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11 Jan 2021, 00:49
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An urn contains five chips, numbered 1 through 5. Three chips are drawn out at random (without replacement). What is the probability the largest chip in the sample is a 4?
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 1/2
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 1/2
11 Jan 2021, 12:29
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In many cases, there are two ways to solve problems of sampling without replacement. The answer is the same from both methods:
1. Order does not matter (This is generally the easier method.)
There are \(^5C_3\) ways to choose three objects out of a set of size 5. So, \(P= ^5C_3\). Let us count the number of outcomes where the largest number is \(i\). We need only consider one case: one of the numbers is \(i\) and the other two are smaller than \(i\). There are \(^{i-1}C_2\)
ways to choose two numbers smaller than \(i\). Hence,
\(P = \frac{^{i-1}C_2 }{ ^5C_3}\) where \(i=4\)
\(P= \frac{^{4-1}C_2 }{ ^5C_3}=\frac{3}{10}\)
2. Order matters
The method here is similar to the method of the previous problem. Suppose we consider the order to be important. Note that this is not to say that we are considering another experiment – this is just another way of looking at the same experiment.
There are \(5\) ways to choose the first ball, \(4\) to choose the second, and \(3\) to choose the third. So \(|P| = 5*4*3\). Let us count the number of the ways that the largest number is \(i\). We have three ways to choose which number is \(i\) and we have \((i − 1) (i − 2)\) choices for the remaining numbers since numbers cannot be repeated. So
\(P = \frac{3(i-1)(i-2)}{5*4*3}\) whereas \(i=4\)
Note that the solution is the same
\(P=\frac{3*3*2}{5*4*3}=\frac{3}{10}\)
Ans C
1. Order does not matter (This is generally the easier method.)
There are \(^5C_3\) ways to choose three objects out of a set of size 5. So, \(P= ^5C_3\). Let us count the number of outcomes where the largest number is \(i\). We need only consider one case: one of the numbers is \(i\) and the other two are smaller than \(i\). There are \(^{i-1}C_2\)
ways to choose two numbers smaller than \(i\). Hence,
\(P = \frac{^{i-1}C_2 }{ ^5C_3}\) where \(i=4\)
\(P= \frac{^{4-1}C_2 }{ ^5C_3}=\frac{3}{10}\)
2. Order matters
The method here is similar to the method of the previous problem. Suppose we consider the order to be important. Note that this is not to say that we are considering another experiment – this is just another way of looking at the same experiment.
There are \(5\) ways to choose the first ball, \(4\) to choose the second, and \(3\) to choose the third. So \(|P| = 5*4*3\). Let us count the number of the ways that the largest number is \(i\). We have three ways to choose which number is \(i\) and we have \((i − 1) (i − 2)\) choices for the remaining numbers since numbers cannot be repeated. So
\(P = \frac{3(i-1)(i-2)}{5*4*3}\) whereas \(i=4\)
Note that the solution is the same
\(P=\frac{3*3*2}{5*4*3}=\frac{3}{10}\)
Ans C
