In the given question by adding the probability for 1 & 2 defective computers to get a total of [tex]23.81+2.38 =26.19\%[/tex] chance of at least two defective laptops.
For point a:
When 5 out of 10 are faulty.
Let X represent the number of defectives among some of the 6 people chosen at random. For parameters [tex](N=10, k=5 ,n=6)[/tex], X follows the Hypergeometric Distribution.
We want to find the probability that [tex]X = 2[/tex].
[tex]\text{P(2 are d\1efective)} = \frac{ \text{(of ways of selecting 4 of 5 non-d\1efective and 2 of 5 de fective)}}{\text{( total of ways of selecting 6 from 10)}}[/tex]
[tex]\to \frac{\binom{5}{4}\binom{5}{2}}{\binom{10}{6}}= \frac{5 \times 10}{210} =\frac{5}{21}= 0.238095 \approx 0.2381[/tex]
Calculating the probability percentage:
[tex]\to 0.2381\times 100\\\\\to 23.81\%[/tex]
For point b:
In this question there we use the "at least" which implies that is greater than 2, we required the probability for 1 defective + 2 defective, for 1 defective that is solved as:
[tex]\to ^5 C_1\times \frac{^5C_5}{^{10} C_6}\\\\\to \frac{5!}{1! \times 5-1!} \times \frac{ \frac{5!}{5! \times 5-5!} }{\frac{10!}{6! \times 10-6!}}\\ \\ \to \frac{5!}{1! \times 4!} \times \frac{ \frac{5!}{5! \times 0!} }{\frac{10!}{6! \times 4!}}\\ \\[/tex]
[tex]\to \frac{5\times 4!}{1 \times 4!} \times \frac{ \frac{5!}{5! \times 1} }{\frac{10\times 9 \times 8 \times 7 \times 6! }{6! \times 4 \times 3 \times 2 \times 1}}\\\\\to 5 \times \frac{ 1 }{10\times 3 \times 7 }\\\\\to \frac{ 1 }{2\times 3 \times 7 }\\\\\to \frac{ 1 }{ 42}\\\\ \to 0.023[/tex]
Calculating the probability:
[tex]\to 0.023 \times 100\\\\\to 2.38\%[/tex]
Add the probability for 1 & 2 defective computers to get a total of [tex]23.81+2.38 =26.19\%[/tex] chance of at least two defective laptops.
Find out more information about the probability here:
brainly.com/question/15409496
