Total number of transistors (n) = 24.
Several defective transistors (r) = 4.
The total number of ways of selecting four transistors is,
[tex]\begin{gathered} ^{24}C_4=\text{ }\frac{24!}{(24-4)!\times4!} \\ ^{10}C_4\text{ = }10626 \end{gathered}[/tex](a) The probability that exactly two are defective is calculated as,
[tex]P(Exactly\text{ 2 are defective) = }\frac{Ways\text{ of selecting defective}\times Ways\text{ of selecting non defective}}{Total\text{ number of ways of selecting }4\text{ transistors}}[/tex]Therefore,
[tex]\begin{gathered} P(Exactly\text{ 2 are defective) = }\frac{^4C_2\times^{20}C_2}{^{24}C_4} \\ P(Exactly\text{ 2 are defective) = }\frac{6\times190}{10626} \\ P(Exactly\text{ 2 are defective) = }\frac{1140}{10626} \\ P(Exactly\text{ 2 are defective) = 0.1073} \end{gathered}[/tex](b) The probability that none of them are defective is calculated as,
[tex]\begin{gathered} P(\text{None of them are defective) = }\frac{No\text{. of ways of selecting }non\text{ defectives}}{\text{Total number of ways}} \\ P(\text{None of them are defective) = }\frac{^{20}C_{^{}4}}{^{24}C_4} \\ P(\text{None of them are defective) = }\frac{4845}{10626} \\ P(\text{None of them are defective) = }0.4560 \end{gathered}[/tex](c) The probability that all of them are defective is calculated as,
[tex]\begin{gathered} P(All\text{ of them are defective) = }\frac{^4C_4}{^{24}C_4} \\ P(All\text{ of them are defective) = }\frac{1}{10626} \\ P(All\text{ of them are defective) = }0.00009 \end{gathered}[/tex](d) The probability that atleast one of them is defective is calculated as,
[tex]\begin{gathered} P(atleast\text{ one is defective) = }1\text{ - P(no defective)} \\ P(atleast\text{ one is defective) = }1-\text{ 0.4560} \\ P(atleast\text{ one is defective) = }0.5440 \end{gathered}[/tex]