Conditional probability distribution question issue

Question

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
@MichaelHardy I'm assuming because hes already asked this exact question once before — XRBtoTheMOON, Dec 11 '17 at 16:48
@btcgrl : It seems the poster improved the question to make it acceptable but posted separately rather than editing the original question. Responding to that with unexplained downvoting rather than by explaining to a newbie what the proper procedures are seems rather like a tantrum. — Michael Hardy, Dec 11 '17 at 17:02
@btcgrl my previous account was blocked and i can't comment or get any answer through that. You told me to give it try and then come back and I have done the same. — himanshu chawla, Dec 11 '17 at 17:02
@MichaelHardy Thanks a lot. You seem to be very nice and humble person. — himanshu chawla, Dec 11 '17 at 17:04
Just to be clear, I didn't downvote it, was merely stating the most likely reason — XRBtoTheMOON, Dec 11 '17 at 17:04
@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
@btcgrl posted a new question. Can you help ? link - https://math.stackexchange.com/questions/2562014/proabability-bayes-theorem-cancer-question — himanshu chawla, Dec 11 '17 at 18:08
@MichaelHardy posted a new question. Can you help ? link - https://math.stackexchange.com/questions/2562014/proabability-bayes-theorem-cancer-question# — himanshu chawla, Dec 11 '17 at 18:09

XRBtoTheMOON · Accepted Answer · 2017-12-11T17:01:21.733

0

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.

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

edited Dec 11 '17 at 17:01

answered Dec 11 '17 at 16:28

XRBtoTheMOON

1,031

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

score 0 · Answer 2 · answered Dec 11 '17 at 16:42

0

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

answered Dec 11 '17 at 16:42

Michael Hardy

1

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

Conditional probability distribution question issue

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