Question 1162700: The fine print on an instant lottery ticket claims that one in nine tickets win a prize. B,what is the probability that you win at least twice if you purchase ten tickets. C,what is the proximate probability that you win more that 120times if you purchase 900 tickets. D,of all awarded prizes, 10% are worth Gh1000,and 70% are worth $10.find the expected winning if you purchase a single tickets.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! 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
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