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Probability Distribution Worksheet

Probability Distribution Worksheet
  • Page 1
 1.  
A fruit basket contains 5 plums, 4 apples and 3 oranges. Each time a fruit is picked at random, it is replaced in the basket. Find the probability that out of 6 fruits picked so, 3 are plums, 2 are apples and 1 is an orange.
a.
25 5184
b.
125 5184
c.
25 27
d.
625 5184


Solution:

Formula for multinomial distribution
P(X = x1, X = x2, . . . , X = xk) = n!x1!x2! . . .xk! . p1x1 p2x2 . . . pkxk
where x1 + x2 + . . . + xk = n
p1 + p2 + . . . + pk = 1

Probability that the fruit is a plum P (Plum) = 5 / 12

P (Apple) = 4 / 12 = 1 / 3

P(Orange) = 3 / 12= 1 / 4

P(3 plums, 2 apples, 1 orange) = [6!(3! 2! 1!)] × (5 / 12)3 · (1 / 3)2 × (1 / 4)
[Substitute.]

= 625 / 5184
[Simplify.]

So, the probability that out of 6 fruits picked so, 3 are plums, 2 are apples and 1 is an orange is 625 / 5184 .


Correct answer : (4)
 2.  
A fair die is rolled 6 times. Find the probability that 1 appears exactly once, 2 appears exactly twice and 3 appears exactly thrice.
a.
5 3888
b.
5 324
c.
5 1944
d.
5 18


Solution:

Formula for multinomial distribution
P(X = x1, X = x2, . . . , X = xk) = n!x1!x2! . . .xk! . p1x1 p2x2 . . . pkxk
where x1 + x2 + . . . + xk = n
p1 + p2 + . . . + pk = 1

Probability for any one number 1 or 2 or 3 to appear = 1 / 6

Setting n = 6 and using the formula for multinomial distribution, the required probability,
P(X = 1, X = 2, X = 3) = [6!(3! 2! 1!)] × (1 / 6)1 · (1 / 6)2 · (1 / 6)3 = 5 / 3888
[Substitute and simplify.]

So, the probability that 1 appears exactly once, 2 appears exactly twice and 3 appears exactly thrice = 5 / 3888 .


Correct answer : (1)
 3.  
In a cycle manufacturing factory, it is observed that chances that a cycle contains 0, 1, 2 defects is 0.88, 0.08, 0.04 respectively. If 16 cycles are chosen at random, then find the probability that 9 will have no defects, 4 have 1 defect and 3 have 2 defects.
a.
[16!(9! 4! 3!)] × (0.88)9 · (0.08)4 · (0.04)3
b.
[16!(9! 4! 3!)] × (0.04)3 · (0.88)9 ·
c.
(0.88)9 · (0.08)4 · (0.04)3
d.
[16!(9! 4! 3!)] × (0.88)9 · (0.08)4


Solution:

Formula for multinomial distribution
P(X = x1, X = x2, . . . , X = xk) = n!x1!x2! . . .xk! . p1x1 p2x2 . . . pkxk
where x1 + x2 + . . . + xk = n
p1 + p2 + . . . + pk = 1

Probability that there is no defect in the cycle = P (0 defect) = 0.88

Similarly, P (1 defect) = 0.08

P (2 defects) = 0.04

Setting n = 16 and using the formula for multinomial distribution, the required probability
P (X = 0, X = 1, X = 2) = [16!(9! 4! 3!)] × (0.88)9 · (0.08)4 · (0.04)3
[Substitute and simplify.]


Correct answer : (1)
 4.  
The probability that a student commits 0, 1, 2, 3 errors in a short assignment is 0.6, 0.3, 0.05 and 0.05 respectively. If 10 assignments are chosen, then find the probability that 6 contain no errors, 2 contain 1 error, 1 contains 2 errors, and 1 contains 3 errors.
a.
(10356)(3107)
b.
(5107)(10336)
c.
(3103)(53106)
d.
(3107)(10356)


Solution:

Formula for multinomial distribution
P(X = x1, X = x2, . . . , X = xk) = n!x1!x2! . . .xk! . p1x1 p2x2 . . . pkxk
where x1 + x2 + . . . + xk = n
p1 + p2 + . . . + pk = 1

Probability that there is no error in the assignment = P (0 error) = 0.6

Similarly, P (1 error) = 0.3

P (2 error) = 0.05

P (3 error) = 0.05

Setting n = 10 and using the formula for multinomial distribution, the required probability
P(X) = [10!(6! 2! 1! 1!)] × (0.6)6 · (0.3)2 · (0.05)1 · (0.05)1
[Substitute.]

P(X) = (3107)(10356)
[Simplify.]

