Monday

March 30, 2015

March 30, 2015

Posted by **carol** on Sunday, March 30, 2014 at 9:32am.

a. Exactly 7

b.All 11

c.At least 4

d. Exactly 2

e. Less than 3

f None

- math -
**bobpursley**, Sunday, March 30, 2014 at 10:04amI wonder how anyone can measure who is capable of college work? Exactly what is "college work" anyway? From my observation of all these online students enrolled in "colleges", being told they are doing "college work", and getting "degrees", one wonders.

I have seen a number of online colleges (want me to list them?) whose students are performing at less than high school level, yet getting "college credit" for it. An it amazes me how these "colleges" backscratch each other in giving each other "accrediation" to one another.

Enough, but it is a worthy question.

Here Pr(each)=.56, so if each kid makes independent decisions..., and the probability for a kid not going to college is .47

Pr(7)=.53^7*.47^4 * ways to do this, or

ways to do this (selecting 7 out of 11)

= 11!/(7!4!)

SO the final answer to your question..

Pr(7 of 11)=.53^7*.47^4*11!/(4!*7!)= nice math problem. Enter this into your google search window:

.53^7*.47^4*11!/(4!*7!)=

check my logic..

- math -
**bobpursley**, Sunday, March 30, 2014 at 10:05amI did the exactly 7, follow my steps to get the others. On the less than three, do zero first, then exactly1, and exactly 2, then add those to get less than three.

- math -
**Steve**, Sunday, March 30, 2014 at 12:54pmGood logic, but I wonder about the .47 in light of P(college) = .56

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