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Probability Distribution Problems

Probability Distribution Problems
  • Page 1
 1.  
There are 5 defective items in a batch of 25 items. 4 are chosen at random. Find the probability distribution of the defective items.

a.
II
b.
IV
c.
I
d.
III


Solution:

Probability of choosing a defective items, p = 5 / 25= 1 / 5

q = 1 - p = 4 / 5

Probability Distribution of defective items = 4Cr (15)r · (45)4-r where r = 0, 1, 2, 3, 4.


Correct answer : (3)
 2.  
Find the mean, variance and standard deviation for the number of heads, when 20 coins are tossed.
a.
10, 5, 5
b.
5, 52, 5 2
c.
10, 5, 5
d.
5, 5 2, 52


Solution:

This is a binomial situation, where n = 20, p = 1 / 2 and q = 1 / 2.

Mean = μ = n · p = 20 × 1 / 2 = 10

Variance = n · p · q = 20 · (1 / 2) · (1 / 2) = 5

Standard deviation = npq = 5

So, the mean is 10, variance is 5, and standard deviation is 5.


Correct answer : (1)
 3.  
If the probability of a defective bolt is 0.05, then find the mean and the standard deviation for the distribution of defective bolts in a total of 400.
a.
19, 20
b.
20, 19
c.
19, 20
d.
20, 20


Solution:

This is a binomial situation where n = 400, p = 0.05 = 1 / 20 and q = 19 / 20.

Mean = μ = n · p = 400 × 1 / 20 = 20

Variance = npq = 400 × (1 / 20) × (19 / 20) = 19

Standard deviation = npq = 19

So, mean is 20, and standard deviation is 19.


Correct answer : (2)
 4.  
The probability of a shooter hitting the target is 1 3. If he tries 9 times, then what is the probability of his hitting the target at least two times?
a.
11(2839)
b.
1- 11(2839)
c.
11(2938)
d.
1- 11(2938)


Solution:

This is a binomial situation where p = probability of hitting the target = 1 / 3, q = 1 - p = 2 / 3

Probability of hitting the target at least two times = P (r ≥ 2) = 1 - P (r < 2) = 1 - [P(r = 0) + P(r = 1)]

P(r = 0) = 9C0 · (2 / 3)9 = (2 / 3)9

P(r = 1) = 9C1 · (1 / 3) · (2 / 3)8 = 9 × 2839

P (r = 0) + P(r = 1) = 11(2839)

Required probability = 1 - 11(2839)

The probability of the shooter hitting the target at least two times is 1 - 11(2839) .


Correct answer : (2)
 5.  
Three defective pencils are mixed with 7 good ones. Two pencils are drawn at random simultaneously. Find the mean and standard deviation of the probability distribution of defective pencils drawn.
a.
0.21, 1.212
b.
0.6, 0.611
c.
0.6, 0.422
d.
0.42, 0.422


Solution:

P(X = 0) = number of ways of selecting 2 out of 7 good pencilsnumber of ways of selecting 2 out of 10 pencils= 7C210C2 = 7 / 15
[Find the probability of drawing 0 defective pencils.]

P(X = 1) = number of ways of selecting 1 out of 7 good pencils × number of ways of selecting 1 out of 3 defective pencilsnumber of ways of selecting 2 out of 10 pencils = 7C1×3C110C2 = 7 / 15
[Find the probability of drawing 1 defective pencil and 1 good pencil.]

P(X = 2) = number of ways of selecting 2 out of 3 defective pencilsnumber of ways of selecting 2 out of 10 pencils= 3C210C2 = 1 / 15
[Find the probability of drawing 2 defective pencils.]

The probability distribution for the defective pencils drawn is:
X012
P(X)715715115


The mean = μ = X · P(X) = 0 + 7 / 15 + 2 / 15 = 3 / 5 = 0.6

The variance = σ2 = [X2 · P(X)] - μ2
[Formula for the variance.]

σ2 = (0 + 7 / 15+ 4 / 15) - (3 / 5)2
[Simplify.]

σ2 = 11 / 15 - 9 / 25 = 28 / 75 = 0.3733
[Simplify.]

Standard deviation σ = 0.3733 = 0.611
[Simplify.]

