SOLUTION: For Adam the probability a bull's eye with a bow and arrow is .4.
Adam will take 5 shots.
a) What type of probability distribution is this?
b) Complete the chart to show
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Question 450570: For Adam the probability a bull's eye with a bow and arrow is .4.
Adam will take 5 shots.
a) What type of probability distribution is this?
b) Complete the chart to show the probability distribution for the random variable X which represents the number of possible bull's eyes its associated probability.
X = # Bull's Eyes P(X)
c) What is the probability that Adam will get at least 2 bull's eyes?
d) How many bull's eyes do you expect him to make?
____________
A box contains 5 dozen heads of lettuce. There are eight spoiled heads in the box. A person chooses a dozen heads at random.
a) What is the expected number of spoiled heads that will be chosen?
b) What is the probability that he did not choose any of the spoiled heads?
Found 2 solutions by stanbon, edjones:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
For Adam the probability a bull's eye with a bow and arrow is .4.
Adam will take 5 shots.
a) What type of probability distribution is this?::Binomial
------------------------
b) Complete the chart to show the probability distribution for the random variable X which represents the number of possible bull's eyes its associated probability.
X = # Bull's Eyes P(X)
c) What is the probability that Adam will get at least 2 bull's eyes?
Ans: 1-binomcdf(5,0.4,1) = 0.6630
--------
d) How many bull's eyes do you expect him to make?
Ans: E(x) = np = 5*0.4 = 2
=====================================================
A box contains 5 dozen heads of lettuce.
There are eight spoiled heads in the box.
A person chooses a dozen heads at random.
a) What is the expected number of spoiled heads that will be chosen?
---
Comment: That is not a binomial problem as the probability
of selecting a spoiled head changes after each selection.
--------
b) What is the probability that he did not choose any of the spoiled heads?
P(no spoiled in choosing 12)
= P(choose 12 good heads)
= 52C12/60C12
= 0.1475
=================================
Cheers,
Stan H.
==============
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
A box contains 5 dozen heads of lettuce. There are eight spoiled heads in the box. A person chooses a dozen heads at random.
a) What is the expected number of spoiled heads that will be chosen?
b) What is the probability that he did not choose any of the spoiled heads?
.
a) 8/60 * 12 = 8/5 the expected number of spoiled heads that will be chosen.
.
b)
52C12/60C12 = .1475
.
Ed
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