Download Repeated Trials Homework Solutions 1. What is the probability of

Repeated Trials Homework Solutions
1. What is the probability of making exactly three 7’s in five tosses of a pair of dice?
Answer: Recall that if we denote pn (k) as the probability of winning exactly k times out of n games,
where the probability of winning a game is p and the probability of losing is q = 1 − p, then
pn (k) = n Ck pk q n−k .
1
In this example, n = 5 and k = 3. Recall that the probability p of making a 7 is p = and so the
6
5
probability of a failure q = 1 − p = .
6
So our answer is
5 C3
5
1
× ( )3 × ( )2 ≈ .0321502
6
6
2. What is the probability of making at most two 7’s in five tosses of a pair of dice?
Answer: We compute the probability of making zero 7’s, then of making one 7, and then of making
two 7’s. These are mutually exclusive so we take the sum:
p5 (0)+p5 (1)+p5 (2) = 5 C0 ×(1/6)0 ×(5/6)5 + 5 C1 ×(1/6)1 ×(5/6)4 + 5 C2 ×(1/6)2 ×(5/6)3 ≈ 0.964506
3. A fair coin is flipped 14 times, what is the probability of it coming up heads exactly 7 times?
Answer:
14 C7
× (1/2)14 = .209472656
4. A 6-sided die is rolled seven times. What is the probability that exactly three or exactly four rolls
come up 1?
Answer: The probability of a 1 coming up on a single throw is 1/6. Thus the probability that
exactly 3 ones come up in seven throws is 7 C3 (1/6)3 (5/6)4 , which is approximately .078, and the
probability of exactly 4 ones coming up is 7 C4 (1/6)4 (5/6)3 , which is approximately .015. Thus the
probability of exactly three or four ones is about .094 (the round off is what causes this to be .094
rather than .093).
5. What is the probability of passing a 10-question true-false exam by random guessing, if a passing
score is 70%?
Answer: You need to get 7,8,9, or 10 problems correct. These are mutually exclusive so we take
the sum of the probabilities:
p = p10 (7) + p10 (8) + p10 (9) + p10 (10)
= 10 C7 ×(1/2)7 ×(1/2)3 +10 C8 ×(1/2)8 ×(1/2)2 +10 C9 ×(1/2)9 ×(1/2)1 +10 C10 ×(1/2)10 ×(1/2)0 ≈ 0.171875
6. You toss a die twelve times. What is the expected number of times that you get a one? What is
the probability that you get a one that many times?
Answer: The probability of having a success (getting a one) is 1/6. So you expect (on the average)
that one out of six times you will get a one, hence (again, on the average) you expect to get a one
two times out of twelve tosses. So the answer is 2. If you wish to use previous material, recall that
if p is the probability of having a success, then 1/p is the expected number of trials until a success.
So you “expect” a success after 6 throws, so you expect two successes after twelve throws (on the
average.) So your expected number of ones is two. Notice that “expected value” is misleading
terminology. Some books use the terminology “most likely value” instead of “expected value.”
An alternate way to solve the problem, although tedious, would be to let E0 be the event that
you get zero ones, E1 the event that you get one one, E2 the event that you get two ones, . . . ,
E12 the event that you get 12 ones. Then v(Ei ) = i and p(Ei ) = 12 C i × (1/6)i × (5/6)12−i for
i = 0, 1, 2, . . . , 12. You would then sum the terms
12 C i
× (1/6)i × (5/6)12−i × i
for i = 0, 1, 2, . . . , 12. The sum is 2.
The second question is answered by computing the probability of getting a one exactly two times
in twelve tries.
Use the formula pn (k) = n Ck pk q n−k . In this example, n = 12, p = 1/6, and k = 2
p12 (2) = 12 C2 (1/6)2 × (5/6)10 ≈ .296094
Notice that the “expected value of two times” has a probability of approximately .296 of actually
occurring. In other words, you expect to see the “expected value” of 2 approximately three out
of every ten times (rounding .296094 . . . to .3) you perform the experiment of tossing a die twelve
times. If an event only occurs three out of ten times, in ordinary usage of the language, we would
not call that event “expected.” However, two times is the most likely number of times that you
would get two ones.