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
R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having a
college education is important to succeed in the business world. If a random sample of five Americans
is s selected, find these probabilities.
(a) Exactly two people will agree with that statement.
(b) At most three people will agree with that statement.
(c) At least two people will agree with that statement.
(d) Fewer than three people agree with that statement.
3 answers
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
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?
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?