Tony Kelly & Kieran Mills) : Aths Roblem O
Tony Kelly & Kieran Mills) : Aths Roblem O
Tony Kelly & Kieran Mills) : Aths Roblem O
MATHS PROBLEM NO. 2 An aeroplane engine fails with probability x. A three-engined plane needs 2 or 3 engines for a successful flight. Find (i) the probability of a successful flight on a one-engined plane in terms of x, (ii) the probability of a successful flight in a three-engined plane,
(iii) the range of values of x for which a one-engined plane is safer than a three-engined plane.
MATHS PROBLEM NO. 2 An aeroplane engine fails with probability x. A three-engined plane needs 2 or 3 engines for a successful flight. Find (i) the probability of a successful flight on a one-engined plane in terms of x, (ii) the probability of a successful flight in a three-engined plane,
(iii) the range of values of x for which a one-engined plane is safer than a three-engined plane. SOLUTION (i) One engined plane: p(Failed flight) = x p(Successful flight) = (1 x) (ii) Three-engined flight: p(Successful flight) = p(3 engines working) or p(2 engines working and one failing)
3! 2!
MATHS PROBLEM NO. 2 An aeroplane engine fails with probability x. A three-engined plane needs 2 or 3 engines for a successful flight. Find (i) the probability of a successful flight on a one-engined plane in terms of x, (ii) the probability of a successful flight in a three-engined plane,
(iii) the range of values of x for which a one-engined plane is safer than a three-engined plane. SOLUTION (iii) p(Successful flight in a one-engined plane) > p(Successful flight in a three-engine flight)
(1 x) > (1 x) 2 (2 x + 1) (1 x) (1 x) 2 (2 x + 1) > 0 (1 x)[1 (1 x)(2 x + 1)] > 0 (1 x)[1 (2 x + 1 2 x 2 x)] > 0 (1 x)[1 ( x + 1 2 x )] > 0
2
NOTE: There are no solutions less than 0 or greater than 1 as the probability of an event must be between 0 and 1.
1 2
x=
1 4
x=
3 4
(1 x)[1 x 1 + 2 x 2 ] > 0 (1 x)[2 x 2 x] > 0 (1 x)[ x(2 x 1)] > 0 x(1 x)(2 x 1) > 0
1 1 (1 4 )(1 4 )( 2( 4 ) 1) 3 1 = (1 4 )( 4 )( 2 1) 3 1 = (1 4 )( 4 )( 2 )
3 3 (3 4 )(1 4 )( 2( 4 ) 1) 3 1 = (3 4 )( 4 )( 2 1) 1 1 = (3 4 )( 4 )( 2 )
<0 False
>0 True
ANSWER:
1 2
< x <1