Answer to Question #86834 in Statistics and Probability for Anand

Question

If 20% of the memory chips made in a certain plant are defective, find the probability
that in a lot of 100 randomly chosen chips for inspection
i) at most 15 chips will be defective,
ii) the number of defectives will be between 15 and 25.

Answer 1

The binomial distribution b(x; n, p)

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$ $$n=100, p=0.2$$

i) at most 15 chips will be defective

$$P(X\le15)=\sum_{\mathclap{0\le i\le 15}} P(X=i)$$

$$P(X\le15)={100! \over {0!(100-0)!}}0.2^0(1-0.2)^{100-0}+{100! \over {1!(100-1)!}}0.2^1(1-0.2)^{100-1}+$$

$$+{100! \over {2!(100-2)!}}0.2^2(1-0.2)^{100-2}+{100! \over {3!(100-3)!}}0.2^3(1-0.2)^{100-3}+$$

$$+{100! \over {4!(100-4)!}}0.2^4(1-0.2)^{100-4}+{100! \over {5!(100-5)!}}0.2^5(1-0.2)^{100-5}+$$

$$+{100! \over {6!(100-6)!}}0.2^6(1-0.2)^{100-6}+{100! \over {7!(100-7)!}}0.2^7(1-0.2)^{100-7}+$$

$$+{100! \over {8!(100-8)!}}0.2^8(1-0.2)^{100-8}+{100! \over {9!(100-9)!}}0.2^9(1-0.2)^{100-9}+$$

$$={100! \over {15!(100-15)!}}0.2^{15}(1-0.2)^{100-15}+{100! \over {16!(100-16)!}}0.2^{16}(1-0.2)^{100-16}+$$

$$+{100! \over {12!(100-12)!}}0.2^{12}(1-0.2)^{100-12}+{100! \over {13!(100-13)!}}0.2^{13}(1-0.2)^{100-13}+$$

$$+{100! \over {14!(100-14)!}}0.2^{14}(1-0.2)^{100-14}+{100! \over {15!(100-15)!}}0.2^{15}(1-0.2)^{100-15}\approx$$

$$\approx0.1285$$

ii) the number of defectives will be between 15 and 25.

$$P(15 \le X \le25)=\sum_{\mathclap{15\le i\le 25}} P(X=i)$$

$$P(15 \le X \le25)=$$

$$={100! \over {15!(100-15)!}}0.2^{15}(1-0.2)^{100-15}+{100! \over {16!(100-16)!}}0.2^{16}(1-0.2)^{100-16}+$$

$$+{100! \over {17!(100-17)!}}0.2^{17}(1-0.2)^{100-17}+{100! \over {18!(100-18)!}}0.2^{18}(1-0.2)^{100-18}+$$

$$+{100! \over {19!(100-19)!}}0.2^{19}(1-0.2)^{100-19}+{100! \over {20!(100-20)!}}0.2^{20}(1-0.2)^{100-20}+$$

$$+{100! \over {21!(100-21)!}}0.2^{21}(1-0.2)^{100-21}+{100! \over {22!(100-22)!}}0.2^{22}(1-0.2)^{100-22}+$$

$$+{100! \over {23!(100-23)!}}0.2^{23}(1-0.2)^{100-23}+{100! \over {24!(100-24)!}}0.2^{24}(1-0.2)^{100-24}+$$

$$+{100! \over {25!(100-25)!}}0.2^{25}(1-0.2)^{100-25}\approx0.8321$$