# Probability that sheepdog performs at least one task successfuly - Am I doing this problem right?

What is the probability that a sheepdog performs at least $$1$$ of these tasks successfully?

My approach is to subtract the probability of performing at most $$1$$ of these tasks successfully from the probability of performing all $$4$$ tasks successfully.

$$P(\text{fetch})=.9, P(\text{drive})=.7, P(\text{herd})=.84, P(\text{separate})=.75$$.

The complement of these four probabilities is, $$.1,.3,.16,$$ and $$,.25$$, respectively.

So the probability that the sheepdog performs all four tasks successfully is simply, $$(.9)(.7)(.84)(.75)$$.

The probability that the sheepdog performs at most $$1$$ task successfully can be split into $$4$$ cases. Either the sheepdog performs the fetch task (and not the other 3) successfully, performs the drive task, performs the herd task, or performs the separate task.

This would look like:

$$(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)$$

Subtracting this from the case in which the sheepdog performs all four tasks would yield:

$$(.9)(.7)(.84)(.75)-[(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)]$$.

Is this correct?

• note: i would obviously simplify my answer at the end by multiplying all of the numbers out, i just want to see if this approach is correct... Sep 27, 2018 at 1:25
• So the probability that he will perform at least one correctly is 60% less than the probability of him doing just the first task correctly? Sep 27, 2018 at 13:00
• I'm so disappointed that this was not a question about a sheep/dog hybrid animal.
– pipe
Sep 27, 2018 at 16:37

The complement of at least $$1$$ is not at most $$1$$.

The complement is if none of the task is perform.

Hence just compute $$1-\prod_{i=1}^4 (1-p_i)$$

• woowwwww...i cant believe i missed that. thank you ! i got it now Sep 27, 2018 at 1:36

By sheepdog, I assume you mean 'good sheepdog' as defined in the question. Let $$X$$ denote the number of things in the list satisfied. If you want the probability the at least one of the 4things listed is satisfied $$P(X\geq 1)$$, you work it out by solving $$1-P(X=0)$$ This is equal to $$1-(0.1)(0.3)(0.16)(0.25) = 0.9988$$

Calculating the compliment of the chance of failing in all trials is undoubtedly the 'correct' approach to this problem, as per Siong Thye Goh's answer.

The same result can be calculated by accumulating the additional chance of success from each trial, given that the trial is needed, like this:

Chance this    Chance of success  Additional      Cumulative chance
trial needed   in this trial      success from    chance of success
this trial      in any trial
--------------------------------------------------------------------
A = 1.0 - D'     B (input data)     C = AB         D = C + D'
--------------------------------------------------------------------
0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1           0.9              0.9             0.9
0.1           0.7              0.07            0.97
0.03           0.84             0.0252          0.9952
0.0048           0.75             0.0036          0.9988