Question

Defective units a shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because each is identically packaged, the selection will be random. What are the probabilities that (a) all four units are good, (b) exactly two units are good, and (c) at least two units are good?

Answer 1

(a) $\dfrac{14}{55}$

(b) $\dfrac{12}{55}$

(c) $\dfrac{54}{55}$

Step-by-step explanation:

Given,

Total units = 12,

Defective units = 9,

So, non defective units = 12 - 9 = 3,

If 4 units are chosen randomly,

Then the total number of ways = $^{12}C_4$

(a) Ways of selecting 4 good units = $^9C_4$

Thus, the probability of selecting 4 good units = $\frac{\text{Ways of selecting 4 good units}}{\text{Total ways of selecting 4 units}}$

$=\frac{^9C_4}{^{12}C_4}$

$=\frac{\frac{9!}{4!5!}}{\frac{12!}{4!8!}}}$

$=\frac{126}{495}$

$=\frac{14}{55}$

(b) Ways of selecting exactly 2 good units = $^9C_2\times ^3C_2$

Thus, the probability of selecting exactly two good units

$=\frac{^9C_2\times ^3C_2}{^{12}C_4}$

$=\frac{\frac{9!}{2!7!}\times \frac{3!}{2!1!}}{495}$

$=\frac{108}{495}$

$=\frac{12}{55}$

(c) Ways of selecting at least 2 good units = Ways of selecting 2 good units + Ways of selecting 3 good units + Ways of selecting 4 good units

$=^9C_2\times ^3C_2+9C_3\times ^3C_1+9C_4$

$=108 + \frac{9!}{3!6!}\times 3 + 126$

$=108 + 84\times 3 + 126$

$=486$

Thus, probability of selecting at least 2 good units = $\frac{486}{495}$

$=\frac{54}{55}$

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