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I got this probability and statistics question when I'm doing an A-level past paper. Question 6 C in 9709/31/O/N/24.

The question asks for the probability of the three Rs being together when separating the word RECORDERS into a group of 5 and a group of 4.

I tried to list the total number of possibilities. For example, groups of 5 can have 3Rs and 1E, while groups of 4 have 1E and the rest. I calculated the number of combinations per item.

However, the answer was incorrect as I had over calculated, getting the answer 16/41 instead of 1/6.

I did not understand why, as in the correct method in the mark scheme, I don't need to consider that the Es and Rs repeat and can directly choose 5 from 9 to get the total combinations. (For instance, 3e R(1) _ and R(2) _ _ _, and 3e R(2) _ and R(1) _ _ _ should be the same combination, so they need to be calculated individually from other permutations.)

The correct method simply uses 9C5 to find the total, but wouldn't the different locations of the 3 Rs and 2 Es mess up the calculations? If we consider only AABC and separate them into groups of 3 and 1, there are only 3 distinct combinations, not 4C3 = 4 combinations.

Why don't I need to consider repeats in the correct method, and what is wrong with my old method(sorry I cannot get a picture)

Justin
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  • The individual $R$s are distinct but their locations are not. "If we consider only AABC and separate them into groups of 3 and 1, there are only 3 distinct combinations, not 4C3 = 4 combinations." That's simply not true. $\color{red}ABC-\color{blue}A$ is distinct (i.e. has a distinct affect upon calculating the probability) from $\color{blue}ABC-\color{red}A$. We can describe them as being for our goal "just one $A$" as the same but their probabilities are effective The probability of getting $ABC$ is not $\frac 13$ but is $\frac 24$. – fleablood May 07 '25 at 17:10
  • that is. The probability of getting $ABC$ = $\frac 24$ because there are four outcomes. $\color{red}A\color{blue}AB$, $\color{red}A\color{blue}AC$, $\color{red}ABC$, and $\color{blue}ABC$. ANd two of them have $ABC$ and two of them do not. – fleablood May 07 '25 at 17:13

2 Answers2

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Let's try a simpler example, with the word ERR split into a group of $2$ and a group of $1$ and ask for the probability that the two Rs are together.

An answer the examiner would see as wrong could be $\frac12$ with the argument that there are two possibilities, either E is alone (so the two Rs together) or an R being alone (so the two Rs separate). What is wrong with this is that the examiner thinks these two outcomes do not have equal probabilities.

The desired answer would be $\frac13$ and there are various ways of stating this:

  • the probability of one of the Rs being alone is $\frac23$ because there are two Rs and only one E
  • the three equally likely splits could be: $\{R_1,R_2\}$ and $\{E\}$; or $\{R_1,E\}$ and $\{R_2\}$; or $\{R_2,E\}$ and $\{R_1\}$; one of the three has the Rs together
  • there are six equally likely split permutations of which two have the Rs together: $(R_1,R_2\mid E)$, $(R_1,E\mid R_2)$, $(R_2,R_1\mid E)$, $(R_2,E\mid R_1)$, $(E, R_1\mid R_2)$, $(E,R_2\mid R_1)$

So to get the desired probability, you are expected to treat the Rs as individual letters which look the same.

Henry
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    Your last sentence implies that we would get a different answer if we didn't do what the examiners expect! That is simply not true. – TonyK May 07 '25 at 17:27
  • @TonyK I would agree with the examiners, particularly if the letters were written on balls in a bag. But there are others here who might say the question does not specify that and does not say what the equally probable possibilities are. So I wrote this from the examiners' point of view. – Henry May 07 '25 at 18:27
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    The probability is not affected by whether the R's are identical or not. To take your simpler example: there is no reasonable interpretation of the question that assigns equal probabilities to your two outcomes. You might just as well apply the same logic to the OP's more complicated setup, and say that there are two possibilities: either the three R's are together, or they are not; so the answer is $\frac12$ in this case too. This is obviously absurd, but no more so than the simpler case. – TonyK May 07 '25 at 19:12
  • @TonyK I do not disagree with you. But a similar question and answer provoked a debate I was trying to avoid here. – Henry May 07 '25 at 19:28
  • I don't see how else to interpret the question. If it's a matter or probability how can we interpret the Rs as a single letter yet have a mechanism for probability distribution? It'd be like coming up with a model of rolling dice where somehow all of the 21 outcomes are equally likely (that is rolling a 3 and a 4 is the same probability, $\frac 1{21}$, as rolling two 3s). I just don't see how a probability model can exist. – fleablood May 07 '25 at 19:48
  • @fleablood: At the risk of sounding pedantic, a probability model basically just has to be a measure that adds up to 1 over the sample space. If we say that the sample space consists of 21 discrete events, then we can arbitrarily assign the same measure to all of them. There is nothing requiring that these events can be "nicely" divided up and recombined according to (e.g.) the value shown on the first die, and in the general case that does not happen. – Kevin May 08 '25 at 03:11
  • @Kevin fair enough. – fleablood May 08 '25 at 23:28
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There are $9\choose4$ ways to choose the group of 4, so $9\choose4$ ways to divide the letters into a group of 4 and a group of 5.

You want to know the probability that the group of 4 letters contains either all or none of the Rs. There are $6\choose4$ ways to pick the group of 4 using only the 6 letters that are not R, and there are $6\choose1$ ways to pick one non-R to add to the three Rs and form a group of 4 that includes all the Rs.

Therefore the required probability is $\displaystyle{{6\choose4}+{6\choose1}\over{9\choose4}}={15+6\over126}={1\over6}$.

Steve Kass
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