Question

question1: A company randomly generates 5-digit passwords for its clients. Each contains 3 unique numbers chosen from {0, 1, ..., 5} & 2 unique letters from {A, ..., G}. Determine the probability that Sally receives a password containing the letters B & E (in either order) & the numbers 2, 4, 5 (in any order).

question2: With probability = .25 , two switches are selected without replacement from box A, & with probability = .75 , two switches are selected without replacement from box B.

box A contains 4 defective switches & 6 good switches (prob = .)

box B contains 8 defective switches & 2 good switches (prob = .)

Determine the probability that the 2nd chosen switch is defective, given that the 1st chosen switch is defective.

Answer 1

1:

Since numbers need to be unique so repetition is not allowed. First number can be selected in 6 ways, second in 5 ways and 3rd in 4 ways. Likewise first can be selected in 7 ways and second can be selected in 6 ways. So possible number of passwords is

6*5*4*7*6 = 5040

The B and E can come in password in 2 ways and 2,4, 5 can come in password in 3! =6 ways so possible number of passwords with B and E and digits 2, 4 and 5 is

2 *6 = 12

The probability that Sally receives a password containing the letters B & E (in either order) & the numbers 2, 4, 5 is

12 /5040 = 0.0024

Answer: 0.0024