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Try to generalize the result of the previous problem to include the case when one part of the vessel contains k out of n balls (k le n)in conditions when (a) the probabilities of a ball being in the left-hand and the right-hand parts of the vessel are different, (b) the probabilities of a ball being in either part of the vessel are equal. |
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Answer» Solution :(a) Let the probability of finding the given ball in one PART of the vessel bep, then the probability of finding it in the other part of the vessel is `q = 1 - p`. The probability of finding kc particular balls in the first part of the vessel is `p^k`, the probability of finding the rest in the other half is `q^(n-k)`. Consequently the probability of the event when k particular balls are found in the first part of the vessel and n - k are found in the second part of the vessel is `p^k q^(n - k)`. However, we assume all the balls to be identical, and because of that this result may be realized in `C_n^h` WAYS in other words, the thermodynamic prob: ability W of this state is `C_n^h`. To obtain the MATHEMATICAL probability w, one should multiply the number of ways by the probability of a favourable combination, Therefore `w_k = C_n^h p^h q^(n-h) = C_n^h (1 - p)^(n-h) p^h` Such a distribution of probabilities is called binomial distribution.  (b) If the probability of finding the ball in both parts of the vessel is the same, then p = q = 1/2, and we have `w_k = C_n^h (1/2)^h (1/2)^(n-h) = C_n^k (1/2)^n` We have generalized the result of PROBLEM 18.4 for the case of n balls. |
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