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The Binomial Distribution Worksheet

The Binomial Distribution Worksheet
  • Page 1
 1.  
In a family of 4 children find the probability of having atleast one boy.
a.
0.5
b.
0.0625
c.
1
d.
0.9375


Solution:

The family has 4 children, so n = 4.

The probability of having a boy is 0.5, so p = 0.5

q = 1 - p = 0.5

p (at least 1 boy) = 1 - p (no boy) = 1 - 4C0 (0.5)0(0.5)4

= 1- (1)(1)(0.0625)

= 0.9375


Correct answer : (4)
 2.  
A group of people gathered for a small party. 15% of them are left handed, find the probability that if 6 people are chosen at random, all of them are left handed.
a.
(0.15)6
b.
0.85
c.
(0.85)6
d.
0.15


Solution:

n = 6, p = 15% = 0.15, q = 1 - p = 0.85

P (6 out of 6 are left handed) = 6C6 p6q0
[Use P(x successes of n trials) = nCx pxqn-x.]

= 6C6 (0.15)6 (0.85)0

= (0.15)6


Correct answer : (1)
 3.  
In a bombing attack there is 40% chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. What is the probability that the target will be destroyed, if 3 bombs are dropped?
a.
0.4
b.
0.56
c.
0.288
d.
0.96


Solution:

n = 6, p = 40% = 0.4, q = 1 - p = 0.6, x = 6

P (2 out of 3 bombs strike the target) = 3C2 (0.4)2 (0.6)
[Use P(x successes of n trials) = nCx pxqn-x.]

= (3) (0.16) (0.6) = 0.288


Correct answer : (3)
 4.  
A player has to cross 10 hurdles in a hurdle race. The probability that he will clear each hurdle is 0.83. What is the probability that he will knock down fewer than 2 hurdles?
a.
48%
b.
50%
c.
83%
d.
80%


Solution:

Each hurdle represents a trial, n = 10.

Clearing a hurdle represents a success.

Probability of clearing a hurdle, p = 0.83

Probability of failure, q = 0.17

P (player knocks 0 hurdles) = P (player clears 10 hurdles) = 10C10 (0.83)10 (0.17)0 = (1) (0.16) (1) = (0.16)
[Using nCx pxqn-x]

P (player knocks 1 hurdle) = P (player clears 9 hurdles.] = 10C9 (0.83)9 (0.17) = (10) (0.19) (0.17) = 0.323

P (player knocks fewer than 2 hurdles) = P (Player knocks fewer than 0 hurdles) + P (player knocks 1 hurdle) = 0.16 + 0.323 = 0.483 or 48.3%.


Correct answer : (1)
 5.  
About 75% of the seeds of a certain kind sold in a shop germinate under normal conditions. Sandra buys a packet containing 10 seeds. What is the probability that 3 of them germinate?
a.
0.001
b.
0.003
c.
0.75
d.
0.4219


Solution:

The number of trials is the number of seeds in the packet, n = 10

If the seed germinates, it is considered a success, x = 3

The probability of a success, p = 0.75

The probability of failure, q = 1 - p = 0.25

P (3 out of 10 seeds germinate) = 10C3 (0.75)3 (0.25)7
[Use P (x successes of n trials) = nCx pxqn-x.]

= (120) (0.4219) (0.00006) = 0.0030


Correct answer : (2)
 6.  
The chance that any one telephone line is busy at a given moment of time is 0.01. What is the chance that out of 5 telephone lines, less than 3 lines are busy?
a.
0.048
b.
.001
c.
1
d.
0.951


Solution:

The number of telephone lines is 5, n = 5.

A line being busy represents a success, x < 3 or x = 0, 1, 2.

The probability of a success, p = 0.01

The probability of failure, q = 1 - 0.01 = 0.99

P (0 lines are busy) = 5C0 (0.01)0(0.99)5 = (1) (1) (0.951) = 0.951.

P (1 line is busy) = 5C1 (0.01)1 (0.99)4 = (5) (0.01) (0.9606) = 0.048

P (2 lines are busy) = 5C2 (0.01)2 (0.99)3 = (10) (0.0001) (0.9703) = 0.001

P (less than 3 lines are busy) = P (0 lines are busy) + P (1 line is busy) + P (2 lines are busy) = 0.951 + 0.048 + 0.001. 1


Correct answer : (3)
 7.  
Evaluate: 6C2 (0.4)2 (0.6)4
a.
0.1296
b.
2.4
c.
0.0207
d.
0.3110


Solution:

6C2 (0.4)2 (0.6)4 = 6!4! 2! (0.4)2 (0.6)4

= (15) (0.16) (0.1296)

= 0.3110


Correct answer : (4)
 8.  
Evaluate: 5C4 (0.5)4 (0.5)
a.
0.0625
b.
0.25
c.
0.1563
d.
0.0313


Solution:

5C4 (0.5)4(0.5) = 5!1! 4! (0.5)4(0.5)

= (5) (0.0625) (0.5)

= 0.1563


Correct answer : (3)
 9.  
Find the probability of 3 successes in 3 trials, if the probability of success on each trial is 0.8.
a.
0.832
b.
0.48
c.
0.163
d.
0.512


Solution:

The probability of x successes in n trials each with a probability of success p and a probability of failure q (with p + q = 1) is: nCx pxqn-x

Here, n = 3, x = 3, p = 0.8, q = 1 - 0.8 = 0.2

The probability of success is given by: 3C3 (0.8)3 (0.2)o

= (1) (0.8)3(1) = 0.512


Correct answer : (4)
 10.  
What is the probability of 1 success in 4 trials, if the probability of success on each trial is 0.3?
a.
0.4116
b.
0.0756
c.
0.0081
d.
1.2


Solution:

n = 4, x = 1, p = 0.3, q = 0.7.

The probability of success is given by: nCx pxqn-x = 4C1 (0.3)1 (0.7)3

= (4) (0.3) (0.343)

= 0.4116


Correct answer : (1)

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