Question

If you roll a 10-sided dice 3 times, what is the probability that they form neither a strictly increasing nor a strictly decreasing series?
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Answer 1

Answer:

0.76

Explanation:

Let the number of strictly increasing arrangements  be = X

Let the number of strictly decreasing arrangements  be = Y

Let the number of neither strictly decreasing nor strictly increasing  be = Z

Thus ,

X+Y+Z=1000

A strictly increasing arrangement from the opposite side, will look like a strictly decreasing arrangement.

Thus, X=Y

= 2X+Z=1000

For a strictly increasing arrangement ,all the three numbers should be different which can be done in 10C3 ways.

Z = 1000-(10C3) × 2

= 760

Therefore the probability

= 760/1000

= 0.76

Answer 2

The probability of obtaining a sequence which is neither strictly increasing nor decreasing in 0.76($\dfrac{760}{1000}$).

Explanation:

All possible outcomes are:

1, 1, 1 / 1,1 ,2 /……….. 10, 10,9 / 10,10,10

Total number of all possible outcomes = 10 x 10 x 10 = 1000

Let 2,6, 1 to 6, 6 denote that where the first roll is 2 and the second is 6, then the third roll can be any of 1 to 6 ( 1,2,3,4,5 or 6) which gives 6 “ successful” outcomes.

For first roll of 1, we have 64 successful outcomes, namely

1, 1, 1 to 10, 10

1, 2, 1 to 2, 2

1, 3, 1 to 3, 3

1, 4, 1 to 4, 4

1, 5, 1 to 5, 5

1, 6, 1 to 6, 6

1, 7, 1 to 7, 7

1, 8, 1 to 8, 8

1, 9, 1 to 9, 9

1, 10, 1 to 10 , 10

For first roll 3, the list begins

3, 1, 1 to 10, 10

3, 2, 2 to 10, 9

3, 3, 1 to 10,10

3, 4, 1 to 4, 4

The total number of successful outcomes is 78

For first rolls 1 to 10 , the total number of successful outcomes are 64 , 72 , 78 , 82, 84 , 84 , 82 , 78, 72 , and 64 respectively. This gives 760 successful outcomes from a total of 1000 outcomes.

∴ P(getting a neither a strictly increasing nor a strictly decreasing series)

= $\dfrac{760}{1000}$

Hence, the probability of obtaining a sequence which is neither strictly increasing nor decreasing in 0.76($\dfrac{760}{1000}$).