Year 12 Methods - Part 2?

1. The probability of winning a single game of chance is 0.15, and whether or not the game is won is independent


of any other game. Suppose Jodie plays a sequence of n games.


If the probability of Jodie winning at least one game is more than 0.95, then the smallest value n can take is


closest to


A. 19


B. 15


C. 8


D. 7


E. 4





2. The number, X, of children in a family is a random variable with the following probability distribution.


x 0 1 2 3


Pr(X = x) 0.4 0.3 0.2 0.1


If two families are selected at random, the probability that they have the same number of children is


A. 0.10


B. 0.20


C. 0.30


D. 0.40


E. 0.50





3.Consider the function f : R → R, f (x) = (x − 1)2(x − 2) + 1.


Find the real values of h for which only one of the solutions of the equation f (x + h) = 1 is positive.

Year 12 Methods - Part 2?
Hi,





1) If the probability of winning at least one game is more than 95%, then the probability of of winning no games would have to be less than 5%. Find these probabilities on a graphing calculator by entering the following equations as an equation under Y=:





If n = 4:


Y1 = 4 nCr X (.15)^X*(.85)^(4 - X)


If you look at the table when X = 0 to 4, it will show the probabilities of winning from 0 to 4 games. Notice that the probability of winning no games is about 52%, so clearly the probability of winning at least one game is far less than 95%.





If n = 7:


Y1 = 7 nCr X (.15)^X*(.85)^(7 - X)


If you look at the table when X = 0 to 7, it will show the probabilities of winning from 0 to 7 games. Notice that the probability of winning no games is about 32%, so clearly the probability of winning at least one game is far less than 95%.





If n = 8:


Y1 = 8 nCr X (.15)^X*(.85)^(8 - X)


If you look at the table when X = 0 to 8, it will show the probabilities of winning from 0 to 8 games. Notice that the probability of winning no games is about 27%, so clearly the probability of winning at least one game is far less than 95%.





If n = 15:


Y1 = 15 nCr X (.15)^X*(.85)^(15 - X)


If you look at the table when X = 0 to 15, it will show the probabilities of winning from 0 to 15 games. Notice that the probability of winning no games is about 9%, so clearly the probability of winning at least one game is less than 95%.





If n = 19:


Y1 = 19 nCr X (.15)^X*(.85)^(19 - X)


If you look at the table when X = 0 to 19, it will show the probabilities of winning from 0 to 19 games. Notice that the probability of winning no games is about 4.56%, so the probability of winning at least one game is about 95.44%





The correct answer is A.





2) The probability that both families will have:


no kids is (.4)(.4) = .16.


1 kid is (.3)(.3) = .09.


2 kids is (.2)(.2) = .04.


3 kids is (.1)(.1) = .01.


The total is .16 + .09 + .04 + .01 = .30


The answer is C.





3) For f(x) = (x-1)²(x-2) + 1, if you want to find where f(x + h) = 1 then look for where (x-1)²(x-2) + 1 = 1. This means to find where does (x-1)²(x-2) = 0. It clearly equals zero when x = 1 or x = 2, so that means you are looking for where x + h = 1 and x + h = 2 so that only one of the solutions of the equation is positive. That means you want to shift the x intercepts to the left so that only one answer is positive. That means that x could be a minimum value of 1 up to, but not including 2. So 1 ≤ h %26lt; 2.





I hope that helps!! :-)
Reply:q1


1-P(lost all) %26gt;0.95


P(lost all) %26lt;0.05


(1-0.15)^n %26lt;0.05


0.85^n %26lt;0.05


take log on both sides


n %26lt; log 0.05/log 0.85


n = 19





answer=A





q2


P=(0.4^2+0.3^2+0.2^2+0.1^2)/1


=0.16+0.09+0.04+0.01


=0.30





answer=C





q3


sorry...