The technology company DataGenetics suggests that 17.8% of all four-digit personal identification numbers, or PIN codes, have a repeating digits format such as 2525 . Assuming this to be true, if the PIN codes of five people are selected at random, what is the probability that at least one of them will have repeating digits? Round your answer to four decimal places.

0.6247

Explanation:

Probability = 1 - P(Do not have repeating digits)⁵

                  = 1 - (0.822)⁵

                  = 0.6247

The probability that at least one of the codes will have repeating digits is 0.6263

Explanation:

The probability of a pin being repeated is 17.8%

i.e 17.8/100

Eliminating the decimal point; 17.8/100 = 178/1000

Reducing 178/1000 to its lowest term;

178/1000 = 89/500

The probability that a pin, if selected will have a repeating digit format = 89/500

The probability of a pin not having repeating digits format = (500/500) - (89/500) = 411/500

If five codes are selected at random (without replacement), the probability that all five of the codes will not have repeating digits format =

(411/500) × (410/499) × (409/498) × (408/497) × (407/496)

= 0.37365052859184

The probability that all selected five won't have repeating digits pattern = 0.37365052859184

The probability of selecting at least one code with repeating digits format out of five codes can be solved by subtracting the probability of all five codes not having repeating digits format from 1

i.e 1 - 0.37365052859

= 0.626349471

= 0.6263 ( rounded off to four decimal places)

no backup memory

continuous information you receive.