Question

Suppose that a computer program randomly generates an 8-letter string from the letters A,B,C,D,E. For example, the program might generate the string CCCCCCCC or DAAEDCBB. The letter in each of the 8 positions is chosen independently of the other positions, and each of the five letters is chosen with equal likelihood. What is the probability that the string contains at least one A or at least one B?

Answer 1

Number of letters available for the computer to generate = 8

Length of the string = 5

First letter in the string can any of the 5 letters

Second letter in the string can be any of the 5 letters...and so on 8th letter in the string can be any of the 5 letters

Therefore number of ways generating 8-letter string from the 5 letters (A,B,C,D,E) = 5x5x5x5x5x5x5x5=5⁸ = 390625

Probability that the string contains at least one A or at least one B = 1-Probability that the string contains no A and no B(Zero As and Zero Bs)

If exclude A and B from the available letters ; then available letters are CDE i.e 3 letters.

using the same logic as above

First letter in the string can any of the 3 letters

Second letter in the string can be any of the 3 letters ...and so on 8th letter in the string can be any of the 3 letters

Therefore number of ways generating 8-letter string from the 3 letters (C,D,E) = 3x3x3x3x3x3x3x3=3⁸ = 6561

Probability that the string contains no A and no B(Zero As and Zero Bs)

= number of ways generating 8-letter string from the 3 letters (C,D,E)/ number of ways generating 8-letter string from the 5 letters (A,B,C,D,E)

= 6561/390625

Probability that the string contains no A and no B(Zero As and Zero Bs) = 6561/390625=0.01679616

Probability that the string contains at least one A or at least one B = 1-Probability that the string contains no A and no B(Zero As and Zero Bs) =1 - 0.01679616=0.98320384

Probability that the string contains at least one A or at least one B = 0.98320384