4
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$4$ integers are randomly selected from the numbers from $1$ to $10$. The chance that there are atleast two successive numbers among those $4$ selected is
$(A)\frac{5}{6}\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{2}{3}\hspace{1cm}(D)\frac{1}{2}\hspace{1cm}$

I calculated answer as $\frac{24}{\binom{10}{4}}$ but this wrong. Please help me find the right answer.

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  • 3
    $\begingroup$ Hint: it is much easier to calculate the complementary probability that all 4 numbers are spaced apart by at least 1. $\endgroup$ – Erick Wong Sep 29 '15 at 5:18
  • $\begingroup$ Please elaborate i am not getting the answer. @ErickWong $\endgroup$ – user1442 Sep 29 '15 at 6:44
  • $\begingroup$ Is the selection with or without replacement? $\endgroup$ – robjohn Sep 29 '15 at 10:03
  • $\begingroup$ See 'math.stackexchange.com/questions/1295252' for a closely related, slightly simpler, problem. $\endgroup$ – BruceET Feb 3 '16 at 0:10
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Take out $4$ numbers, 6 numbers (N) are left with $7$ possible gaps including ends

$_ N _ N _ N _ N _ N _ N _$

We can replace the $4$ numbers in forbidden way in any of 7 gaps (including ends) in $\dbinom{7}{4}$ ways

thus indicated $Pr = 1 - \dfrac{\dbinom{7}{4}}{\dbinom{10}{4}}$

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0
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Here is a simulation in R with 100,000 performances of the experiment (choices without replacement), which verifies the Answer $5/6$ of @trueblueanil within simulation error.

 m = 10^5;  d = numeric(m)
 for(i in 1:m) {
   x = sample(1:10, 4);  sort(x)
   d[i] = min(diff(sort(x)))  }
 mean(d == 1)  # 'd==1' is logical vector of T's and F's
 ## 0.83222    #  its 'mean' is proportion of T's
 1 - choose(7,4)/choose(10,4)  # exact analytic answer
 ## 0.8333333
 2*sqrt((.83)*(.17)/m)  # 95% margin of simulation error
 ## 0.002375710

To get greater accuracy, use m=10^6, but run shown is sufficient to distinguish among the four answers proposed.

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I came up with the following reasoning :

Pattern 1: At least two successive numbers, $$ \eqalign { & (**)(*)(*) \cr & (*)(**)(*) \cr & (*)(*)(**) \cr & choices \ = 3 \times (2!\cdot 9) \times 8 \times 7 = 3024 } $$ Pattern 2: At least three successive numbers, $$ \eqalign { & (***)(*) \cr & (*)(***) \cr & choices \ = 2 \times (3!\cdot 8) \times 7 = 672 } $$

Pattern 3: At least four successive numbers, $$ \eqalign { & (****) \cr & choices \ = 1 \times (4!\cdot 7) \times 6 = 1008 } $$ The total number of legal possible choices, using the inclusion exclusion principle is : $$ 3024-672+1008 = 3360 $$ and the required probability is $$ {3360 \over 10 \cdot 9 \cdot 8 \cdot 7} = {2 \over 3} $$

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