91影视

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

(a):

Since there are no restrictions, and we have total of \({\rm{10}}\) digits, by using the rule:

Product Rule for k-Tuples

Assume that we can select first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways, and that for cash selected clement we can select the second element in \({{\rm{n}}_{\rm{2}}}\)ways, then we have the number of pairs to be\({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\).

Similarly, for ordered collection of \({\rm{k}}\)elements, where the \({{\rm{k}}^{{\rm{th }}}}\)can be chosen in \({{\rm{n}}_{\rm{k}}}\)ways, then there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\)possible \({\rm{k}}\)-tuples.

There are\({\rm{10*10 \ldots *10 = 1}}{{\rm{0}}^{\rm{4}}}{\rm{ = 10,000}}\)possible four-digit PINs.

b)

To calculate the probability that it will be a legitimate PIN, we want to use

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\)

where we need to find # of favorable outcomes in \({\rm{A}}\) and \({\rm{\# }}\) of outcomes in the sample space.

Number of outcomes in the sample space is\({\rm{10,000}}\), as we have calculated in (a).

To calculate the number of favorable outcomes, we will subtract from \({\rm{10,000}}\)the number of prohibited PINs.

(i): all four digits identical - there are \({\rm{10}}\) PINs with all digits identical:

\({\rm{0000,1111, \ldots ,9999}}\)

(ii): sequences of consecutive ascending or descending digits - there are

\({\rm{0123,1234,2345,3456,4567,5678,6789,7890,3210,4321,5432,6543,7654,8765,9876,0987,}}\)which makes it total of \({\rm{14}}\)PINs.

(iii): any sequence starting with - all PINs from \({\rm{1900}}\) to\({\rm{1999}}\), there are total of\({\rm{100}}\).

The number of prohibited PINs is

\({\rm{10 + 14 + 100 = 124}}\)

which makes the number of favorable outcomes (PINs)

\({\rm{10,000 - 124 = 9,876}}\)

The probability is therefore

\(\begin{array}{c}{\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\\{\rm{ = }}\frac{{{\rm{9,876}}}}{{{\rm{10,000}}}}\\{\rm{ = 0}}{\rm{.9876}}\end{array}\)

c)

Notice that even with the restrictions, PINs starting with \({\rm{8}}\) and ending with \({\rm{1}}\) in between could have any of the \({\rm{10}}\) digits, therefore there are

\({\rm{10*10 = 100}}\)

not prohibited PINs which starts with \({\rm{8}}\) and ends with \({\rm{1}}\) . Since we select only \({\rm{3}}\) of those \({\rm{100}}\)combinations the probability of guessing right is

\(\frac{{\rm{3}}}{{{\rm{100}}}}{\rm{ = 0}}{\rm{.03}}\)

Therefore, the probability is \({\rm{0}}{\rm{.03}}\).

d)

There are a couple of prohibited PINs:

\({\rm{1111,1901,1911,1921, \ldots 1991}}\)

which makes it total of \({\rm{1 + 10 = 11}}\) prohibited PINs, or \({\rm{100 - 11 = 89}}\) possibilities. Since we assume that the individuals know about the restrictions, out of three guess he has

\(\frac{{\rm{3}}}{{{\rm{89}}}}{\rm{ = 0}}{\rm{.0337}}\)

probability to guess right.