The probability that Abiola will be late to the office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will:
(a) only be late for 3 days.
(b) not be late in the week:
(c) be late throughout the six days.
Explanation
he will be late to office = \(\frac{2}{5}\) = p (success)
he will not be late to office is = 1 - \(\frac{2}{5}\) = \(\frac{3}{5}\) = q( failure)
n = total number of trials, k = number of successful trial(s)
If he will be late for 3 days only, he will also not be late for 3 days
P(x = 3) = \(^n\)C\(_k\) p\(^k\)q\(^{n - k}\) = \(^6C_3\) x (\(\frac{2}{5})^3\) x (\(\frac{3}{5})^3\)
p ≈ 0.2765
(b) not be late to office is = 1 - 2/5 = 3/5
\(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\)
p = 0.0467
(c) be late to office = 2/5
\(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\)
p = 0.0041