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Anita505
Suppose that we have a white urn containing four white balls and one red ball and have a red urn containing one white ball and four red balls. An experiment consists of selecting at random a ball from the white urn and then (without replacing the first ball) Selecting at random a ball from the urn having the colour of the first ball. Find the probability that the second ball is red. The probability that the second ball is red is ______.
There are two situations to consider: (1) If a white ball is selected on the first draw then the second draw will be from the white urn. Probability of white on the first draw is 4/5 and the probability of red on the second draw is 1/4. P(red second ball) = 4/5 * 1/4 = 4/20 ...............(1) (2) If a red ball is selected on the first draw then the second draw will be from the red urn. Probability of red on the first draw is 1/5 and the probability of red on the second draw is 4/5. P(red second ball) = 1/5 * 4/5 = 4/25 ...............(2) The probability that the second ball is red is the sum of the fractions (1) and (2).
so the sum of (4/20)+(4/25) So in this case the answer would be 0.36 correct?
Correct, or alternatively 9/25.
thank you for showing me the process
An urn contains 3 one-dollar bills, 1 five- dollar bill and 1 ten dollar bill. A player draws bills one at a time without replacement from the urn until a ten dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine: A) The probability of winning $15 B)The probability of winning all bills in the urn C) The probability of the game stopping at the second draw.
can you help me with this and show me the process?
as far as i can see there is only one way to win $15: first draw the five, then draw the ten
since there are 5 bills, and one is a five, the probability of drawing the five first is \(\frac{1}{5}\) then there are 4 bills of which one is a ten, the probability of drawing a ten second given that the first bill was a five is \(\frac{1}{4}\) the probability of both things occurring is \[\frac{1}{5}\times \frac{1}{4}\]
so for part a) it would be 1/20? the answer
okay thank you but i need help with part b and c
for b) it means you pick the ten dollar bill last right?
the probability you pick the ten dollar bill last is the same as the probability you pick the ten dollar bill first, namely \(\frac{1}{5}\)
so b then in this case is 1/5
and for the last one, that means you pick something other then the ten dollar bill, and then you pick the ten dollar bill that is also \(\frac{1}{5}\) via \[\frac{4}{5}\times \frac{1}{4}=\frac{1}{5}\]
more simply put, the probability you pick the ten first, second, third, fourth or fifth are all the same, namely \(\frac{1}{5}\)
Okay thank you for your assistance i have one last question to ask,
do me a favor and post in a new thread this is hard so scroll down to
A grade 11 art class is offering students two choices for a project: a pottery project and a mixed media project. Of the 46 students in the class, 23 have selected to do the pottery project and 33 have selected to do the mixed media project (notice some students have decided to do both) It two students are selected at random from the class to show their finished project(s), what is the probability that at least one pottery project and at least one mixed media project will be shown? Probability (given to three decimal places)