Answer to Question #119235 in Statistics and Probability for Prob

Question

8. A package of 6 fuses are tested where the probability an individual fuse is defective is 0.05. (That is, 5% of all fuses manufactured are defective). What is the probability
that more than one fuse will be defective, given that at least one is defective?

9. An archer shoots arrows at a circular target where the central portion of the target inside is called the bull. The archer hits the bull with probability 1/32. Assume that the archer shoots 96 arrows at the target, and that all shoots are independent. What
is an approximated probability that an archer hit not more than one bull?

10. Emmanuel is given a multiple-choice exam with ten questions and each question with five possible answers. He decided to guess randomly for each question. What is the probability that he will get at least six questions correct?

Answer 1

8. Let $X=$  the number of defective fuses:$X\sim Bin(n,p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given $n=6,p=0.05$. What is the probability that more than one fuse will be defective, given that at least one is defective? 

$$P(X>1,X\geq1)=P(X>1)=$$

$$=1-P(X=0)-P(X=1)=$$

$$=1-\binom{6}{0}(0.05)^0(1-0.05)^{6-0}-\binom{6}{1}(0.05)^1(1-0.05)^{61}\approx$$

$$\approx1-0.735092-0.232134\approx0.032774$$

$$P(X\geq1)=1-P(X=0)\approx0.264908$$

$$P(X>1|X\geq1)={P(X>1,X\geq1)\over P(X\geq1}\approx$$

$$\approx{0.032774\over 0.264908}\approx0.1237$$

9. Let $X=$ the number of bulls hitted by an archer: $X\sim Bin (n, p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given $n=96>20, p=1/32$

Using the Poisson approximation with $\lambda=np=96(\dfrac{a}{b})=3<5$

$$P(X\leq1)=P(X=0)+P(X=1)=$$

$$={e^{-\lambda}\cdot\lambda^0 \over 0!}+{e^{-\lambda}\cdot\lambda^1 \over 1!}=$$

$$=e^{-3}(1+3)=4e^{-3}\approx0.1991$$

10. Let $X=$ the number of correctly solved questions: $X\sim Bin(n,p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given $n=10,p=0.2$

$$P(X\geq 6)=P(X=6)+P(X=7)+P(X=8)+$$

$$+P(X=9)+P(X=10)=\binom{10}{6}(0.2)^6(1-0.2)^{10-6}+$$

$$+\binom{10}{7}(0.2)^7(1-0.2)^{10-7}+\binom{10}{8}(0.2)^8(1-0.2)^{10-8}+$$

$$+\binom{10}{9}(0.2)^9(1-0.2)^{10-9}+\binom{10}{10}(0.2)^{10}(1-0.2)^{10-10}=$$

$$=210(0.2)^6(0.8)^4+120(0.2)^7(0.8)^3+45(0.2)^8(0.8)^2+$$

$$+10(0.2)^9(0.8)+(0.2)^{10}=$$

$$=0.005505024+0.000786432+0.000073728+$$

$$+0.000004096+0.0000001024\approx0.0064$$