A shipment of 10 microwaves contains 3 defective units. In how many ways can a vending company purchase 4 of these units and receive (a) all good units (b) two good units and (c) at least two good units?

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A shipment of 10 microwaves contains 3 defective units. In how many ways can a vending company purchase 4 of these units and receive (a) all good units (b) two good units and (c) at least two good units?

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Answer 1 · 2017-07-31T10:58:27

Out of 10 units of microwave there are 7 good units and 3 defective units.

(a)The number ways of receiving all 4 units perchased by the vending company to be good ones are
$^{7} C_{4} = \frac{7!}{3! \times 4!} = 35$ $""^7C_4=(7!)/(3!xx4!)=35$

(b)The number ways of receiving 2 good units out of 4 units perchased by the vending company are
$^{7} C_{2} \times^{3} C_{2} = \frac{7!}{2! \times 5!} \times \frac{3!}{2! \times 1!} = 63$ $""^7C_2xx""^3C_2=(7!)/(2!xx5!)xx(3!)/(2!xx1!)=63$

(c)The number ways of receiving at least 2 good units out of 4 units perchased by the vending company are
$^{7} C_{2} \times^{3} C_{2} +^{7} C_{3} \times^{3} C_{1} +^{7} C_{4}$ $""^7C_2xx""^3C_2 +""^7C_3xx""^3C_1+ ""^7C_4$

$= \frac{7!}{2! \times 5!} \times \frac{3!}{2! \times 1!} + \frac{7!}{4! \times 3!} \times \frac{3!}{2! \times 1!} + \frac{7!}{3! \times 4!}$ $=(7!)/(2!xx5!)xx(3!)/(2!xx1!)+(7!)/(4!xx3!)xx(3!)/(2!xx1!) + (7!)/(3!xx4!)$

$= 63 + 105 + 35 = 203$ $=63+105+35=203$

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A shipment of 10 microwaves contains 3 defective units. In how many ways can a vending company purchase 4 of these units and receive (a) all good units (b) two good units and (c) at least two good units?

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