. Of the 7-digit passwords formed by mixing 3 numbers (chosen from 1 to 9, without repeats) and 4 letters (chosen from A to Z, allowing repeats), I pick one at

Question

(a) Has four A’s.
(b) Has at least two A’s.
(c) Begins and ends with an A. 5 points
(d) Has two numbers next to each other with the third one not next to them.

 

 

Leave your probabilities in terms of mn, Pnm, Cnm, and factorials.

Answer 1

(a) Has four A’s.

 

P(A) = 1/26

 

P(4A + any 3 numbers) = (1/26)^4

 

(b) Has at least two A’s.

 

P(at least 2 A) = P(2A) + P(3A) + P(4A)

 

You can use the binomial distribution to solve. 

 

P(2A) = 4C2 * (1/26)^2 * (25/26)^(4-2) = 12 * (1/26)^2 * (25/26)^2

P(3A) = 4C3 * (1/26)^3 * (25/26)^1 = (100/26) * (1/26)^3

P(4A) = (1/26)^4

 

Add them all together to get the answer.

(c) Begins and ends with an A. 

 

P(A) * P(non A) * P(non A) * P(A)

= (1/26) * (25/26) * (25/26) * (1/26)

= (1/26)^2 * (25/26)^2

(d) Has two numbers next to each other with the third one not next to them.

 

I'm not exactly sure what you mean here. Is this the same number next to itself or consecutive numbers next to each other?