Post: #1

A Bag of Marbles

The Miracle Marble manufacturing Company manufactures orange marbles and purple marbles.

A Bag of their marbles may contain any combination of orange and purple marbles (including all orange or all purple) and all combinations are equally probable.

One of my friend bought a bag of their marbles and pulled one out at random. It was Purple.

What is the probability that if he pulled out a second marble at a random?????

Please try to provide the solution for this !!!!!

Thanks in Advance.

Cheers,

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Post: #2
chandu garu this problem will solve on the probability factor the formula for probability is n!/(n-r)!

Let me explain in detail

A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?

Outcomes: The possible outcomes of this experiment are red, green, blue and yellow.

Probabilities:
P(red) = # of ways to choose red = 6 = 3
----------------------------- ------ ----
total # of marbles 22 11

P(green) = # of ways to choose green = 5
------------------------------ ----
total # of marbles 22

P(blue) = # of ways to choose blue = 8 = 4
------------------------- ----- -----
total # of marbles 22 11

P(yellow) = # of ways to choose yellow = 3
--------------------------- -----
total # of marbles 22

I think u r able to understand this

Thank you very much for your question
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Post: #3

మిత్రమా మనోహర్ గారు,

నన్ను క్షమించ గలరు.

నాకు అర్థము అయ్యిందో కాలేదో కూడా అర్థం కాని పరిస్థితిలో ఉన్నానండీ!!! (అన్యధా భావించవద్దని నా విన్నపము)

జవాబు ఇవ్వగలరని నా ఆశిస్తున్నాను.

ధన్యవాదములు

మీ...

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Post: #4

Solution:

There is a probability of 2/3 that the second marble would also be purple. This result is independent of the number of marbles in the bag.

This proof is as follows:

Let n = number of marbles in bag.

Each  of the following n+1 combinations has a probability of  1/(n+1)

No. of Bags                                        No. of Purple Marbles

0                                                                       n

1                                                                      n-1

2                                                                      n-2            

....                                                                    ........

(n-k)                                                                k

....                                                                   ..........

n                                                                       0

There are a total number of n(n+1)/2 purple marbles in all of the above combinations, each of which has an equal probability, 2/n(n+1),  of being picked up first, if the first one picked is specified as being purple. Thus, the probability that the first marble came from the bag with a given number, k,  of purples is 2k/n(n+1).  After the first one is picked, there are  k-1 purples among the  n-1 marbles left in the bag, and the probability  of the second pick being purple is (k-1)/(n-1). The probability that the first purple marble came from that bag and that the second marble was also purple is the product of the individual probabilities, 2k(k-1)/n(n-1).

The Overall probability of all the possible combinations is

n

Σ     2k(k-1)/n(n+1)(n-1) = 2(1/3)(n-1)(n+1)/n(n-1)(n+1) = 2/3

k=0

Thanks,

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Post: #5
ofcourse u r right my dear friend, i tried with my basic knowledge, any way thank you very much for giving explanation
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