Consider two letters Q and Y and two numbers 5 and 8. There 24 passwords of length 4 that uses these characters without repetitions. Show that when a password is chosen at random from this set, the probability of getting passwords in which Q comes earlier than both the numbers (written from left to right) is 1 3 .

The probability of selecting a password such that Q comes earlier than both the numbers is .

Step-by-step explanation:

The options to form a password of length 4 are: Q, Y, 5 and 8.

The total number of passwords that can be formed is, 4! = 24.

The condition applied is: Q comes earlier than both the numbers.

The sample space satisfying this condition is:

S = {QY58, QY85, YQ58, YQ85, Q58Y, Q85Y, Q5Y8, Q8Y5}

  = 8 possible passwords.

The probability of selecting a password such that Q comes earlier than both the numbers is:

Thus, the probability of selecting a password such that Q comes earlier than both the numbers is .

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)

37 = ending with 7

unit's digit in the cube..3

56 = ending with 6

unit's digit in the cube..2

133 = ending with

unit's digit in the cube..7

hope it : ( : )