This is a case of binomial distribution because any account would either be correct or would have an error.

The probability of error is 6% Or 0.06, probability of any account being correct would be 0.94.

If the 5^th Account has to be the first with an error, The first four accounts would be correct, probability of which would be (0.94)⁴ = 0.7807. The probability of first four accounts being correct and the 5^th Account having an error would be (0.7807) (.06) or 0.0468

Out of 100 Accounts, the likely number of accounts being defective is 6. Thus one account being defective would occur in 100/6 or 16.66.. or 17 accounts.

The probability of one account being defective out of six accounts would mean the probability of 5 accounts being correct and 1 account being with error. This would be ⁶C₁ (.06)(.94)⁵ = 0.2462

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