Binomial distribution - failed tires

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steller

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In testing a certain kind of truck tire over a rugged terrain, it is found that 25 % of the trucks fail to complete the test run without a blowout. Of the next 15 trucks tested, find the probability that

a. none will fail the test?
b.at least 2 will fail the test?
c. more than 2 will fail the test?

Here is my attempt:

For all problems

\(\displaystyle n= 15 \)
\(\displaystyle p= .25 \)

For problem a

a. \(\displaystyle P(X = 0) \)

\(\displaystyle P(X = 0) = b(0; 15, .25) \)

I looked this up in the appendix in my book and got an answer. I think i did this correctly!

For problem b

\(\displaystyle P(X \ge 2) = 1 - P(x\le 2) \)

I am not sure if r= 1?

\(\displaystyle 1 - \sum\limits_{x=0}^1 b(1; 15, .25) \)


For problem c

\(\displaystyle P(X \ge 2) = 1 - P(x\le 2) \)

I am not sure if r= 2?

\(\displaystyle 1 - \sum\limits_{x=0}^2 b(2; 15, .25) \)

Are these set up correctly? I understand i still need to look them up in the appendix table in my book.
 
steller said:
In testing a certain kind of truck tire over a rugged terrain, it is found that 25 % of the trucks fail to complete the test run without a blowout. Of the next 15 trucks tested, find the probability that

a. none will fail the test?
b.at least 2 will fail the test?
c. more than 2 will fail the test?

Here is my attempt:

For all problems

\(\displaystyle n= 15 \)
\(\displaystyle p= .25 \)

For problem a

a. \(\displaystyle P(X = 0) \) This is correctly formulated.

\(\displaystyle P(X = 0) = b(0; 15, .25) \)

I have no idea what this notation means, but probably others will. You can do this without tables using a hand calculator

\(\displaystyle P(X = 0) = \dbinom{15}{0} * (0.25)^{0} * (1 - 0.25)^{15} = \dfrac{15}{15! * (15 - 15)!} * 0.25^{0} * (0.75)^{15}.\)

I looked this up in the appendix in my book and got an answer. I think i did this correctly!

For problem b

\(\displaystyle P(X \ge 2) = 1 - P(x\le 2) \)

The right hand side of your equation is wrong. \(\displaystyle P(X < 2) + P(X \ge 2) = 1 \implies P(X \ge 2) = 1 - P(X < 2).\)

I am not sure if r= 1?

\(\displaystyle 1 - \sum\limits_{x=0}^1 b(1; 15, .25) \)


For problem c

\(\displaystyle P(X \ge 2) = 1 - P(x\le 2) \)

The left hand side of your equation is wrong: what you want is P(X > 2). What about the right hand side?

I am not sure if r= 2?

\(\displaystyle 1 - \sum\limits_{x=0}^2 b(2; 15, .25) \)

Are these set up correctly? I understand i still need to look them up in the appendix table in my book.
Click to expand...
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I was aware of the "less than or equal to" sign
I just didnt know how to do it in latex.

So it appears that aside from "that" minor error everything is correct.
I am still not sure if the r values are correct but you failing to mention them tells me they are.

Thank you
 
steller said:
I was aware of the "less than or equal to" sign

I am still not sure if the r values are correct but you failing to mention them tells me they are.

Thank you
Click to expand...
You seemed perfectly capable to use those symbols in your original post.

As for your notation, I explained that I am not familiar with it. I have no idea whether it is correct or not.
 
steller said:
I was aware of the "less than or equal to" sign
I just didnt know how to do it in latex.

So it appears that aside from "that" minor error everything is correct.
I am still not sure if the r values are correct but you failing to mention them tells me they are.

Thank you
Click to expand...
I would rather compute binary probabilities from the formula, rather than looking them up.

\(\displaystyle b(m; n, p) = p^m\ (1-p)^{n-m}\ _n\mathrm C_m \)

I suspect your \(\displaystyle r\) is what I usually \(\displaystyle m\) and which is \(\displaystyle x\) in your probability statements.

JeffM said:
\(\displaystyle P(x < 2) + P(x \ge 2) = 1 \implies P(x \ge 2) = 1 - P(x < 2)\).
Click to expand...
The summation over all \(\displaystyle x<2\) has exactly two terms, and you have already found one of those:


\(\displaystyle 1 - \sum\limits_{x=0}^1 b(x; 15, .25) = 1 - \left[b(0; 15, .25) + b(1; 15, .25)\right] = \cdot \cdot \cdot\)
 
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steller said:
In testing a certain kind of truck tire over a rugged terrain, it is found that 25 % of the trucks fail to complete the test run without a blowout. Of the next 15 trucks tested, find the probability that

a. none will fail the test?
Click to expand...
For this first one, at least, you should not have needed to "look up" anything. Since 25% fail the test, 75% do not fail so the probability for any one to pass is 0.75. The probability that all 15 pass is \(\displaystyle .75^{15}\).
 
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