Question

Bob sets two alarm clocks (battery-powered) to be sure he arises for his Monday 8:00 a.m. accounting exam. There is a 90 percent chance that either clock will wake Bob.

(a) What is the probability that Bob will oversleep? (Round your answer to 4 decimal places.)

Probability            

(b) If Bob had three clocks, would he have at least a 85 percent chance of waking up?

Yes

No

Answer 1

a) This is a binomial distribution question with
n = 2
p = 0.9
q = 1 - p = 0.1
where

P(X = 1) =


P(X = 0), n = 2, p= 0.9 X-Binomial(n = 2, p=0.9, X = 0) P(X = 0) = ()0.900.12-0 P(X = 0) = 10.9°0.12 = 1(1.0) (0.01) P(X = 0) = 0.01


b) This is a binomial distribution question with
n = 3
p = 0.9
q = 1 - p = 0.1
$\\ P(X \ge 1), n = 3, p = 0.9 \\ P(X \ge 1) = P(1 \le X \le 3) \\ P(X \ge 1) = 1 - P(X = 0) \\ P(X \ge 1) = 1 - ( \binom{3}{0} 0.9^{0} 0.1^{3-0} ) \\ P(X \ge 1) = 1 - ( 1 (0.9^{0}) (0.1^{3}) ) \\ P(X \ge 1) = 1 - ( 1 (1.0) (0.001) ) \\ P(X \ge 1) = 1 - ( 0.001 ) \\ P(X \ge 1) = \textbf{0.999} \\ \par\noindent\rule{\textwidth}{0.4pt}$
Yes
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