Assume a jar has 7 red marbles and 5 black marbles. Draw out 2 marbles with and without replacement. Find the requested probabilities. (a) P(2 red marbles) With replacement , without replacement . (b) P(2 black marbles) With replacement , without replacement . (c) P(one red and one black marble) With replacement , without replacement . (d) P(red on the first draw and black on the second draw) With replacement , without replacement .
a. With replacement. for the first draw how many red marbles are there in the jar?
9
The jar has 7 red marbles and 5 black marbles. Explain how you got 9?
I have to find out with replacement and without replacement.....sorry i meant 12 i looked at it wrong
When you reach in the first time. you are looking for a red marble. How many red marbles are in the jar on the first pick?
7?
Yes 7 red and 12 total marbles... so you use 7/12 for the first draw. Now. we are replacing it (putting it back in) For the second pick. How many red marbles are there in the jar and how many total marbles are in the jar?
so for the 2nd it would be 6 red marbles and 11 total marbles...correct?
Not when we replace it... we put the red marble we just picked back in so there would still be 7 red and 12 marbles when we pick our second marble. Do you understand this?
ok. Somewhat....
So first draw is 7/12 second draw is 7/12 P(2 reds) WITH REPLACEMENT is 7/12 * 7/12 = 49/124 Pretend that you put your hand in the jar and pull out a red marble, then you put it back in and reach your hand in to draw another one out.. there are 7 in there for the second pick because you put it back in.
ok I see what you are saying now.
wait....is it 49/124 or 144?
Now without replacement. First draw 7/12 then you put that red marble on a shelf and reach back in trying to pull out another red marble. This time there are only 6 reds in the jar because one of them is sitting on the shelf. There are also only 11 marbles left because of the one sitting on the shelf. So without replacement we have First draw = 7/12 Second draw = 6/11 P(2 reds) without replacement = 7/12 * 6/11 = 42/132 which reduces down to 7/22
OOPS 144
ok so the next one would be with rep....25/144 and with out 20/132?
Yes except the 20/132 should reduce down to 5/33 Very good.
33 or 22?
I divided both the top and bottom by 4 20 / 4 = 5 132 / 4 = 33
how would i got about the next ones?
Let's do d first. I have to think about c. With replacement First draw red 7/12 Second draw black 5/12 7/12 * 5/12 = 35/144
Without replacement First draw red 7/12 Second draw black 5/11 7/12 * 5/11 = 35/132
it wouldnt be 7/12 * 5/12? if its with replacement?
oh ya sorry i didnt see that
how about c?
Just a minute.. Looking that one up to make sure I am correct.
ok
Not sure I am correct. Why don't you post just c again. Someone will answer it. I don't want to lead you wrong.
ok i will do that. can u try to help me with another question real quick?
Sure
Suppose that people's heights (in centimeters) are normally distributed, with a mean of 175 and a standard deviation of 6. We find the heights of 60 people. (a) How many would you expect to be between 167 and 183 cm tall? (b) How many would you expect to be taller than 170 cm?
I think on a) x = 183 - 175 8 -------- = - = 1 8 8 x = 167 - 175 -8 -------- = -- = -1 8 8 34% of the population are within 1 standard deviation from the mean. So 34% below 175 and 34 % above 175 which would be 68% of the people. 68% of 60 = 40.8 which I would say about 41 people
b.) x = 170 - 175 -------- 8 x = -5/8 x = - .625 about 22% would be between 170 and 175 and then 50% of the people above 175 so ~72% of the 60 people about 43 people
I used a standard deviation table... it shows you when you have the x, it tells you what percent that would represent.
oh ok i see i was a lil confused on the percents but that makes sense
are u still available?
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