Question

Question 3: a. A computer password is 6 characters long. A character can be either a digit or an uppercase English letter. A valid password must contain exactly one digit. How many valid passwords are there? (hint: you do not know the position of this one digit. It might be in the beginning or the end or the middle of the password. You may use the combination low to select the position of this digit and then assigno value to it) b. What is the probability of an adversary guessing this password correctly (with the specification mentioned above in question 3)? Question 4 Q2: Are the events E, that a family with three children has children of both sexes, and F, that this family has at most one boy, independent? Assume that the eight ways a family can have three children are equally likely. (find the probabilities of E and Fusing counting principle)

Answer 1

3.a. No.of valid pwds =

3.a. No.of valid pwds = N1 * N2 * N3; where Ni : No.of ways to choose position of a digit, N2 : no.of possible digits choices, N3 : No.of ways to fill remaining 5 positions with Upper case letters N1 = C1; N2 = 10; N3 = 26"; So no.of valid pwds = 60 * 265 b. Clearly among total valid pwds, there is exactly 1 correct pwds, so probty = 60 * 265

4. PE = P(B1B263) OR (B1G2B3) OR (G1B2B3) or (G1G2B3) OROR (G1B2G3) OR (B1G2G3)) = 6/8 = 3/4; P(F) = P((G1G2G3) OR (G1G2B3) OR (G1B2G3) OR (BIG2G3)) = 4/8 PENF) = P((G1G2B3) OR (G1B2G3) OR (B1G2G3)) = 3/8; So P(E)P(F) = (3/4)(1/2) = 3/8 = P(En F) so yes, E and Fare independent events.