Thread: Probability

*this is not a homework question, my exams are coming up on the 29th, and i found this on my revision papers.

-Box A contains 12 balls, 6 of which are blue, 2 are red, and the remaining are yellow. Box B contains 10 balls, 4 of which are blue and the remaining are red. one ball is removed at random from each box.
find the probability that
i)one red ball and one blue ball are removed
ii)the blue ball from Box B is blue
iii)balls of the same colors are removed
iv)balls of different colors are removed

appreciate any explanations that is clear to understand, *this topic is puzzling me a bit, i would really appreciate your help

There are two things that are easy to do in probability. The first is finding the probability that two independent events will happen together, such as drawing a red ball from the first box and a blue ball from the second (just multiply the probabilities). The second is finding the probability that one of two mutually exclusive events will happen, such as drawing a red ball from the first box or a blue ball from the first box (just add the probabilities). The goal then is to take your givens, turn them into probabilities, then use these two things to find the probability of more complicated questions.

How far have you gotten on this? I could help more if I knew where you were so far.

thank you for the response, i so far studied the two fundamental parts of probabilty
the mutually exclusive events: P(A or B)=P(A)+P(B)
and the independent events:P(A and B)=P(A) multiplied by P(B)
but i still dont get how to apply this on the my question, can you please elaborate the question and answers please?
thank you

Some comments to help you.

Always write down what you know immediately from the given information.
This makes answering more complicated problems easier.
Here we know that for Box A; P[blue] = 6/12 P[red] = 2/12 P[yellow] = 4/12
and for Box B P[blue] = 4/10 P[red] = 6/10
Now interpret the questions as a function of these basic facts.

Also note that P[iii] + P[iv] = 1

yes i have done up to that part, but do i have to draw a tree diagram for this?

Originally Posted by Heinsbergrelatz

yes i have done up to that part, but do i have to draw a tree diagram for this?

You make it sound as if it is a chore rather than an aide to the solution.
Answer is yes if you would like to.
But you do not have to.

The difficult part is interpreting what the statements mean and writing the Prob expression.
Take statement number 2.
ii)the blue ball from Box B is blue
It says you have a blue ball from Box B.
I assume it means a blue ball has been picked and it came from Box B
So it is mearly P[blue ball picked from Box B] = 4/10 = 2/5.

But suppose it said ..... "the blue ball in Box B is blue".
Then P[ ] = 1

Or suppose it said .... "one blue ball is picked, what is the P[it came from Box B]
That would mean you not only picked a blue ball from Box B but you also
did not get a blue ball from Box A. So P[ ] = P[pick blue from B] P[pick a non blue ball from A]
Note that ...... P[pick a non blue ball from A] = 1 - P[pick a blue ball from A]

Tree diagrams are useful for sequential probability problems.

am i asking too much, but can you try and draw a tree diagram for this question, i want to check if its same with mine.
thank you

Without resorting to diagrams, you need to break each question down into a series of independent and/or mutually exclusive parts. Since one ball is drawn from each box, each pair of (color of ball from A, color of ball from B) is one of a set of mutually exclusive events. (You can't draw red from A and blue from A.) You can use the formula for independent events to find the probability for each of these. Then, you can see which of these satisfy each question and use mutually exclusive events to find the probability for that question.

Take question 1. Which ordered pairs of colors satisfies that question? What's the probability for each of those pairs? What's the probability for all of those pairs?

(Also, the probabilities for the second box should be 4/10 and 6/10.)

Originally Posted by MagiMaster

(Also, the probabilities for the second box should be 4/10 and 6/10.)

Thanks for catching this.
Yes, careless errors from assuming the 2nd box also had 12 balls.
The correct numbers have been edited in for the previous posts.

thank you for the responses
i now get the idea of this probability chapter

Hello,

Remember these formulas and wordings when you solving, When the word And is used we use P(A intersection B) = P(A) x P(B)x P(A/B) and When Or is used use P(A U B)= P(A) + P(B) - P(A intersection B).

i see to;"vantheo"
thankyou for your concern and your time, yes that rule had been confusing me, now i seem to understand the concept