The probability that Meena is on time to catch
the bus to her office is 0.8. Find the probability
that she is late
(a) exactly twice in a 6-day week, and
(b) at least once in a 6-day week.
I have tried it, Please correct it if i am wrong or tell if there is any other easy way to solve it:
$P(\text{Meena on time}): 0.8$
$P(\text{Meena is late}): 1 – 0.8 = 0.2$
i) probability that she is late exactly twice: $\binom{6}{2} (0.8)^2 (0.2)^4$
ii) probability that she is late at least once: $\binom{6}{1} (0.8)^5 (0.2)^1 + \binom{6}{2} (0.8)^4 (0.2)^2 + \binom{6}{3} (0.8)^3 (0.2)^3 + \binom{6}{4} (0.8)^2 (0.2)^4 + \binom{6}{5} (0.8)^1 (0.2)^5 + \binom{6}{6} (0.8)^0 (0.2)^6$
$ \Rightarrow 6(0.32768)(0.2) + 15(0.4096)(0.4) + 20(0.512) (0.8) + 15 (0.64) (0.16) + 6 (0.8)(0.32) + (0.000064)$
$ \Rightarrow 3.93 + 2.46 + 8.19 + 1.536 + 0.000064$
$\Rightarrow 16.116064$
Please tell me If I am wrong during this question ?
Thanks. Help is appreciated
Best Answer
If we are doing a specific number of trials, and the probability each time is the same, we can use a binomial random variable to solve it.
For $\text{Binomial}(n, p)$, where $n$ is the number of trials and $p$ is the probability of success on any trial. The general formula is
$$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$$
So in your example for part a, the probability she is late on any day is always .2, and we want the probability over 6 days. So we can use $\text{Binomial}(6, .2) $and solve for $P(X=2)$.
$$P(X=2) = \binom{6}{2}.2^2(1-.2)^{6-2} = .246$$
For part B, we use the same Binomial except this time, we want
$$P(X=1) +P(X=2) + \cdots + P(X=6)$$
Since that would give us the odds of her being late at least once. Instead of doing all that tho, we can just find $1 - P(X=0)$, because her being late at least once is going to be the compliment of her never being late.
$$1 - P(X=0) = 1 - \binom{6}{0}.2^0(1-.2)^{6-0} = 1- .262 = .738$$