Question

For each of the following, define the random variable(s) needed to answer the questions,tell what kind of distribution each has, and what probabilites need to be calculated to answer the question.



2) Patty has basketball practice everyday of the week at 4:00pm (M-F). Every morning, Patty has a 20% chance of going to class.





a)What is the probability that in a given week (5 days), Patty goes to class at least once.





b) What is the probability that the Exam day (22nd day) is the first day Patty goes to his class?

X=# of days he goes to class on first day

X~Geom(22)

P(X=22)

P(x)= (1-p)^(x-1)p

= (1-0.20)^(22-1)(0.20)

= 0.0018 -answer


c) What is the probability that the Final Exam day(84th day) is the third day Patty goes to his class?


d) If Patty goes to class at least two times in a given week, Patty will get himself a doughnut. What is the probability that he doesn't get a doughnut until the fourth week.

Answer 1

a) $X=$ the number of days when Patty goes to class: $X\sim B(n, p)$

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

Given that $n=5, p=0.2$

Find the probability that in a given week (5 days), Patty goes to class at least once. 

$$P(X\geq1)=1-P(X=0)=$$

$$=1-\binom{5}{0}0.2^0(1-0.2)^{5-0}=0.67232$$

b) $X=$the number of days until Patty goes to class for the first time: $X\sim Geom(p)$

$$P(X=x)=(1-p)^{x-1}p, x\geq1$$

Given that $p=0.2$

$$P(X=22)=(1-0.2)^{22-1}(0.2)\approx0.0018$$

c) $X=$ number of days until $r^{th}$ day Patty goes to his class: $X\sim NegBin(r, p)$  

Given that $r=3, p=0.2$

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

$$P(X=84)=\binom{84-1}{3-1}(1-0.2)^{84-3}0.2^3\approx0.0000003848$$

d) $X=$ days when Patty goes to class in a given week (5 days):: $X\sim B(n, p)$

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

Given that $n=5, p=0.2$

Find the probability that in a given week (5 days), Patty goes to class less than two times

$$P(X<2)=P(X=0)+P(X=1)=$$

$$=\binom{5}{0}0.2^0(1-0.2)^{5-0}+\binom{5}{1}0.2^1(1-0.2)^{5-1}=$$

$$=0.32768+0.4096=0.73728$$

Find the probability that he doesnot get a doughnut until the fourth week.

$$P(\text{no doughnut})=0.73728^3\approx0.400772$$