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A code consists of two digit numbers chosen from 00-99 (i.E 00, 01, 02…..99) followed by two different letters of the alphabet. What is the probability that the code is 43MZ?

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Answer: 1/67600

Step-by-step explanation:

probability that the code is 43MZ:

Probability = (Total Required outcome / total possible outcomes)

Possible outcomes (digit) = (00,.............. ..., 99)

2.) possible outcomes (alphabet) = (A, B,............... .., Z)

P(43MZ) = P(43) × P(M) × P(Z)

P(43) = 1/100

P(M) = 1/26

P(Z) = 1/26

THEREFORE ;

P(43MZ) = (1/100) × (1/26) × (1/26) = 1/67600

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