Download 6.2 The Binomial Distribution - Solutions 6.2.9 Coin Toss A fair coin

6.2 The Binomial Distribution - Solutions
6.2.9 Coin Toss
A fair coin is flipped three times. For each of the following, determine the probability of getting
A. The probability of getting exactly three heads.
Let X = the event of getting a head on the flip.
n=3
p = 0.5
q = 1 − p = 1 − 0.5 = 0.5
Ã
!
3
3
P (X = 3) =
(0.5)3 (1 − 0.5)3−3
3!
(0.5)3 (0.5)3−3
3!(3 − 3)!
= 0.125
=
B. The probability of getting exactly two heads.
Let X = the event of getting a head on the flip.
n=3
p = 0.5
q = 1 − p = 1 − 0.5 = 0.5
!
Ã
P (X = 2) =
3
2
(0.5)2 (1 − 0.5)3−2
3!
(0.5)2 (0.5)3−2
2!(3 − 2)!
= 0.375
=
1
2
C. The probability of getting exactly two or more heads.
Let X the event of getting a head on the flip.
n=3
p = 0.5
q = 1 − p = 1 − 0.5 = 0.5
P (X ≥ 2) = P (X = 2) + P (X = 3)
à !
à !
3
3
3
3−3
=
(0.5) (1 − 0.5)
+
(0.5)2 (1 − 0.5)3−2
3
2
3!
3!
(0.5)3 (0.5)3−3 +
(0.5)2 (0.5)3−2
3!(3 − 3)!
2!(3 − 2)!
= 0.125 + 0.375
= 0.5
=
D. Exactly three tails.
Let X = the event of getting a tail on the flip.
n=3
p = 0.5
q = 1 − p = 1 − 0.5 = 0.5
!
Ã
P (X = 3) =
3
3
(0.5)3 (1 − 0.5)3−3
3!
(0.5)3 (0.5)3−3
3!(3 − 3)!
= 0.125
=
E. The expected values and standard deviation for this distribution.
Expected Value
E(X) = µ = np = (3)(0.5) = 1.5
Standard Deviation
σ=
√
npq =
p
(3)(0.5)(0.5) =
√
0.75 = 0.866
3
6.2.10 Multiple Choice
Richard has just been given a 10 - question multiple choice quiz in his statistics
course. Each question has 5 answers, of which only 1 of them is correct. Since
Richard has not been attending classes, he does not know any of the answers. Suppose he guesses on all 10 questions
A. What is the probability that he will answer all of the questions correctly?
Let X = the number of questions Richard guesses correctly.
n = 10
p=
1
5
= 0.2
q = 1 − p = 1 − 0.2 = 0.8
Ã
P (X = 10) =
=
!
10
10
(0.2)10 (1 − 0.2)10−10
10!
(0.2)1 0(0.8)0
10!(10 − 10)!
= 1.024 × 10−7
= 0.0000001024
B. What is the probability that he will answer all of the questions incorrectly?
Ã
P (X = 0) =
!
10
0
(0.2)0 (1 − 0.2)10−0
10!
(0.2)0 (0.8)10
0!(10 − 0)!
= 0.107
=
C. What is the probability that he will answer at least one question correctly?
P (X ≥ 1) = P (X = 1) + P (X = 2) + (P X = 3) + P (X = 4)
+ P (X = 5) + P (X = 6) + (P X = 7) + P (X = 8) + P (X = 9)
+ P (X = 10)
= 1 − P (X = 0)
Ã
!
10
=1−
(0.2)0 (1 − 0.2)10−0
0
10!
(0.2)0 (0.8)10
0!(10 − 0)!
= 0.893
=1−
4
D. What is the probability that he will answer at least half of the questions correctly?
P (X ≥ 5) = P (X = 5) + P (X = 6) + (P X = 7) + P (X = 8) + P (X = 9) + P (X = 10)
Ã
!
10
5
=
Ã
!
Ã
+
Ã
7
10−7
9
10−9
(0.2) (1 − 0.2)
Ã
(0.2) (1 − 0.2)
+
(0.2)6 (1 − 0.2)10−6
!
10
8
+
!
10
9
!
10
6
(0.2)5 (1 − 0.2)10−5 +
10
7
+
Ã
(0.2)8 (1 − 0.2)10−8
!
10
10
(0.2)10 (1 − 0.2)10−10
=
10!
10!
10!
(0.2)5 (0.8)5 +
(0.2)6 (0.8)4 +
(0.2)7 (0.8)3
5!(10 − 5)!
6!(10 − 6)!
7!(10 − 7)!
+
10!
10!
10!
(0.2)8 (0.8)2 +
(0.2)9 (0.8)1 +
(0.2)1 0(0.8)0
8!(10 − 8)!
9!(10 − 9)!
10!(10 − 10)!
= 0.026 + 0.006 + 7.86 × 10−4 + 7.37 × 10−5 + 4.096 × 10−6 + 1.024 × 10−7
= 0.033
6.9.12 Ethics
The phenomena known at the one- time fling is where people purchase a clothing
item (pants, hats, shirts, etc) wear it once, and then return the purchase. About
10% of all adults reported that they feel no guilt when deliberately doing a one-time
fling. In a group of 7 people, determine the probability that
A. No one has done a one-time fling.
Let X = be the number of people who engage in this type of shopping behavior.
n=7
p = 0.1
q = 1 − p = 1 − 0.1 = 0.9
5
Ã
!
7
0
P (X = 0) =
(0.1)0 (1 − 0.1)7−0
7!
(0.1)0 (0.9)7
0!(7 − 0)!
= 0.478
=
B. At least one person has done a one-time fling.
P (X ≥ 1) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5)
+ P (X = 6) + P (X = 7)
= 1 − P (X = 0)
à !
7
=1−
(0.1)0 (1 − 0.1)7−0
0
7!
(0.1)0 (0.9)7
0!(7 − 0)!
= 1 − 0.478
= 0.522
=
C. No more than two people have done a one-time fling.
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
à !
à !
7
7
=
(0.1)0 (1 − 0.1)7−0 +
(0.1)1 (1 − 0.1)7−1
0
1
à !
7
(0.1)2 (1 − 0.1)7−2
+
2
7!
7!
(0.1)0 (0.9)7 +
(0.1)1 (0.9)6
0!(7 − 0)!
1!(7 − 1)!
7!
+
(0.1)2 (0.9)5
2!(7 − 2)!
= 0.478 + 0.372 + 0.124
= 0.974
=
D. What is the expected value and standard deviation for this distribution?
Expected Value: E(X) = µ = np = (7)(0.1) = 0.7
Standard Deviation: σ =
√
npq =
p
7(0.1)(0.9) = 0.794