Two Probability Questions pls

[mooch]

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.

 

[arunbg]

What are your thoughts on solving the problem? You have to post your work if you want help.

 

[mooch]

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

 

[Rogerio]

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"

 

[arunbg]

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.

 

[mooch]

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?

 

[arunbg]

Looks good to me :)

 

[mooch]

cool thanks man

 

[mooch]

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 ?

 

[mooch]

any1 know if my method is correct

 

[tiny-tim]

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 … ?

 

[mooch]

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 … ?

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