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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

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  • $\begingroup$ 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 $\endgroup$ – XRBtoTheMOON Dec 11 '17 at 16:36
  • $\begingroup$ Why this was down-voted seems anything but clear. $\endgroup$ – Michael Hardy Dec 11 '17 at 16:43
  • $\begingroup$ @MichaelHardy Thanks $\endgroup$ – himanshu chawla Dec 11 '17 at 16:47
  • $\begingroup$ @MichaelHardy I'm assuming because hes already asked this exact question once before $\endgroup$ – XRBtoTheMOON Dec 11 '17 at 16:48
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    $\begingroup$ @btcgrl No issues. You seem to be very active too. Thanks for helping and correcting me out always. Thanks $\endgroup$ – himanshu chawla Dec 11 '17 at 17:05
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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.

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

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  • $\begingroup$ your calculation for second part is correct ? I am getting a different answer for this. can you verify this ? $\endgroup$ – himanshu chawla Dec 11 '17 at 16:57
  • $\begingroup$ @himanshuchawla its just $1 - .8^6$ which is $1 - .262 = .738$. Edited answer to make it a little more clear $\endgroup$ – XRBtoTheMOON Dec 11 '17 at 16:59
  • $\begingroup$ @himanshuchawla np, cheers $\endgroup$ – XRBtoTheMOON Dec 11 '17 at 17:02
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Your way of finding the probability that she is late at least once is correct, but there's a simpler way: $$ \Pr(\text{late at least once}) = 1 - \Pr(\text{on time every day}) = 1 - \left( 0.8 \right)^6. $$

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

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