Question

There are three events X,Y and Z, one of which must and only can happen. If the odds are 7:4 against X, 5:3 against Y, the odds against Z must be:

Answer 1

SOLUTION

GIVEN

There are three events X,Y and Z, one of which must and only can happen. If the odds are 7:4 against X, 5:3 against Y,

TO DETERMINE

The odds against Z

EVALUATION

Here it is given that the odds are 7 : 4 against X

$\displaystyle \sf{ \frac{P(X')}{P(X)} = \frac{7}{4} }$

$\displaystyle \sf{ \implies \frac{1 - P(X)}{P(X)} = \frac{7}{4} }$

$\displaystyle \sf{ \implies 4 - 4P(X) = 7P(X) }$

$\displaystyle \sf{ \implies 11P(X) =4 }$

$\displaystyle \sf{ \implies P(X) = \frac{4}{11} }$

Again it is also stated that odds are 5 : 3 against Y

$\displaystyle \sf{ \frac{P(Y')}{P(Y)} = \frac{5}{3} }$

$\displaystyle \sf{ \implies \frac{1 - P(Y)}{P(Y)} = \frac{5}{3} }$

$\displaystyle \sf{ \implies 5 P(Y) = 3 - 3P(Y) }$

$\displaystyle \sf{ \implies 8 P(Y) = 3 }$

$\displaystyle \sf{ \implies P(Y) = \frac{3}{8} }$

Since there are three events X , Y and Z, one of which must and only can happen.

$\therefore \: \: \sf{P(X) + P(Y) + P(Z) = 1}$

$\displaystyle \: \sf{ \implies \: \frac{4}{11} + \frac{3}{8} + P(Z) = 1}$

$\displaystyle \: \sf{ \implies \: \frac{32 + 33}{88} + P(Z) = 1}$

$\displaystyle \: \sf{ \implies \: \frac{65}{88} + P(Z) = 1}$

$\displaystyle \: \sf{ \implies \: P(Z) = 1 - \frac{65}{88} }$

$\displaystyle \: \sf{ \implies \: P(Z) = \frac{88 - 65}{88} }$

$\displaystyle \: \sf{ \implies \: P(Z) = \frac{23}{88} }$

Hence the odds against Z

$\displaystyle \sf{ = P(Z') : P(Z) }$

$\displaystyle \sf{ = \frac{P(Z')}{P(Z)} }$

$\displaystyle \sf{ = \frac{1 - \frac{23}{88} }{ \frac{23}{88} } }$

$\displaystyle \sf{ = \frac{ \frac{88 - 23}{88} }{ \frac{23}{88} } }$

$\displaystyle \sf{ = \frac{ \frac{65}{88} }{ \frac{23}{88} } }$

$\displaystyle \sf{ = \frac{65}{23} }$

$= 65 : 23$

FINAL ANSWER

The odds against Z = 65 : 23

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