If each coded item in a catalog begins with 3 distinct letters followed by 4 distinct nonzero digits, ﬁnd the probability of randomly selecting one of these coded items with the ﬁrst letter a vowel and the last digit even.

The probability of randomly selecting one of the considered coded items with the ﬁrst letter a vowel and the last digit even is 0.04 approx

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

What is the rule of product in combinatorics?

If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in [tex]p \times q[/tex] ways.

Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.

Thus, doing A then B is considered same as doing B then A

There are 52 alphabets(A,B,..., Z, a,b,c,.... z) (26+26 = 52) and 10 digits (0,1,2,3,4,5,6,7,8, 9)

For first coded item, total 52 choices, for second total 51 (as they're distinct, so can't use what first item used), and for third there are total 50 choices (as third item can't use what 2 distinct choices first and second item chose). Similarly, for fourth item, there are 10 choices, for fifth there are 9, for sixth, there are total 8 choices and for seventh item, there are total 7 choices left).

Thus, total number of ways these seven item coded items is:

[tex]52 \times 51 \times 50 \times 10\times 9\times 8\times 7 = 668304000[/tex] ways. This is the size of the sample space, made by these seven coded items.

For the favorable choice, we have only 10 choices for first item(as 5 upper case vowel and 5 lower case vowel are allowed). For second to sixth items, the choices will be same, but for last, option, there are either 5 (if no even used by previous 3 digits), or 4 (if one even used), or 3 (if two evens used), or 2 choices(if 3 evens used by earlier items).

Thus, there favorable number of cases is addition of such cases:

Case 1: 0 evens used earlier

So only 4 odds used earlier, and thus 5 even choices for last digit

[tex]10 \times 51 \times 50 \times 4\times 3\times 2\times 5[/tex] =3060000

Case 2: 1 even used earlier

4 odds + 1 even = 5 choices, and thus 4 even choices for last digit

[tex]10 \times 51 \times 50 \times 5\times 4\times 3\times 4[/tex]= 6120000

Case 3: 2 evens used earlier

4 odds + 2 evens = 6 numbers, and thus 5-2=3 even choices for last digit

[tex]10 \times 51 \times 50 \times 6\times 5\times 4\times 3[/tex] = 9180000

Case 4: 3 evens used earlier

4 odds + 3 evens = 7 numbers, and thus 5-3=2 even choices for last digit

[tex]10 \times 51 \times 50 \times 7\times 6\times 5\times 2[/tex] = 10710000

Total counts of these cases is 29070000

Thus, the probability of getting such coded item with first item vowel and last item even is: [tex]\dfrac{29070000}{668304000} \approx 0.04[/tex]

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