# Probability at least one of two buses arrive on time

I want to make sure my answer here is correct, I am extremely confident that it is but I am taking no chances :)

Question: The probability that a random bus arrives on time is $1 - (0.5)^5$. What is the probability that at least one of two buses arriving independently will arrive on time?

Solution:

P(At least one arrives on time) = 1 - P(Neither arrive on time)

= 1 - P(Bus 1 not on time $\cap$ Bus 2 not on time)

(By independence...)

= 1 - P(Bus 1 not on time) $*$ P(Bus 2 not on time)

Now the probability that a random bus is not on time

= 1 - P(Bus is on time)

= $1 - (1 - (0.5)^5)$

= $1 - (1 - \frac{1}{32}$)

= $\frac{1}{32}$

So

1 - P(Bus 1 not on time) $*$ P(Bus 2 not on time)

= 1 - $\frac{1}{32}\frac{1}{32}$

= $\frac{1023}{1024}$

Correct, yes?

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Good on you...... – Ross Millikan Nov 5 '12 at 14:19