Question

A factory $$A$$ produces $$10$$% defective valves and another factory $$B$$ produces $$20$$% defective. A bag contains $$4$$ valves of factory $$A$$ and $$5$$ valves of factory $$B$$. If two valves are drawn at random from the bag. Find the probability that at least one valve is defective. Give your answer upto two places of decimals.

A
0.30 approximately
B
0.28 approximately
C
0.29 approximately
D
0.27 approximately

Solution

The correct option is B $$0.29$$ approximatelyProbability of producing defective valves by factor $$\displaystyle A=\dfrac { 10 }{ 100 } =\dfrac { 1 }{ 10 }$$Probability of producing defective valves by factory $$\displaystyle B=\dfrac { 20 }{ 100 } =\dfrac { 1 }{ 5 }$$Bag $$A$$ contain $$9$$ valves, $$4$$ of factory $$a$$ and $$5$$ of factory $$b$$.Two valves are $$0$$ be drawn at random.$$P$$(at least one defective) $$=$$$$1-P$$(both are non defective)$$P$$(both are non defective)$$=$$$$P$$(both valves of factory $$B$$)$$\times P$$(both are non defective) $$+$$$$P$$(both valves of factory $$B$$)$$\times P$$(both are non defective)$$+$$$$P$$(one valve of factory $$A$$ and other of factory $$B$$)$$\times P$$(Both are non defective)$$\displaystyle ={ \dfrac { { ^{ 4 }{ C } }_{ 2 } }{ ^{ 9 }{ C }_{ 2 } } \left( \dfrac { 9 }{ 10 } \right) }^{ 2 }+\dfrac { { { ^{ 5 }{ C } } }_{ 2 } }{ ^{ 9 }{ C }_{ 2 } } { \left( \dfrac { 4 }{ 5 } \right) }^{ 2 }+\dfrac { ^{ 4 }{ C }_{ 1 }.^{ 5 }{ C }_{ 1 } }{ ^{ 9 }{ C }_{ 2 } } .\dfrac { 9 }{ 10 } .\dfrac { 4 }{ 5 }$$$$\displaystyle =\dfrac { 1 }{ 6 } { \left( \dfrac { 9 }{ 10 } \right) }^{ 2 }+\dfrac { 10 }{ 36 } .\dfrac { 16 }{ 25 } +\dfrac { 4.5.9.4 }{ 3.6.10.5 }$$$$\displaystyle =\dfrac { 27 }{ 200 } +\dfrac { 8 }{ 45 } +\dfrac { 2 }{ 5 } =\dfrac { 1283 }{ 1800 }$$$$\therefore P$$(at least one defective) $$\displaystyle =1-\dfrac { 1283 }{ 1800 } =0.2872=0.29$$(approx)Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More