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Math 360: Introduction to Probability and Statistics, Spring
2004, Section 01, CRV 71243
Homework 01 Key for Thursday, 27 Jan 2005, 3pm PST. Please see the accompanying Excel
workbook for my calculations.
- The table below gives a probability mass function for a random variable X.
x 0 1 2 3 4 5 p(x) 0.35 0.1 0.15 0.3 0.05 0.
Verify that the table does in fact define a probability mass function. Find the mean. Find the variance. Find the standard deviation. Sketch a graph of this probability mass function.
Find P ( X = 2 ).
Find P ( X > 2 ).
Find P ( X ≤ 2 ).
Part a. 0.35+0.10+0.15+0.30+0.05+0.05 = 1. All p(x)’s are between 0 and 1. Yep, it’s a pmf.
Part b. ∑
=
5
0
x
μ xp x
Part c.
2 2 2
2 2 2
5
0
2 2
x =
σ x px
Part d. 2. 3875 1. 5452
2
Part e:
Problemn 1 pmf
0
0 1 2 3 4 5
x values
p(x) values
Problemn 1 pmf
Part f. P ( X = 2 )= p ( 2 )= 0. 15
Part g. P ( X > 2 )= P ( X = 3 or X = 4 or X = 5 )= p ( 3 )+ p ( 4 )+ p ( 5 )= 0. 40
Part h. P ( X ≤ 2 )= 1 − P ( X > 2 )= 0. 60
( 2 ) ( 0 ) ( 1 ) ( 2 ) 0.
OR you can
sum : P X ≤ = p + p + p = 60
- A jar contains 6 fair coins and 2 biased coins. The two biased coins have a probability of
heads of 0.90. We reach into the jar, pull out a coin, and flip it six times. Find the probability of flipping at least four Hs.
The experiment involves two basic steps: drawing the coin out of the jar, and then flipping
the drawn coin six times. Were you to detail the sample space, it would have to contain
outcomes of the form Here C denotes the coin you drew (there
are eight different C’s two of which are biased, and six of which are fair). The F’s denote the
flip outcomes. For each F, the value is H or T for heads or tails. You Computer Scientists
may wish to use 0 and 1 in place of H and T. Rather than write out all the possible outcomes
and count the number that have at least 4 H’s in the six flips, we’re going to you conditional
probabilities.
( C , F 1 , F 2 , F 3 , F 4 , F 5 , F 6 ).
Let X denote the number of H’s in 6 flips. Let Y be =0 if the coin is biased, and Y=1 if the
coin is fair.
( 4 and 0) ( 4 and 1 )
( 4 ) ( 4 and 0 )or( 4 and 1 )
PX Y PY P X Y P Y
PX Y P X Y
P X P X Y X Y
In the first line of the computation, we are looking at the outcomes that produce at least 4
H’s, and we are grouping them together into two events, that the coin is biased, or that the
coin is fair. Noting that the coin cannot be both fair and biased at the same time, the
probability of the “or” becomes the sum of the probabilities, leading to the second line. The
third line applies the definition of conditional probability. At this point, we may use exactly
what we know in the problem statement. 0. 75 8
P ( Y = 0 )= = PY = = = are
the probabilities associated with the coin draw. Given that the coin is fair, the probabilities
for X are determined with the binomial, using n=6 and p=0.5. Given that the coin is biased,
the probabilities for X are determined with the binomial, using n=6 and p=0.9.
6
6
4
6
6
4
= + =
−
=
−
=
x x
x
x x
x x x
P X
The above number and the associated computations constitute the solution of this problem.
However, some discussion is in order. There are two important principles in play here:
P ( A )= P ( A and B )+ P ( A and"not B ")for any events A and B. Here “not
B ” means all outcomes that are not in the event B.
P ( A and B )= P ( A | B ) P ( B )for any events A and B.
We have one last discussion point, before the next problem. It is important to understand the
distinction between. Here the probability that was “easy” to
compute is for y=0,1. The important distinction is that
P ( A and B )and P ( A | B )
P ( X ≥ 4 | Y =y ) given the
knowledge of the coin type (fair of biased), the binomial can be used to predict the
probabilities of the flips. Think about dealing two cards from a deck. What’s the probability
of dealing an Ace and then a second Ace? What’s the probability of dealing an Ace for the
second card given that you dealt an Ace on the first card?
- The Poisson pmf is used to model random events occurring in sequence for a fixed duration. For example, when we are monitoring the number of arrivals into a line at the bank, we would use the Poisson to predict probabilities for number of arrivals. If the average arrival rate is 25 customers per hour, find the probability that we receive exactly 5 customers in a 12 minute period. Find the probability that we receive exactly 3 customers in a 12-minute period. Find the probability that we receive exactly 3 customers in a 10-minute period. Use the formulas on the handout.
Part a. Exactly 5 customers in a 12-minute period. Here λ = 25 *( 12 / 60 )= 5. The Poisson
formula is 0. 175467 5!
5 = =
λ (^) −λ p e
Part b. Exactly 3 in a 12-minute period. Again λ = 25 *( 12 / 60 )= 5 , and the Poisson
computation is 0. 140374 3!
3 = =
λ (^) −λ p e
Part c. Exactly 3 in a 10-minute period. Here λ = 25 *( 10 / 60 )= 25 / 6 ≈ 4. 166667 , and the
Poisson computation is 0. 18692 3!
3 = =
λ (^) −λ p e
- The Geometric probability mass function is used to model the number of trials required to conduct until the first success occurs. For example, a certain product is effective 95% of the times it is used. Use the geometric pmf to determine the probability that the first failure occurs on the 10
th try. Find the probability that the first failure occurs on or before the 10
th try.
The probability of success is p=0.95.
Part a. ( 10 ) ( 0. 95 ) ( 0. 05 ) 0. 031512
9 p = =
Part b. On or before 10 means 1 or 2 or 3 or … or 9 or 10. Add up the geometric probs.
∑^ (^ )
=
−
10
1
1
- 95 ( 0. 05 ) 0. 401263 x
x
