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The probability that we pick any number for the first time is $\dfrac{1}{5}$

the sample space of sample spaces after the first event is then

{2,3,4,5}

{1,3,4,5}

{1,2,4,5}

{1,2,3,5}

{1,2,3,4}

prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$

prob. to pick an odd from the 2nd sample space is $\dfrac{3}{4}$

prob. to pick an odd from the 3rd sample space is $\dfrac{1}{2}$

prob. to pick an odd from the 4th sample space is $\dfrac{3}{4}$

prob. to pick an odd from the 5th sample space is $\dfrac{1}{2}$

The final result is:

$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$

Is this reasoning correct? Are there any simpler ways to solve this problem?

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  • $\begingroup$ The probability that we pick any number for the first time is technically 1. $\endgroup$
    – bjb568
    Mar 26, 2015 at 1:11

5 Answers 5

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There is a simpler way. No digit should have an inherent advantage of being picked second. So since 3 of the 5 digits are odd, there is a $\frac{3}{5}$ chance of the second number picked being odd.

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    $\begingroup$ Great insight. Its these un-intuitive insights that makes probability hard for most (including me). @StevenClontz had updated his answer to provide little more explanation. $\endgroup$
    – leesei
    Mar 26, 2015 at 4:03
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The probability we pick an odd number first is $\frac{3}{5}$. Then the probability of another odd number being chosen from the remaining numbers is $\frac{2}{4}=\frac{1}{2}$. Put together, the probability is $\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$.

The probability we pick an even number first is $\frac{2}{5}$. Then the probability of an odd number being chosen from the remaining numbers is $\frac{3}{4}$. Put together, the probability is $\frac{2}{5}\times\frac{3}{4}=\frac{3}{10}$.

The probability that the second number being chosen is odd is the sum of these two probabilities. So $\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$ is the correct answer.


As others have noted, picking randomly could also be thought of as shuffling the numbers like a deck of cards, and then choosing top to bottom. Of course, there's a $\frac{3}{5}$ chance that the second "card" would be an odd number, but it's nice to see that the conditional probability works out as well.

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There are simpler ways. There is absolutely no preference between the numbers, so each number is equally likely to be the second one drawn. Since 3 of 5 numbers are odd, the probability is 3/5 that an odd number is drawn.

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You are probably familiar with "tree diagrams" to solve these types of problems. For the first pick you have a (2/5) chance to remove an even, and a (3/5) chance to remove an odd.

When you remove an odd first, there are only (2/4) odds left. Meanwhile, if you remove an even first, there are now (3/4) odds left.

Finally (2/5)(3/4)+(3/5)(2/4) = (6/20)+(6/20) = (12/20) = (3/5)

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There's another (exhaustive) way of seeing it, and it's counting.

Let us represent the act of picking a number $a$ and then a number $b$ by $(a,b)$, we then have $5\times 4=20$ of this possible "actions". Now, the cases that favors your situation are:

$$(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3)\ .$$

Since there are $12$ favor cases, the probablity is $12/20=3/5$. It's not easier, it's just another way of seeing it.

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