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Olin [163]
3 years ago
6

Suppose that 2% of cereal boxes contain a prize and the other 98% contain the message, "sorry, try again." consider the random v

ariable x, where x = number of boxes purchased until a prize is found. (round all answers to four decimal places.) (a) what is the probability that at most three boxes must be purchased? p(at most three boxes) = (b) what is the probability that exactly three boxes must be purchased? p(exactly three boxes) = (c) what is the probability that more than three boxes must be purchased? p(more than three boxes) =
Mathematics
1 answer:
user100 [1]3 years ago
5 0
<span>a) This is the sum of the probabilities of winning in 1, 2 or 3 boxes. 0.02 + (0.98*0.02) + (0.98*0.98*0.02) = 0.0588 b) Exactly three boxes requires the first two to lose. (0.98*0.98*0.02) = 0.0192 c) This is the complement of answer for a). 1-.0588 = .9412</span>
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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude
EleoNora [17]

Answer:

a) P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

b) P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

c) P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

d) E(X) = 20*0.2= 4

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=20, p=0.2)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Part a

We want this probability:

P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

Part b

We want this probability:

P(X=4)

And using the probability mass function we got:

P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

Part c

We want this probability:

P(X>3)

We can use the complement rule and we got:

P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

Part d

The expected value is given by:

E(X) = np

And replacing we got:

E(X) = 20*0.2= 4

3 0
3 years ago
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