Binomial Distribution-Probability Density Function
The pdf(such as in binompdf) stands for probability density function. It is a function that tells you the probability of certain events occurring.
EXAMPLE 1
In a particular population, the
probability that people test positive for the COVID-19 virus infection is 0.1,
therefore the probability of having a negative result is 0.9. Four patients from
this population test for COVID-19, what is the probability that the results of
the four are all positive?
Answer:
In how many patterns can all four results be positive? Only one, i.e. A,B,C,D are all positive?
|
Patient |
Test result |
p |
|
A |
+ |
0.1 |
|
B |
+ |
0.1 |
|
C |
+ |
0.1 |
|
D |
+ |
0.1 |
Independence check: whether one person’s result is positive or negative is not influenced by any other person’s result, we can multiply individual probabilities of being positive to find the probability that all the results are positive: (0.1)x( 0.1)x( 0.1)x( 0.1) = (0.1) 4 = 0.0001
What is the probability that exactly two out of three people have positive results?
Answer:
In how many different ways (arrangements, combinations) can two out of three people test positive? Altogether there are three ways, listed in the table below.
|
Patient |
combo1 |
combo2 |
combo3 |
|
A |
+ (p=0.1) |
+ (p=0.1) |
-
(p=0.9) |
|
B |
+ (p=0.1) |
-
(p=0.9) |
+ (p=0.1) |
|
C |
-
(p=0.9) |
+ (p=0.1) |
+ (p=0.1) |
|
|
0.1x0.1x0.9 = 0.009 |
0.1x0.9x0.1= 0.009 |
0.9x0.1x0.1=0.009 |
The probability that patient A and B test positive and C tests negative is (0.1)( 0.1)( 0.9) = 0.009.
The probability that patient A and C test
positive and B tests negative is (0.9)( 0.1)( 0.1) = 0.009.
Finally, the probability that ptient B and C test
positive and A tests negative is (0.1)( 0.9)( 0.1) = 0.009.
Summing, we find that the probability
that exactly two out of three patients test positive is:
0.009 + 0.009 + 0.009 = 3(0.009) =
0.027.
What is the probability that exactly three out of eight people test positive?
Answer:
The probability of any particular
arrangement of three positive and five negative results is (0.1) 3(0.9)
5 = 0.00059049. We need to know the number of such arrangements.
The answer is given by combinations:
8!/3!x5!
8! = 8*7*6*5*4*3*2
3! = 3*2
5!=5*4*3*2
|
Patient |
Combo1 |
Comobo 2 |
Combo…. |
Combo56 |
|
A |
1(0.1) |
1 |
|
|
|
B |
1(0.1) |
1 |
|
|
|
C |
1(0.1) |
0 |
|
|
|
D |
0(0.9) |
0 |
|
|
|
E |
0(0.9) |
0 |
|
|
|
F |
0(0.9) |
0 |
|
|
|
G |
0(0.9) |
0 |
|
|
|
H |
0(0.9) |
1 |
|
|
|
|
(0.1) 3(0.9) 5 = 0.00059049 |
|
|
|
The probability that exactly three out of eight patient test positive is 56 × 0.00059049 = 0.03306744.
On the TI-84, binompdf(8,0.1,3) = 0.03306744.
How to find the number of arrangements to have exactly 2 test positive out of 4 patients tested:
4!/2!(4-2)! = 4!/2!x2! = 4x3x2/2x2 = 6
|
Patient |
Combo1 |
Combo2 |
Combo3 |
Combo4 |
Combo5 |
Combo6 |
|
A |
+ |
+ |
+ |
- |
- |
- |
|
B |
+ |
- |
- |
+ |
- |
+ |
|
C |
- |
+ |
- |
+ |
+ |
- |
|
D |
- |
- |
+ |
- |
+ |
+ |
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