Question 1186837: Problem VI - [14 points] A “fair,” “relatively thick” coin, when flipped, can
show heads (H) or tails (T) with equal probabilities, but it is also probable to
come to rest on its edge (E) with a probability of 4.5% (0.045). Showing your
work, answer the following questions:
1. Set up a probability distribution table for the three possible events.
2. If the coin were flipped twice, what would be the probability that it
would show tails the first time and would come to rest on its edge the second
time? Provide a brief explanation to support your computation(s). Show the
probability value in 4 decimal places.
3. If the coin were flipped 14 times, what would be the probability that the
coin would come to rest on its edge (“getting edge”) at most twice? Showing
your work, report the probability value in 4 decimal places. Treat the problem
as a binomial experiment.
4. Determine the probability of getting heads between 4 to 6 times (both
inclusive), i.e., 4 times, 5 times, or 6 times, when the coin is flipped 14
times. Showing your work, report the probability value in 4 decimal places.
Treat the problem as a binomial experiment.
5. What is the probability of getting “heads or tails” at least once when
the coin is flipped 14 times? Showing your work, report the probability value in
4 decimal places. Treat the problem as a binomial experiment.
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website! Problem VI - [14 points] A “fair,” “relatively thick” coin, when flipped, can
show heads (H) or tails (T) with equal probabilities, but it is also probable to
come to rest on its edge (E) with a probability of 4.5% (0.045). Showing your
work, answer the following questions:
1. Set up a probability distribution table for the three possible events.
P(H)+P(T)+P(E) = 1
Let P(H)=P(T)=x and we are given P(E)=0.045, so
x + x + 0.045 = 1
2x + 0.045 = 1
2x = 1-0.045
2x = 0.955
x = 0.4775
P(H) = P(T) = 0.4775
X P(X)
----------
H 0.4775
T 0.4775
E 0.045
----------
1.0000
2. If the coin were flipped twice, what would be the probability that it would
show tails the first time and would come to rest on its edge the second time?
Provide a brief explanation to support your computation(s). Show the probability
value in 4 decimal places.
Since the results of the flips are independent events, we can multiply their
probabilities to find the probability that both occur.
P(H&E) = P(H)P(E) = (0.4775)(0.045) = 0.0214875, round to 0.0215
3. If the coin were flipped 14 times, what would be the probability that the
coin would come to rest on its edge (“getting edge”) at most twice? Showing your
work, report the probability value in 4 decimal places. Treat the problem as a
binomial experiment.
On your TI-84, press ON, CLEAR 2ND VARS, use down arrow key to scroll down to
B:binomcdf (the "c" is for "cumulative", there is also an A:binompdf (the "p"
is for particular"). You are accumulating for 0,1, or 2 E's.
Press ENTER
Make the next screen (if your calculator is a newer model) read:
binomcdf
trials:14
p:0.045
x value:2
Paste
Use down arrow to scroll down to Paste
Press ENTER
read this (or if you calculator is old, you will have to type this in yourself:
binomcdf(14,0.045,2)
read the answer 0.9771592698
4. Determine the probability of getting heads between 4 to 6 times (both
inclusive), i.e., 4 times, 5 times, or 6 times, when the coin is flipped 14
times. Showing your work, report the probability value in 4 decimal places.
Treat the problem as a binomial experiment.
I'll leave this to you. But you have to use A:binompdf (not B:binomcdf), because
you are not accumulating from 0 up to x, but want particular values of x, once
with x=4, once with x=5 and once with x=6. Then add the three values you get.
5. What is the probability of getting “heads or tails” at least once when the
coin is flipped 14 times? Showing your work, report the probability value in 4
decimal places. Treat the problem as a binomial experiment
This is the same as getting 'Edge' at most 1 time. It's done the same way as
problem 3 except x=1 instead of x=2.
I'll also leave this one up to you. Good luck.
Edwin
|
|
|
