Question 1162700
Given:
p = 1/9 = known and constant probability of success (winning)
N = number of trials
x = number of successes
Binomial distribution applies, with
P(x,N,p) = C(N,x)*p^x*(1-p)^(N-x)


Solution:


B. Win at least twice with purchase of 10 tickets
P(>=2, 10, p) 
= P(2,10,p)+P(3,10,p)+...P(10,10,p)
= 0.21651+0.072175+0.015788+0.00024667+0.0000176+0.00000+...
= 0.30712 (to 5 decmals)
Alternatively, and more advantageously,
P(>=2, 10, p) 
= 1 - P(<2, 10, p) 
= 1 - (P(0,10,p) + P(1,10,p))
= 1 - (0.30795+0.38493)
= 0.30712 (to 5 decimals as before)


C. Win at least 120 times with purchase of 900 tickets
Repeating the exercise as in part B, we find that
P(>=120,900,p)
=P(<120,900,p)
=1 - ( P(0,900,p)+....+P(119,900,p) )
=1 - 0.97868
=0.02132


However, above calculations are very long unless a software is used.  The practical manual procedure is to apply the normal approximation.
We calculate
N = 900
p = 1/9
mean, mu = n*p = 100
standard deviation, sigma = sqrt(np(1-p)) = 9.42809
Z=(x-mu)/sigma = (119.5-100)/9.4289 = 2.06811
Looking up the normal probability table for Z = 2.06811,
N(2.06811,0,1) = 0.980685 (probability of 0-119 successes)
P(>=120,900,1/9) = 1-0.980685 = 0.01931
(approximation is within about 10% of theoretical answer)


D. Expected winning if 10% are wirth $1000, 70% are worth $10.
(assuming the remaining 20% are worth nothing)
E(winning) = 0.1*1000 + 0.7*10 = $107