So, the mean is 0.6, and the standard deviation is 0.611.


Correct answer : (2)
 6.  
6 dice are thrown 729 times. How many times do you expect at least 3 dice to show a 4 or 5?
a.
323
b.
322
c.
233
d.
232


Solution:

Let the probability of occurrence of the event exactly in 3 dice be referred to as P (E, 3).

Let the event be, getting 4 or 5 when a die is thrown.

Probability of occurrence of the event in a single throw is p = 1 / 6 + 1 / 6= 1 / 3

q = 1 - p = 1 - 1 / 3= 2 / 3

Probability of occurrence of the event in at least 3 dice = P (E, 3) + P (E, 4) + P (E, 5) + P (E, 6)

This is a binomial situation and hence using the formula P (E, r) = nCr · pr · qn - r

The required probability = (6C3)(1 / 3)3 (2 / 3)3 + (6C4)(1 / 3)4(2 / 3)2 + (6C5)(1 / 3)5(2 / 3) + (6C6)(1 / 3)6

Required probability = (1 / 729) [(20 × 8) + (15 × 4) + (6 × 2) + 1] = 233 / 729
[Simplify.]

Out of 729 throws, the number of times atleast 3 dice will show a 4 or 5 is 729 × 233 / 729 = 233.

So, we can expect at least 3 dice to show a 4 or 5 for 233 times.


Correct answer : (3)
 7.  
n = 100, p = 0.25. Find the mean and variance when the conditions for the binomial distribution are met.
a.
75, 18.75
b.
25, 12.25
c.
75, 12.25
d.
25, 18.75


Solution:

Given n = 100, p = 0.25

q = 1 - p = 1 - 0.25 = 0.75

Mean μ = n p


Variance s2 = n p q


μ = 100 × 0.25 = 25
[Substitute n = 100, p = 0.25 .]

s2 = 100 × 0.25 × 0.75
[Substitute n = 100, p = 0.25, q = 0.75 .]

s2 = 18.75
[Simplify.]

So, the mean is 25, and variance is 18.75 .


Correct answer : (4)
 8.  
n = 300, p = 0.55. Find the mean and standard deviation when the conditions for the binomial distribution are met.
a.
165, 8.62
b.
74.25, 8.62
c.
165, 74.25
d.
165, 12.85


Solution:

Given n = 300, p = 0.55

q = 1 - p = 1 - 0.55 = 0.45

Mean μ = n p


μ = 300 × 0.55 = 165
[Substitute n = 300, p = 0.55 .]

Standard deviation s = n p q

s = 300× 0.55 × 0.45
[Substitute n = 300, p = 0.55, q = 0.45 .]

s = 74.25 = 8.62
[Simplify.]

So, the mean is 165, and standard deviation is 8.62.


Correct answer : (1)
 9.  
30% of the people in a community went out on a vacation. If 600 people are selected at random, then find the standard deviation of the number of people who went out on vacation.
a.
126
b.
180
c.
11.23
d.
13.4


Solution:

Probability that a person selected went out on a vacation is p = 0.30

Probability that a person selected did not go on a vacation is q = 1 - p = 0.70

Number of people selected n = 600.

Standard deviation σ = n p q

σ = 600 × 0.30 × 0.70
[Substitute n = 600, p = 0.30, q = 0.70 .]

σ = 126 = 11.23
[Simplify.]

The standard deviation of the number of people who went out on vacation is 11.23 .


Correct answer : (3)
 10.  
In a school, 40% of the students play basketball. If 400 students are selected at random, then find the mean and variance of the number of students who play basketball.
a.
160, 9.8
b.
96, 160
c.
96, 9.8
d.
160, 96


Solution:

Probability that a student selected plays basketball is p = 0.40

Probability that a student selected does not play basketball is q = 1 - p = 0.60

Number of students selected n = 400.

Mean μ = n p


Variance s2 = n p q


μ = 400 × 0.40 = 160
[Substitute n = 400, p = 0.4 .]

s2 = 400 × 0.40 × 0.60 = 96
[Substitute n = 400, p = 0.40, q = 0.60 and simplify.]

So, the mean and variance of the number of students who play basketball is 160, 96 respectively.


Correct answer : (4)

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