NEED TO FIGURE OUT THE VALUE OF T. AFTER THEY MEET, THE FAST TRAIN TRAVELS ONE MORE HOUR AT SPEED F AND COVERS THE SAME DISTANCE THE SLOW TRAIN COVERED IN T HOURS: F1 = ST OR F = ST AFTER THEY MEET, THE SLOW TRAIN TRAVELS FOR 4 MORE HOURS AND COVERS THE SAME DISTANCE THE FAST TRAIN COVERED IN T HOURS: S4 = FT SUBSTITUTING ST FROM THE FIRST EQUATION IN FOR F IN THE 2ND EQUATION: 4S = STT 4 = TT 2 = T SUBSTITUTE 2 IN FOR T IN THE (T+4) / (T+1) EQUATION TO GET 6/3 OR 2. THE FAST TRAIN IS GOING TWICE AS FAST AS THE SLOW TRAIN