Math Is Fun Forum

wiz

Member

Registered: 2008-05-26

Posts: 3

help me to solve statistics

hi everyone
i am a new user. can anyone there help me with the right solution
Given that 6 computers of type A,12 computers of type B are in a office, if 25%of the computers are connected to color printers and 33 1/3% of the computers that are connected to color printers are type B then how many type A computers are connected to color printers?

thanks

wiz

Member

Registered: 2008-05-26

Posts: 3

Re: help me to solve statistics

Real challenge to solve this!
Urn A contains 6 red and 4 black balls and Urn B contains 4 red and 6 black balls.
                      One ball is picked at random from Urn A and placed in Urn B. Then one ball is picked from Urn B et random and placed in Urn A.
                      If one ball is now picked at random from A now,What is the probability of its being red?

Can anyone help me with the right solution

wiz

Member

Registered: 2008-05-26

Posts: 3

Re: help me to solve statistics

Real challenge to solve this!
Urn A contains 6 red and 4 black balls and Urn B contains 4 red and 6 black balls.
                      One ball is picked at random from Urn A and placed in Urn B. Then one ball is picked from Urn B et random and placed in Urn A.
                      If one ball is now picked at random from A now,What is the probability of its being red?

Can anyone help me with the right solution

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: help me to solve statistics

Question 1:
You don't really need to know how many of each type of computer there are, the important part is that there are 18 in total.

25% (or 1/4) are connected to colour printers, and 1/3 of those are type B (which means 2/3 of them are type A).

Therefore, 1/4 * 2/3 = 1/6 of the computers are type A and connected to a colour printer.
18 * 1/6 = 3, so there's your answer.

Question 2 is a bit time-consuming, but not too complicated once you understand the process.
There are 8 different things that can happen, 4 of which are relevant to you.

The 4 'interesting' outcomes are RRR, RBR, BRR, BBR. (Each letter denotes what colour of ball is picked at each stage)

I'll work out the probability of the first one.
Urn A initially has 6 red and 4 black balls, so there's a 6/10 = 3/5 chance of getting a red ball from it.

After this red ball goes into B, there are 5 red and 6 black balls in B. There is therefore a 5/11 chance of getting a red ball from there.

A red ball got taken out of A and another red ball got put in, so the probability of getting a red from A is now the same as it was before - 3/5.

Multiply all these together to get the probability of RRR happening: 3/5 * 5/11 * 3/5 = 9/55.

You can work out the chances for RBR, BRR and BBR in a similar way. Once you have all those, add them all together to get your final answer.

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Why did the vector cross the road?
It wanted to be normal.