If five numbers are selected at random from the set {1,2,}, what is the probability that their minimum is larger than 5? (a number can be chosen more than once, and the order in which you select the numbers matters)

the probability that their minimum is larger than 5 is 0.2373

Step-by-step explanation:

For calculate the probability we need to make a división between the total ways to selected the 5 numbers and the ways to select the five numbers in which every number is larger than 5.

So the number of possibilities to select 5 numbers from 20 is:

20                 *      20         *          20     *        20         *       20

First number  2nd number   3rd number  4th number   5th number

Taking into account that a number can be chosen more than once, and the order in which you select the numbers matters, for every position we have 20 options so, there are   ways to select 5 numbers.

Then the number of possibilities in which their minimum number is larger than 5 is calculate as:

15                 *      15         *             15     *        15          *       15

First number  2nd number   3rd number  4th number   5th number

This time for every option we can choose number from 6 to 20, so we have 15 numbers for every option and the total ways that satisfy the condition are  

So the probability P can be calculate as:

Then the probability that their minimum is larger than 5 is 0.2373

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47