ECSE-2500 Engineering Probability HW#5 Solutions
Due 9/15/14
1. (8 points) Our chip manufacturing facility ships chips in boxes of 100 chips. Assume that
defective chips are independent from one chip to the next and that on average one chip in 100 is
defective.
1.a. (4 points) What is the probability that a random box of 100 chips contains only zero or one
defective chips?
Solution
Defective chips in one box follow a Binomial Law with n  100 and p  0.01. So we have
100
99
100
100
P 0 or 1 errors   0  1  p    1  1  p  p




 1  p 
100
 100 1  p  p
99
 1  p  1  p   100 p 
99
  0.99   0.99   1
99
 0.3697 1.99  0.7357
1.b. (4 points) We have a very picky customer who checks every box. If any one box has more
than one bad chip, he sends back the whole order. What is the probability that he sends back
an order of n boxes?
Solution
In order for him not to send back an order, each box must contain 0 or 1 defective chips. The
probability of this happening is the answer to part a raised to the nth power, that is
pn   0.7357
n
for n  1,2,3,
So the probability he sends back an order of size n is
1  pn  1   0.7357
n
for n  1,2,3,
Page 1 of 2
ECSE-2500 Engineering Probability HW#5 Solutions
Due 9/15/14
2. (4 points) For a sequence of Bernoulli Trials, let k  #heads in n tosses. Let p  0.4. Use
Bernoulli’s Theorem to compute an upper bound on P  nk  p  0.01 for n  104 and n  106.
Solution
Applying P  nk  p     npq 2 to this case, we get
k
0.6
P  10,000
 0.4  0.1  100.4
2  0.24
4
0.01
k
0.6
P  1,000,000
 0.4  0.1  100.4
2  0.0024
6
0.01
3. (6 points) In some sports, two teams play a series of 5 games or matches. The winner of the
series is the first team to win 3 games. Assume the probability that Team A wins a game is p.
Derive a formula for the probability that Team A wins such a series? Note that if say Team A
wins the first 3 games, we don’t bother to play Games 4 or 5. Evaluate your formula assuming
that p  0.6.
Solution
If we did play all 5 games, then once Team A won 3 games, we would not care about the
outcome of any remaining games, so we can think of this as 5 Bernoulli Trials where we have
3 or 4 or 5 successes and the probability of success is p. Thus the answer is
5
5
5
P 3 or 4 or 5 successes   3  p3q2   4  p4 q1   5  p5q 0
 
 
 
 10 p3q2  5 p4 q1  p5q 0 .
For p  0.6, we have
P 3 or 4 or 5 successes  10  0.6   0.4   5  0.6   0.4    0.6 
3
2
4
5
 10  0.216  0.16   5  0.1296   0.4    0.07776 
 0.68256.
Page 2 of 2