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Chapter 19: Problem 25

For a gas of 6 molecules confined to a box, find the probability that (a) all the molecules will be found on one side of the box and (b) half the molecules will be found on each side.

Short Answer

Expert verified
The probability for case (a) is \(2 \cdot (0.5)^6\), and for case (b) it is \(\binom{6}{3} \cdot (0.5)^3 \cdot (0.5)^3\).

Step by step solution

01

Identify the Total Number of Molecules

The total number of molecules is given as 6. This total number will be useful when calculating the binomial coefficient.
02

Calculate the Probability for Case (a)

For all molecules to be on one side of the box, we are basically choosing 6 molecules out of 6, which can be represented as the binomial coefficient \(\binom{6}{6}\). However, there are two 'sides' to the box, so we also need to account for the cases where all six molecules are on the 'other side'. The binomial probability formula is \(\binom{n}{k} p^k (1-p)^{n-k}\), where \(n\) is the total number of trials, \(k\) is the number of 'successful' trials, and \(p\) is the probability of success on each trial. Because there are an equal number of molecules, and they can be on either side, \(p\) is \(0.5\). Therefore, the probability is \(2 \cdot \binom{6}{6} \cdot (0.5)^6 \cdot (0.5)^0 = 2 \cdot (0.5)^6\).
03

Calculate the Probability for Case (b)

To get half of the molecules on each side (3 on one side, 3 on the other), we again use the binomial formula with \(k = 3\). In this case, we don't need to multiply by 2 because the question doesn't care which 'side' is which. Therefore, the probability is \(\binom{6}{3} \cdot (0.5)^3 \cdot (0.5)^3\).

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