What Is the Probability of Tire Defects and Passing Rates for Students?

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Discussion Overview

The discussion revolves around two probability problems: the likelihood of defective tires in a batch and the odds of a student passing math while failing English. The first problem involves the application of the binomial distribution, while the second concerns conditional probabilities based on given passing rates.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Some participants identify the first problem as suitable for the binomial distribution, suggesting that calculating the probability of more than 4 defective tires involves summing probabilities for 0 to 4 defective tires.
  • One participant proposes a method using the complement rule to find the probability of more than 4 defective tires by calculating the probability of 0 to 4 being non-defective.
  • Another participant provides specific calculations for the probabilities of having 0 to 4 defective tires, leading to a proposed total probability of 0.9735532 for more than 4 defective tires.
  • In the second problem, participants discuss the passing rates for math and English, with one participant using a Venn diagram to visualize the relationships between students passing each subject.
  • Some participants calculate the percentage of students passing math but failing English, arriving at a probability of 0.1116 based on their assumptions about student enrollment in both subjects.
  • There is a mention of uncertainty regarding the assumption that all students take both subjects, which is not explicitly stated in the problem.

Areas of Agreement / Disagreement

Participants generally agree on the approach to the first problem but have not reached a consensus on the exact calculations. In the second problem, there are differing interpretations of the assumptions regarding student enrollment, leading to uncertainty in the calculations.

Contextual Notes

Participants express limitations in their assumptions, particularly regarding the enrollment of students in both subjects for the second problem, which affects the probability calculations.

mooch
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the question is.
5% of tires are defective. There are 200 tires. Probability more than 4 are defected.

2nd question is.
At a university, 90% of students pass math, 85% pass english, and 78% pass both. What are the odds in favor of a student passing math and failing english.
 
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What are your thoughts on solving the problem? You have to post your work if you want help.
 
for the first question i know we are using the binomial distribution and that if i did it the long way i would have to add up all the cases from 5-200. this would take to long.

What i have come up with is using

1- 200C5x(0.05^5)x(0.95^195) (1- 5 tires are defected)
= 0.9641
 
mooch said:
the question is.
5% of tires are defective. There are 200 tires. Probability more than 4 are defected.

What about

"Probability more than 4 are defected" = "Probability less then 5 are OK"

And

"Probability less then 5 are OK" = "Probability exactly 4 are OK" + "Probability exactly 3 are OK" + "Probability exactly 2 are OK" + "Probability exactly 1 is OK" + "Probability exactly 0 is OK"

:smile:
 
Your idea for the first question is correct, but you have only taken the probability for exactly five bulbs to be defective. You need to find probabilities for 4-0 bulbs defective and add them all up before subtracting from 1.
 
it should be then

200C0x(0.05^0)x(0.95^200)=0.000035053
200C1x(0.05^1)x(0.95^199)=0.000368975
200C2x(0.05^2)x(0.95^198)=0.001932266
200C3x(0.05^3)x(0.95^197)=0.006712082
200C4x(0.05^4)x(0.95^196)=0.017398424

add them up i get 0.0264468
1-0.0264468 = 0.9735532

correct?
 
Looks good to me :)
 
cool thanks man
 
mooch said:
2nd question is.
At a university, 90% of students pass math, 85% pass english, and 78% pass both. What are the odds in favor of a student passing math and failing english.

any1 know how to solve this question? What I've done so far is made a venn diagram consisting of math and english. I worked out that 12% (90-78) pass math, and 7%(85-78) pass english. Therefore 93% fail english.

.12 x .93 = 0.1116 ?
 
Last edited:
  • #10
any1 know if my method is correct
 
  • #11
mooch said:
any1 know how to solve this question? What I've done so far is made a venn diagram consisting of math and english. I worked out that 12% (90-78) pass math, and 7%(85-78) pass english. Therefore 93% fail english.

.12 x .93 = 0.1116 ?

hmm … assuming everybody takes both subjects (which the question doesn't say), out of 100 students, 90 pass maths and 78 of them also pass english - so 12 of them fail english.

So the probability is … ? :smile:
 
  • #12
tiny-tim said:
hmm … assuming everybody takes both subjects (which the question doesn't say), out of 100 students, 90 pass maths and 78 of them also pass english - so 12 of them fail english.

So the probability is … ? :smile:

yes it does 78% take both...78 students and the 85 pass english
 

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