SOLUTION: A box contains four quarters. Exactly two of the quarters have the American eagle on the back. Suppose you draw two quarters with replacement at random from the box.
A. What
Algebra ->
Probability-and-statistics
-> SOLUTION: A box contains four quarters. Exactly two of the quarters have the American eagle on the back. Suppose you draw two quarters with replacement at random from the box.
A. What
Log On
Question 617919: A box contains four quarters. Exactly two of the quarters have the American eagle on the back. Suppose you draw two quarters with replacement at random from the box.
A. What is the probability that both have the American Eagle on the back?
B. What is the probability that neither has the American Eagle on the back?
C. What is the probability that at least one has the American eagle on the back?
(I think you have to use the equation but I am not sure how to use it) Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! there are 4 quarters in the box.
2 of them have the american eagle on the back.
you draw 2 quarters with replacement at random from the box.
you replace the quarter in the box after each draw and prior to the next draw (with replacement is assumed).
the probability of getting a quarter with an american eagle on the back is equal to 1/2 because there are a total of 2 quarters with an american eagle on the back out of a total of 4. 2/4 = 1/2.
the probability of getting a quarter without an american eagle on the back is equal to 1/2 because there are a total of 2 quarters without an american eagle on the back out of a total of 4. 2/4 = 1/2.
-----
A. What is the probability that both have the American Eagle on the back?
the probability of getting a quarter with an eagle on the back twice in a row is equal to 1/2 * 1/2 which is equal to 1/4.
-----
B. What is the probability that neither has the American Eagle on the back?
the probability of getting a quarter without an eagle on the back twice in a row is equal to 1/2 * 1/2 which is equal to 1/4.
-----
C. What is the probability that at least one has the American eagle on the back?
the probability of getting at least 1 quarter with an american eagle on the back is equal to 1 minus the probability of getting 0 quarters with an american eagle on the back. that probability is equal to 1 minus 1/4 which is equal to 3/4.
-----
Y equals quarters with an american eagle on the back.
N equals quarters without an american eagle on the back.
number of possible ways you can draw 2 quarters from the 4 is:
YY = 1/4
NN = 1/4
YN = 1/4
NY = 1/4
total of all probabilities is equal to 1 as it should be.
probability of getting at least one eagle would be the probability of getting YY or YN or NY which is equal to 3/4.
that's equal to the probability of not getting none which is equal to 1 - 1/4 = 3/4
-----
the equation of nCr * p^r * q^(n-r) can be used for this type of problem but you have to be careful how to use it.
n would have to be equal to 2 (the number of draws)
p would have to be equal to .5 (probability of getting a quarter with an eagle on the back).
q would have to be equal to .5 (probability of not getting a quarter with an eagle on the back).
r would be the probability of getting 0, 1, or 2 quarters with an eagle on the back.
the formula distribution would be shown as follows:
p(0) = 2c0 * (.5)^0 * (.5)^2 = 1 * 1 * .25 = 1 * .25 = .25
p(1) = 2c1 * (.5)^1 * (.5)^1 = 2 * .5 * .5 = 1 * .5 = .5
p(2) = 2c2 * (.5)^2 * (.5)^0 = 1 * .25 * 1 = .25 * 1 = .25
add up all the probabilities and you get 1 as you should.
probability of getting 0 eagles is equal to .25 (YY)
probability of getting 2 eagles is equal to .25 (NY)
probability of getting 1 eagle is equal to .5 (YN) or (NY)
you do get the same answer but, as stated before, you have to be careful how you use the formula.
in this case, you're not picking 2 out of 4.
you're picking 2 out of 2.
the probabilities, however, are determined from the 4.
you can also have only 2 possible outcomes.
you have either a success (eagle on the back), or you have a failure (eagle not on the back).
the probabiiity of a success and the probability of failure have to add up to 1.
