5

A die is randomly rolled five times, the probability that outcome of the fifth throw is one of the outcomes of the first 4 throws is?

Attempt:

For first 4, there are following possibilities: (after that we will multilply the possibilities for the fifth one)

  1. All different

  2. 2 same 2 different

  3. 3 same 1 different

  4. All same

$1.$ All different:

Number of ways: $(6*5*4*3)\times\dbinom 41 = 1440$

$2$. $2$ same, $2$ different:

Number of ways: $\dfrac{4!}{2!}\dbinom 63 \times \dbinom 31 = 720$

$3$. $3$ same, $1$ different:

Number of ways: $\dfrac{4!}{3!}\dbinom 62 \times \dbinom 21 = 120$

$4$. All same:

Number of ways: $\dbinom 61 \times \dbinom 11 = 6$

Thus, $p(E) = \dfrac{@1+@2+@3+@4}{6^5} = \dfrac{381}{6^4}$

But answer given is: $\dfrac{671}{6^4}$

What's my mistake?

Edit:

Forgot to use: (as pointed out by @anryvian in comments)

$5.$ 2 same, 2 same:

Number of ways: $\dfrac{4!}{2!2!} \times \dbinom 6 2 \times 2 = 180$

Which gives the answer as: $\dfrac{411}{6^4}$ (still incorrect :( )

Community
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Archer
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  • You forgot "two pairs" which has $\frac{4!}{2!2!} \binom{6}{2} \cdot 2 = 180$ more outcomes. But this fix does not seem to yield the given answer either. – angryavian Mar 28 '19 at 05:53
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    Order of which are same and which are different matters. – Diatche Mar 28 '19 at 06:07

2 Answers2

6

An easier approach: the complementary event is "none of the first four throws is the same as the fifth throw."

$$1-\left (\frac{5}{6}\right )^4$$

I haven't yet found how to fix your attempt (which is a valid approach), but hopefully I (or someone else) catches it soon.

Corrected computations:

  1. $1440$ is correct.
  2. $\frac{4!}{2!} 6 \cdot \binom{5}{2} \cdot 3 = 2160$ (You have to pick which number is the pair and which are the singles)
  3. $\frac{4!}{3!} 6 \cdot 5 \cdot 2 = 240$. (You have to pick which number is the triple and which is the single)
  4. $6$ is correct.
  5. $180$ is correct.
N. F. Taussig
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angryavian
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-1

1)1440

2)2160

3)240

4)6

5)180

This will be the correct order of Values

Considering your Answer, the 3rd part has mistakes it should be 240 bcz you need to take in account the possibility of mirror values,

for eg: 1112 and 2111 are of the type AAAB and involve the same two numbers you took from the set of 6 numbers but you did not account for this case

Aditya Pratap Singh
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    This seems to be much the same as the second half of the accepted answer from three years ago. But there are plenty of questions that don't have good answers yet. – David K Aug 22 '22 at 12:02