Explanation:
1.) A certain brand of cereal has a toy in every box. There are 10 different toys and the toys are distributed among the boxes so that any box purchased is equally likely to contain any one of the 10 toys.
1 a.) A boy has collected 7 different toys. Find the probability that he will get the toys he needs to complete the set if he opens only three boxes.
P(X=3)=(310)(210)(110)=.006
The probability of success changes when a new toy is obtained, which is why the chance of getting a new toy in the next attempt goes down by one-tenth.
1 b.) Another boy only needs one toy to complete the set. Find the probability that he gets the remaining toy by opening no more than 5 boxes.
P(X=1)=1−(910)5=0.40951
1 - dbinom(0, 5, .1)
## [1] 0.40951
The probability for getting the new toy is easily found by subtracting the probability of not getting the toy after five attempts from 1 as seen above. R was used to check the answer.
2.) Suppose that S is the sample space associated to a random process and that E,F are two subsets of S. Prove that if E is a subset of F then P(E)≤P(F) using only the 3 axioms of probability.
Proof: If E⊆F then,
F=E+(F−E)F=E∪(F∩Ec)
Axiom 3 can then be applied due to E and (F∩Ec) being mutually exclusive,
P(F)=P(E)+P(F∩Ec)
Axiom 1 can now be applied to P(F∩Ec) to state:
P(F∩Ec)≥0
Since P(F∩Ec) cannot be less than 0, P(F) must be greater than or equal to P(E)
3.) Suppose that 30% of all drivers stop at an intersection having flashing red lights when no other cars are visible. You decide to secretly monitor this intersection until two cars stop at the intersection when no other cars are visible. Let X denote the number of cars who fail to stop at the intersection while you are watching.
a.) Compute P(X≤3) and P(X≥2) by hand.
P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)
(choose(1,0) * .3^2) + (choose(2,1) * .7 * .3^2) + (choose(3,2) * .7^2 * .3^2) + (choose(4,3) * .7^3 * .3^2)
## [1] 0.47178
P(X≥2)=1−P(X≤1)
1 - ((choose(1,0) * .3^2) + (choose(2,1) * .7 * .3^2))
## [1] 0.784
3 b.) What type of random variable is X: binomial, geometric, negative binominal, or Poisson? Negative Binominal
3 c.) Now compute P(X???3) and P(X???2) using appropriate R commands for the distribution you chose in part b).
#Probability of X less than equal to 3
sum(dnbinom(0:3, 2, .3))
## [1] 0.47178
#Probability of X greater than equal to 2
1 - sum(dnbinom(0:1, 2, .3))
## [1] 0.784
4.) A child’s game includes a spinner with four colors on it. Each color is one quarter of the circle. You spin the spinner 5 times and record the number X of blues.
4 a.)Create a table (by hand) displaying the probability mass function of the random variable X if the spinner is fair.
X f(x)
0 (.75)^5
1 (51)(.75)4(.25)
2 (52)(.75)3(.25)2
3 (53)(.75)2(.25)3
4 (54)(.75)(.25)4
5 (.25)5
4 b.) You spin the spinner 5 times and get 4 blues. Perform a hypothesis test (at the Alpha = 0.05 level of significance) to determine if there is enough evidence to reject the game manufacturer’s claim that the spinner is fair, carefully showing each of the four steps. You may use R for computations, but I do not want you to simply use the binom.test() command. (Hint: Your answer to part a) should be helpful when computing your p-value.)
Step 1.) H_o : The spinner is fair; π=14
Step 2.) H_a : The spinner is not fair; π≠14
choose(5,4)*(.75)*(.25)^4 + (.25)^5
## [1] 0.015625
Step 3.) P-test: p-value = 0.015625; Which is less than 0.05.
Step 4.) Since the p-value is less than the alpha value of 0.05, we must reject the null hypothesis meaning there is statistical evidence of the spinner being unfair.
4 c.) What would a Type II error be in the context of the hypothesis test you just performed?
A Type II error would be when we fail to reject the null hypothesis of the spinner being fair, when it is really unfair.
5.) Find the expected value and variance of this discrete random variable when n=5. Justify your answer.
E(X)=(15)+(25)+(35)+(45)+1=(155=3
The expected value is 3, which tells us the weighted average of our random variable X.
Var(X)=(−2)2(15)+(−1)2(15)+0(15)+(1)2(15)+(2)2(15)=(45)+(15)+(15)+(45)=2
The variance is 2, which is the measure of how spread out the random variable X is from its expected value.
Bonus.) Exercise 2.9 in the textbook.
a.) n = 10; and P(X≤1) represents the probability of no one having the same birthday out of the ten people.
P(X≥2)=1−P(X≤1)=1−((365365)∗(364365)∗(363365)∗(362365)∗(361365)∗(360365)∗(359365)∗(3583655)∗(357365)∗(356365))=0.11695
b.) Kept increasing n until P(X≥2)=0.5 and obtained that the smallest group of random people would have to be 23
Collaborated with Maria and Akeem