Posted by FAKAAPO on Wednesday, September 21, 2011 at 7:15am.
Calculate the probabilities that there will be 0,1,2,3,4 and 5 persons that agree.
P(5) = (0.4)^5*[5!/(0!*5!]= 0.4^5
= 0.01024
P(0) = (0.6)^5*[5!/(0!*5!]= (0.6)^5
= 0.07776
P(1) = (0.4)*(0.6)^4*[5!/(4!*1!)]
= 0.25920
P(2) = (0.4)^2*(0.6)^3*[5!/(3!*2!)]
= 0.34560
etc. Calculate P(3) and P(4) using the same formula, and use the results to answer the questions for (a) through (d).
For (a), the answer is P(2) = 0.3456
For (d), the answer is P(0) + P(1) + P(2) = 0.68256
I= important
N=not important
P(i) = 60/100 = 3/5
P(N) = 2/5
a) P(2out of5 for I) = C(5,2)(3/5)^2 (2/5)^3 = 10(9/25)(8/125) = 144/625 or .2304
..
d) fewer than three agree --> 0outof5 or 1outof5 or 2outof5 will agree
= C(5,0)(3/5)^0 (2/5)^5 + C(5,1)(3/5)^1 (2/5)^4 + C(5,2)(3/5)^2 (2/5)^3
= 1(1)(32/3125 + 5(3/5)(16/625) + 10(9/25)(8/125)
= 32/3125 + 48/625 + 144/625
= 992/3125 or .31744
do b) and c) the same way
there are 4 dimes, 4 nickels,
and 2 quarters.
In how many possible ways can the selection be made so that
the value of the coins is at least 25 cents?
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