Probability: More than One Scenario

Let $$x = 36{,}721$$. If $$y$$ is formed by rearranging the digits in $$x$$ at random (such that $$x \ne y$$), then what is the probability that $$y > x$$?

Incorrect. An answer of $$\frac{3}{5}$$ could be the result of including all $$y$$-values whose first digit is 3 when finding the number of ways that $$y > x$$. There are some $$y$$-values whose first digit is 3 that are _not_ greater than $$x$$. For instance, if $$y = 32{,}670$$. [[snippet]]

Incorrect. You might have gotten $$\frac{65}{119}$$ if you found the probability that $$y < x$$, instead of the probability that $$y > x$$. [[snippet]]

Now find the number of possible $$y$$-values. There are $$5! = 120$$ different ways of arranging the digits 3, 6, 7, 2, and 1. However, the problem states that $$x \ne y$$, and this calculation includes $$y = 36{,}721$$. So you need to subtract 1 from the number of arrangements to account for this overcounting: $$120 - 1 = 119$$. Therefore, the probability that $$y > x$$ is $$\frac{54}{119}$$.

Incorrect. It's possible to arrive at $$\frac{9}{20}$$ if you included the case when $$y = x$$—which is explicitly prohibited in the problem—when calculating the total number of $$y$$-values. [[snippet]]

Incorrect. You may have gotten $$\frac{2}{5}$$ if you didn't consider the case in which 3 is the first digit of $$y$$ when calculating the number of $$y$$-values greater than $$x$$. [[snippet]]

Incorrect. It might be tempting to think that half of the options are greater than $$x$$ because half of the other possibilities for the first digit are greater than 3, but that isn't the case. [[snippet]]

That's right. You can find the probability by dividing the number of $$y$$-values that are greater than $$x$$ by the number of possible $$y$$-values. To find these values, remember that the number of ways of arranging $$k$$ items is $$k!$$. Start with the number of $$y$$-values greater than $$x$$. There are two ways that $$y > x$$: 1. The first digit is greater than 3, so it is either 6 or 7. 2. The first digit is 3 and the second is 7. If the first digit is 3 and the second is 6, then $$y$$ can't be greater than $$x$$, so you don't need to go any further.

Begin by considering the number of possible $$y$$-values in each case. __(1)__: If the first digit of $$y$$ is 6, then the other four digits can be anything, so the number of possible values of $$y$$ is >$$\stackrel{\mbox{First}}{\fbox{1}} \times \stackrel{\mbox{Second}}{\fbox{4}} \times \stackrel{\mbox{Third}}{\fbox{3}} \times \stackrel{\mbox{Fourth}}{\fbox{2}} \times \stackrel{\mbox{Fifth}}{\fbox{1}} = 24$$. Similarly, there are another 24 possible values if the first digit is 7, for a total of $$24+24 = 48$$ different possibilities in case (1). __(2)__: If the first two digits are 3 and 7, respectively, then the other three digits can be anything, so the number of possible $$y$$-values is >$$\stackrel{\mbox{First}}{\fbox{1}} \times \stackrel{\mbox{Second}}{\fbox{1}} \times \stackrel{\mbox{Third}}{\fbox{3}} \times \stackrel{\mbox{Fourth}}{\fbox{2}} \times \stackrel{\mbox{Fifth}}{\fbox{1}} = 6$$. Thus, there is a total of $$48+6 = 54$$ possible ways that $$y$$ can be greater than $$x$$.

$$\frac{2}{5}$$

$$\frac{54}{119}$$

$$\frac{1}{2}$$

$$\frac{65}{119}$$

$$\frac{3}{5}$$

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