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 = [tex]{65}C_ 5[/tex]
[tex]=\frac{65!}{(65-5)!5!}[/tex]
[tex]=8259888[/tex]
b) The number of samples that contains exactly 3 defective transistors = the possible combination of exactly 3 defective transistors = [tex]{6}C_3\times {59}C_2[/tex]
[tex]=\frac{6!}{(6-3)!3!}\times \frac{59!}{(59-2)!\times 2!}[/tex]
[tex]=20\times 1711[/tex]
[tex]=34220[/tex]
c) The number of sample without any defective transistor = The possible combination of 0 defective transistor = [tex]^6C_0\times ^{59}C_5[/tex]
[tex]=1\times 45057474[/tex]
[tex]=45057474[/tex]
