# Conditional probability distribution question issue

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

• This is incorrect, you mix up your probabilities in the first one, and Im not sure where you went wrong in the second but I got a different asnwer – XRBtoTheMOON Dec 11 '17 at 16:36
• Why this was down-voted seems anything but clear. – Michael Hardy Dec 11 '17 at 16:43
• @MichaelHardy Thanks – himanshu chawla Dec 11 '17 at 16:47
• @MichaelHardy I'm assuming because hes already asked this exact question once before – XRBtoTheMOON Dec 11 '17 at 16:48
• @btcgrl No issues. You seem to be very active too. Thanks for helping and correcting me out always. Thanks – himanshu chawla Dec 11 '17 at 17:05

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.

$$1 - P(X=0) = 1 - \binom{6}{0}.2^0(1-.2)^{6-0} = 1- .262 = .738$$

• your calculation for second part is correct ? I am getting a different answer for this. can you verify this ? – himanshu chawla Dec 11 '17 at 16:57
• @himanshuchawla its just $1 - .8^6$ which is $1 - .262 = .738$. Edited answer to make it a little more clear – XRBtoTheMOON Dec 11 '17 at 16:59
• @himanshuchawla np, cheers – XRBtoTheMOON Dec 11 '17 at 17:02

Your way of finding the probability that she is late at least once is correct, but there's a simpler way: $$\Pr(\text{late at least once}) = 1 - \Pr(\text{on time every day}) = 1 - \left( 0.8 \right)^6.$$

• Thanks a lot for support. I have implemented this. – himanshu chawla Dec 11 '17 at 16:53
• I am getting a different answer for this. Can you verify your answer after calculation ? – himanshu chawla Dec 11 '17 at 16:59