Good one for all

The probability that a student passes at least in one of the three examinations A,B,C is 0.75.

The probability that he passes in at least two of the exams is 0.5.

And the prabability he passes in exactly two of the exams is 0.4.

Denote p , q , r p,q,r as the probabilities of the student passing in A,B,C respectively.

Let p + q + r = x y p+q+r=\frac xy for coprime positive integers x , y x,y , find the value of x y x-y .


The answer is 7.

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2 solutions

Alex Zhong
Apr 21, 2015

From the conditions we have :

{ ( 1 p ) ( 1 q ) ( 1 r ) = 1 0.75 , ( 1 ) p q + q r + p r 2 p q r = 0.5 , ( 2 ) p q + q r + p r 3 p q r = 0.4. ( 3 ) \begin{cases} (1-p)(1-q)(1-r) = 1-0.75, (1)\\ pq+qr+pr - 2pqr = 0.5, (2)\\ pq + qr + pr -3pqr = 0.4. (3) \end{cases}

(2)-(3) we have p q r = 0.1 , pqr = 0.1, and p q + q r + p r = 0.7. pq+qr+pr=0.7.

Therefore, ( 1 p ) ( 1 q ) ( 1 r ) = p q r + p q + q r + p r ( p + q + r ) + 1 = 0.1 + 0.7 + 1 + ( p + q + r ) = 0.25 , (1-p)(1-q)(1-r) \\= -pqr + pq+qr+pr -(p+q+r) +1 \\= -0.1 + 0.7 +1 + (p+q+r) \\= 0.25,

p + q + r = 27 20 = x y . p+q+r = \dfrac{27}{20} =\dfrac xy.

Thus, x y = 27 20 = 7 . x-y= 27-20 = \boxed{7}.

Janine Yu
May 23, 2015

The equation you'll get if he passes at least one of the tests is: p(1-q)(1-r) + q(1-p)(1-r) + r(1-p)(1-q) + pq(1-r) +pr(1-q) + qr(1-p) + pqr = 0.75 Simplifying... (p+q+r) - 2(pq+pr+qr) +3pqr = 0.25 (1)

The equation you'll get if he passes at least two of the tests is: pq(1-r) + pr(1-q) + qr(1-p) + pqr = 0.5 (2) Simplifying... (pq+pr+qr) - 3(pqr) = 0.5 (3)

The equation you'll get if he passes exactly two tests is: pq(1-r) + pr(1-q) + qr(1-p) = 0.4 (4)

(2)-(4) we have pqr = 0.5-0.4 pqr= 0.1 (5)

Plugging in (5) to (3), we have pq+pr+qr= 0.4 + 3(0.1) pq+pr+qr = 0.7 (6)

Plugging in (6) and (5) to (1), we have p+q+r= 0.25 + 2(0.7) - 3(0.1) p+q+r= 1.35

1.35= 135/100= 27/20 = x/y Therefore, x-y= 27-20 =7

In equation 3 there should be -2pqr

Dhruv Joshi - 4 years, 2 months ago

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