Probability of identifying both defective fuses in four or less tests

[boosewell]

This question is driving me crazy.

According to the textbook, the answer is 7/15, but I get 2/5. If anyone can tell me where I am going wrong I would be much obliged

Here is the question

Six fuses, of which two are defective and four are good, are to be tested one after another in random order until both defective fuses are identified.
Find the probability that the number of fuses that will be tested is four or less.

And this is my answer

Assuming two fuses need to be tested, the combination should be

Bad – bad = 2/6 x 1/5 = 2/30 = 1/15
Assuming three fuses need to be tested, the combinations should be
Bad – good – bad = 2/6 x 4/5 x 1 /4 = 8/120 =2/30 = 1/15
Good - bad – bad = 4/6 x 2/5 x 1 /4 = 8/120 =2/30 = 1/15

Assuming four fuses need to be tested, the combinations should be
Bad – good -good – bad = 2/6 x 4/5 x 3 /4 x 1/3 =24/360 = 1/15
Good - bad -good – bad = 4/6 x2/5 x 3 /4 x 1/3 =24/360 = 1/15
Good - good -bad – bad = 4/6 x3/5 x2 /4 x 1/3 =24/360 = 1/15

1/15 + 1/15 + 1/15 + 1/15 + 1/15 + 1/15 = 6/15

And I know that it is wrong because the textbook says so, but I can't think why.

Any help would be greatly appreciated.

 

[Jameson]

Re: Discrete ramdom variables

Hi boosewell,

Welcome to MHB! Glad you found us! Please let me know if you have any questions.

Your calculations look good, but maybe incomplete. What about the case of testing 3 fuses? :)

 

[MarkFL]

Re: Discrete ramdom variables

We could take another approach, and use the complement's rule. This way we need only consider two case: that it takes 5 tests or it takes 6 tests.

In order for it to take 6 tests, we would need to have 2 groups of fuses...the first group would consist of 5 fuses with one bad among them, and the second group would contain 1 bad fuse. How many ways can we arrange the fuses like this? We'll call this $N_6$.

In order for it to take 5 tests, we would need to have 3 groups of fuses...the first group would have 3 good and 1 bad, the second group would have 1 bad, and the third group would have 1 good. How many ways can we arrange the fuses like this? We'll call this $N_5$.

Then we consider how many total ways there are to arrange the 6 fuses. We'll call this $N$.

Then we can use the complement's rule...we know it is certain it will either take 4 or fewer tests or it will take more than 4 tests...and so we may state:

\(\displaystyle P(X\le4)+\frac{N_5+N_6}{N}=1\)

Hence:

\(\displaystyle P(X\le4)=1-\frac{N_5+N_6}{N}\)

Correctly calculating $N_5,\,N_6,\,N$ will yield:

\(\displaystyle P(X\le4)=\frac{7}{15}\)

I wanted to leave calculating those values for you to try first. :)

 

[boosewell]

Re: Discrete ramdom variables

Bingo !

Thanks MarkFL

and, because my earlier thank you post appears to have gone astray, thanks as well to Jameson.

I think that I am going to enjoy this site.

 

[Jameson]

Re: Discrete ramdom variables

Doh. Well my reading comprehension could use some work because you wrote about the 3 fuse case in your post. I missed it in between the lines breaks. MarkFL's way definitely works but so should your way. Let me see where the error is. You definitely have the right idea, so it's something small...

 

[Jameson]

Thanks to the help of I like Serena, here is what we missed.

There is also the 4 case of GGGG. If you identify all 4 good fuses, then the remaining 2 must be bad. The probability of this is $\frac{4}{6}\frac{3}{5}\frac{2}{4}\frac{1}{3}=\frac{1}{15}$. Now the answer adds up to 7/15 as the book indicates. :)