Question

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 .

Answer 1

Answer:

The probability of selecting a password such that Q comes earlier than both the numbers is $\frac{1}{3}$.

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:

$P (Q\ before\ both\ number)=\frac{8}{24} =\frac{1}{3}$

Thus, the probability of selecting a password such that Q comes earlier than both the numbers is $\frac{1}{3}$.