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 ?
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?
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
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.