Question

From a shipment of 65 transistors, 6 of which are defective, a sample of 5 transistors is selected at random.

Answer 1

Answer:

a) 8259888

b) 34220

c) 45057474

Step-by-step explanation:

Given,

The total number of transistor = 65,

In which, the defective transistor = 6,

So, the number of non defective transistor = 65 - 6 = 59,

Since, out of these transistor 5 are selected,

a) Thus, the number of ways = the total possible combination of 5 transistors = ${65}C_ 5$

$=\frac{65!}{(65-5)!5!}$

$=8259888$

b) The number of samples that contains exactly 3 defective transistors = the possible combination of exactly 3 defective transistors = ${6}C_3\times {59}C_2$

$=\frac{6!}{(6-3)!3!}\times \frac{59!}{(59-2)!\times 2!}$

$=20\times 1711$

$=34220$

c) The number of sample without any defective transistor = The possible combination of 0 defective transistor = $^6C_0\times ^{59}C_5$

$=1\times 45057474$

$=45057474$