1) The document discusses various probability distributions including binomial, normal, uniform, and Poisson distributions. It provides examples of how to calculate probabilities using these distributions. 2) Key radiation measurement terms are defined including activity units, absorbed dose, equivalent dose, and effective dose. Equivalent dose considers the biological damage of different types of radiation while effective dose provides a single value that accounts for radiation exposure across the entire body. 3) Examples are provided for calculating probabilities using binomial, normal, and uniform distributions. Assignments are given including calculating probabilities for dice rolls and coin tosses using binomial distributions.

1. Probability Distributions and
Radiation Quantities
Dr. Munir Ahmad

2. Few Basic Terms
TERM Definition
Random Experiment experiments having random outputs
Random variables variables presenting outcomes of random experiments
Events outcome of a random experiment
Trial A single run of the random experiment
Outcome Result of the random experiment
Sample space Combined set of all events
Equally likely events Outcomes having same probability
Mutually exclusive events Outcome having independent probability
Mutually exhaustive events All events in the sample space

3. Random Variables
Random Variable
Random variables are those variables
whose values are determined by a
random experiment.
• Discrete Random Variables
• Continuous Random Variables
Discrete Random Variable
• Members in family
• Phone calls in a given time
• Houses in a street
• Red marbles in a jar
• Number of heads while flipping a coin
Random Variable
• Height of students in a class
• Time it takes to get to office daily
• Distance traveled between classes in a block

4. Probability Distribution
• Table or formula that lists the probabilities of the random variable X
X 0 1 2 3 4 5 6 7 8
P(X=x) 0.2 0.4 0.6 0.7 0.8 0.7 0.6 0.4 0.2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7 8
P(X=x)
X

5. Example
• Let us toss a coin 3 times and take number of heads appearing as our random
variable. Our sample space will be, HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
• In this sample space, if we take number of heads as the random variable, then
the possible outcomes are, X=0, 1, 2, 3.
• Their probability distribution in tabular form will be,
X 0 1 2 3
P(X=x) 1/8 3/8 3/8 1/8
0
1
2
3
0 1 2 3
P(X=x)
With property of 𝑛=1
∞
𝑃(𝑥𝑛) = 1.0
mean = 𝜇 = 𝑛=1
∞
𝑥𝑛𝑃(𝑥𝑛)
variance = σ2 =
𝑛=1
∞
𝑥𝑛
2𝑃 𝑥𝑛 − 𝜇2

6. Example
• X = sum of two dice thrown and sum is > 2.
X 2 3 4 5 6 7 8 9 10 11 12
P(X=x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
0
1
2
3
4
5
6
2 3 4 5 6 7 8 9 10 11 12
P(X=x)
• Assignment II: Find P(x) of X, where X is
sum of 2 rolled dice and comes out to be > 5.

7. Example
• A box contains 4 white and 6 red balls. 4 balls are draw at random.
Find the probability of number of white balls.
X 0 1 2 3 4
P(X=x) 𝐶4
6
/𝐶4
10
(𝐶1
4
𝐶3
6
)/𝐶4
10
(𝐶2
4
𝐶2
6
)/𝐶4
10
(𝐶3
4
𝐶1
6
)/𝐶4
10
(𝐶4
4
𝐶0
6
)/𝐶4
10
• Assignment II: Find P(x) of X, where X is sum of 2 rolled dice and comes out to be > 5.

8. Example
• Find P(x ≥ 5) and find mean µ (4.66).
X 0 1 2 3 4 5 6 7
P(X=x) 0 k 2k 2k 3k k2 2k2 7k2+k
• Assignment: Find variance?

9. Binomial Distribution
• Bernoulli events
• Bernoulli events are outcomes of a
Bernoulli trial, a trial having only two
outcomes, success (with probability p)
and failure (with probability q=1-p).
• Examples are flipping a coin and if a
person is family member or not.
• Binomial Distribution
• BD is a distribution probability of a
sequence of identical Bernoulli events.
For example having same numbers
while throwing 2 dice at same time.
Mean = 𝐸 𝑋 = 𝑥𝑖 𝑝 𝑥𝑖 = np, Variance = σ2 = 𝐸 𝑦2 − (𝐸(𝑦))2 = npq

10. Example
• If a coin is tossed 5 times, using binomial distribution find
the probability of: (a) Exactly 2 heads (b) At least 4 heads.
(a) The repeated tossing of the coin is an
example of a Bernoulli trial.
Number of trials: n=5
Probability of head: p= 1/2 and hence the
probability of tail, q =1/2
For exactly two heads: x=2
P(x=2) = 5C2 p2 q5-2 = 5! / 2! 3! × (½)2× (½)3
P(x=2) = 5/16
(b) For at least four heads,
x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5)
Hence,
P(x = 4) = 5C4 p4 q5-4 = 5!/4! 1! × (½)4× (½)1 = 5/32
P(x = 5) = 5C5 p5 q5-5 = (½)5 = 1/32
Answer: Therefore, P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16

11. Example
• Find the binomial distribution of the data having mean of
np=4 and variance npq=4/3.
Solution:
np/npq = 4/(4/3)=3
q = 1/3
p = 1 – q = 1 – 1/3 = 2/3
np = 4
n = 4/p = 4/(2/3) = 6
P(x=r) = 6Cr (4)r (1/3)6-r

12. Assignment II
• For the same question given above, find the probability of
getting at most 2 heads.
Solution: P(at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1)
P(X = 0) = (½)5 = 1/32
P(X=1) = 5C1 (½)5.= 5/32

13. Normal Distribution
• Most things follow this distribution normally..

14. Standardized Normal Distribution
• Most of the natural data follow ND, better to use standardized normal
distribution.
Properties: Symmetrical, Described by mean and variance, Area under
the curve is probability.

15. Standardized Normal Distribution

16. Standard Normal Distribution

17. Standard Normal Distribution
Assignment II:

18. Bayes Theorem
• Probability of dependent variables….

19. Bayes Theorem

20. Uniform Distribution

21. Uniform Distribution

22. Uniform Distribution

23. Poisson Distribution

24. Poisson Distribution

25. Radiation Measurements and Quantities

26. Radioactivity

27. Half Life of Decay

28. Activity Units

29. Radiation Exposure

30. Exposure

31. Absorbed Dose

32. Equivalent Dose

33. Equivalent Dose

34. Effective Dose

35. Effective Dose

36. Effective Dose

37. Effective Dose

39. THANKS