So, the required probability is (3107)(10356)


Correct answer : (4)
 5.  
If 3% of the electric bulbs manufactured by a firm are defective, then find the probability that a sample of 100 bulbs has no defective bulb.
a.
e- 3
b.
1 - e- 3
c.
3e- 3


Solution:

The probability that a bulb is defective is p = 0.03

Number of samples n = 100.

The situation is appropriate for making use of the Poisson distribution.

The Poisson variate mean, λ = n · p = 100 × 0.03 = 3

Substitute x = 0 (No defective bulbs) in the formula for the Poisson distribution
P (X = x) = e-λλxx!

P (X = 0) = e-3300! = e-3
[Substitute and simplify.]

So, a sample of 100 bulbs having no defective bulb has a probability of e- 3.


Correct answer : (1)
 6.  
If 3% of the electric bulbs manufactured by a firm are defective, then what is the probability that exactly 5 bulbs out of a sample of 100 are defective?
a.
8140e3
b.
2740e3
c.
8140e2
d.
940e2


Solution:

The probability that a bulb is defective is p = 0.03

Number of samples n = 100.

The situation is appropriate for making use of the Poisson distribution.

The Poisson variate mean, λ = n · p = 100 × 0.03 = 3

Substitute x = 5 (5 defective bulbs) in the formula for the Poisson distribution
P (X = x) = e-λλxx!

P (X = 5) = e-3355! = 8140e3
[Substitute and simplify.]

So, the probability that exactly 5 bulbs out of a sample of 100 are defective is 8140e3.


Correct answer : (1)
 7.  
Six coins are tossed 3200 times. Find the approximate probability of getting 5 heads 2 times.
a.
3002e-3002
b.
300e-300200
c.
300e-300
d.
300e-3002


Solution:

Probability of getting a head when a coin is tossed = 1 / 2

Probability of getting a tail when a coin is tossed = 1 / 2

Probability of getting 5 heads with 6 coins p = 6C5 · (1 / 2)5 · (1 / 2)1 = 3 / 32

The situation is appropriate for applying the Poisson distribution with n = 3200

The Poisson variate mean, λ = n · p = 3200 × 3 / 32 = 300

Substitute x = 2 in the formula for the Poisson distribution
P (X = x) = e-λλxx!

P (X = 2) = e-30030022!
[Substitute and simplify.]

So, the probability of getting 5 heads 2 times is 3002e-3002.


Correct answer : (1)
 8.  
A manufacturing concern employing a large number of workers finds that over a period of time, the average absentee rate is 2 workers per shift of 100. Calculate the probability that exactly 2 workers will be absent in a shift.
a.
1e2
b.
4e2
c.
2e2
d.
2e2


Solution:

Probability of absentee worker is p = 0.02

The situation is appropriate for use of Poisson distribution with n = 100

The Poisson variate mean, λ = n · p = 100 × 0.02 = 2

Substituting x = 2 in the formula for the Poisson distribution
P (X = x) = e-λλxx!

P (X = 2) = e-2222! = 2e2
[Substitute and simplify.]

So, the probability that exactly 2 workers will be absent in a shift is 2e2.


Correct answer : (4)
 9.  
A company knows on the basis of its past experience that 2% of its blades are defective. Find the probability of having 3 defective blades in a sample of 100 blades. [Use e- 2 = 0.1353]
a.
0.1804
b.
0.2804
c.
0.4804
d.
0.3804


Solution:

Probability that a blade is defective is p = 0.02

Applying the Poisson distribution, n = 100.

The Poisson variate mean, λ = n · p = 100 × 0.02 = 2

Substituting x = 3 in the formula for the Poisson distribution
P (X = x) = e-λλxx!

P (X = 3) = e-2233! = (0.1353) · (4 / 3) = 0.1804
[Substitute and simplify.]

So, the probability of having 3 defective blades in a sample of 100 blades is 0.1804.


Correct answer : (1)
 10.  
The accidental drownings per year in a country is 3 per 100,000 population. Find the probability that in a city of 200,000 population, there will be fewer than 3 accidental drownings per year. [Use e- 6 = 0.002479]
a.
0.0149
b.
0.0025
c.
0.062
d.
0.0446


Solution:

Probability of drowning is p = 3 / 100000

With n = 200000, the Poisson variate mean, λ = n · p = (3 / 100000) · (200000) = 6

If there will be fewer than 3 accidental drownings per year, then substitute x = 0, 1, 2 in the formula for the Poisson distribution
P (X = x) = e-λλxx!

Required probability = P (X = 0) + P (X = 1) + P (X = 2)

Required probability = e- 6 (600!+611!+622!)
[Substitute.]

Required probability = (1 + 6 + 18)(e- 6) = 25 × (0.002479) = 0.062
[Simplify.]

So, the probability that there will be fewer than 3 accidental drownings per year is 0.062.


Correct answer : (3)

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