Suppose each laptop of a certain type is assigned a series number, which consists of a sequenceof eight symbols: number, letter, letter, letter, number, number, number, number, where aletter is any one of 26 letters and a number is one among 0, 1,. . ., 9. Assume that all series numbers are equally likely.
(a) What is the probability that all symbols are different if one laptop is picked at random with equal probability?
(b) What is the probability that all symbols are different and the first number is the largest among the numbers?

Step-by-step explanation:

There are 5numbers and 3 letters. Each number can be one among the 10 possible.

Each letter is 1 among the 26 available.

a)The number of possible unique serial numbers are

10^5x 26³

The number of 5 number combinations each different (order important) is 5!(10/5)= 30240

The number of 3 letter combinations each different (order important) is 3!(26/3)=15600

The required probability (that all symbols are different if one laptop is picked at random with equal probability) is

15600x3240/ 10^5x 26³=0.2684

b) The first letter must be at least 4 to be largest among the 5 numbers. So the first letter can be one of . 4,5,6,7,8,9..

The number of 5 number combinations each different and starting with 4 (order important) is .

4!=24

The number of 5 number combinations each different and starting with 5 (order important) is .

4!(5/4)= 120

The number of 5 number combinations each different and starting with 6 (order important) is .

4!(6/4)= 360

The number of 5 number combinations each different and starting with 7 (order important) is .

4!(7/4)= 840

The number of 5 number combinations each different and starting with 8 (order important) is

4!(8/4)= 1680

The number of 5 number combinations each different and starting with 9 (order important) is .

4!(9/4)= 3024

Thus, there are

The required probability ( that all symbols are different and the first number is the largest among the numbers) is

15600* 6040/10^5x 26³= 0.0537

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step-by-step explanation: