From the side where the man is located in the 10' by 12' rectangular room, we have;
A) 0.046
B) The average time to escape in first attempt is approximately 605.02 seconds
<h3>Which concept can be used to find the odds and time of escape?</h3>
A) From The given diagram, we have;
Width of the door = 10 - 6 - 2 = 2
Therefore, the door is 2 feet wide
Distance, d1, from the man to the side of the door closer to him is given by Pythagorean theorem as follows;
From the furthest side of the door to the man, we have;
According to the law of cosines, we have;
2² = 85+113 - 2×√(85)×√(113) cos(A)
Where angle <em>A </em>is the angle formed by the lines from the man to the closer and farther sides of the door.
2×√(85)×√(113) cos(A) = 85+113 - 2²
A = arccos((85+113 - 2²)/(2×√(85)×√(113)))
A = arccos (194/(2×√(9605)) ≈ 8.213°
The angle of the possible directions in which the man can turn is 180°.
Therefore;
The probability, <em>P(</em><em>E)</em>, of escaping in the first move is therefore;
- P(E) = 8.213°/180° ≈ 0.046
B) Given that the man escapes in the first move, the shortest and longest distances the man moves are;
d1 = √(85) and d2 = √(113)
Speed of the man, <em>v </em>= 0.5 cm/sec
Therefore by unit conversion function of a graphing calculator;
The times taken are;
t1 = √(85)/(25/1524) ≈ 562.023
t2 = √(113)/(25/1524) ≈ 648.014.
Average time, <em>t </em>= (t1 + t2)/2
Which gives;
t = (562.023 + 648.014)/2 ≈ 605.02
The average time it takes the man to escape in first move is <em>t </em><em>≈</em> 605.02 seconds.
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