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Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. Find the probability that the numbers in AP.

In the above problem, my teacher told that the total number of ways will be $21C2$. I understood this point. But there is one more problem given in the book:

Let two numbers are selected at random from set ${1,2,....,50,51}$, then find the probability that the sum of two numbers is even.

In this problem, the solution manual is taking the total number of ways of select two numbers as $51×51$.

I am not able to distinguish between these two problems and why aren't we using the combination in the second one too just like the first one. Any help would be appreciated.

Thanks.

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In the first case, tickets are drawn without replacement -- once a particular ticket is drawn, it cannot be drawn again. This leads to the use of the combinations operation.

In the second case, the numbers are drawn "with replacement" -- once a particular number is drawn, nothing prevents drawing it again since we can have as many copies of a particular number as we like. So we can draw the same number twice.

It is generally better for problems to indicate explicitly whether drawings are with or are without replacement. So if you ever construct a problem like this (or intend to communicate it efficiently to others), explicitly say whether replacement occurs.

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In the first one we are selecting a pair among 21 object that is $\binom{21}{2}$ and can be viewed as a selection without replacement, that is $\frac{21\cdot20}2$.

In the second one the selection is with replacement (or repetition) thus we have 51 option for the first and 51 for the second number.

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