Probability Distribution Function Practice Questions

A probability distribution is a rule or function that shows the probability of every possible value of a random variable. It describes how the total probability is distributed among all possible outcomes of a random experiment.

Question 1: Suppose we toss two dice. Make a table of the probabilities for the sum of the dice. The possibilities are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Solution:

Probability Distribution Table

X	P(x)
2	1/36
3	2/36
4	3/36
5	4/36
6	5/36
7	6/36
8	5/36
9	4/36
10	3/36
11	2/36
12	1/36

Question 2: The number of old people living in houses on a randomly selected city block is described by the following probability distribution.

What is the probability that 6 or more old people live in a randomly selected house?

Solution:

Sum of all the p(probability) is equal to 1

Probability that six or more old peoples live in a house,

= 1 - (0.50 + 0.25 + 0.10)
= 0.15

Thus, probability that six or more old peoples live in a house is equal to 0.15

Example 3: When a fair coin is tossed 8 times, then the Probability of:

• Exactly Four Heads
• At least Four Heads

Solution:

Every coin tossed can be considered as the Bernoulli trial. Suppose X be the number of heads in this experiment,

n = 8
p = 1/2

So,
P(X = x) = ⁿC_x p^{n - x} (1 - p)^x, x =  0, 1, 2, 3,...n
P(X = x) = ⁸C_xp^{8 - x}(1 - p)^x

P(Exactly 4 Heads)
= P(x = 4)
= ⁸C₄ p⁴ (1 - p)⁴
= 8!/4!4!(1/2)⁴(1/2)⁴
= (8 × 7 × 6 × 5/2 × 3 × 4) × (1/16) × (1/16)
= 420/1536
= 35/128

Thus, the probability of Exactly Four Heads in a Eight Coin Toss experiment is 35/128

P(At Least 4 Heads)

= P(X >= 4)
= P(X = 4) + P(X = 5) + P(X = 6)+ P(X = 7) + P(X = 8).
= ⁸C₄ p⁴ (1 - p)⁴ + ⁸C₅ p³ (1 - p)⁵ + ⁸C₆ p² (1 - p)⁶ + ⁸C₇ p¹(1 - p)⁷ + ⁸C₈(1 - p)⁸
= 8!/4!4!(1/2)⁸ + 8!/5!3!(1/2)⁸ + 8!/6!2!(1/2)⁸ + 8!/7!1!(1/2)⁸ + 8!/8!(1/2)⁸
= 8 × 7 × 6 × 5/4 × 3 × 2 × 256 + 8 × 7 × 6/3 × 2 × 256 + 8/256 + 1/256
= 1680/6144 + 336/1536 + 9/256
= 70/256 +  56/256 + 9/256
= 135/256

Thus, the probability of Atleast Four Heads in a Eight Coin Toss experiment is 135/256.

Example 4: Calculate the probability of getting 10 heads, if a coin is tossed 12 times.

Solution:

Given,

• Number of Trials(n) = 12
• Number of Success(r) = 10 (getting 10 heads)
• Probability of Single Head(p) = 1/2 = 0.5

To find ⁿC_r =  n!/r!(n – r)!

=  12!/10!(12 – 10)!
=  (12 × 11 × 10!)/10!2!
= 66

To find p^r = (0.5)¹⁰ = 0.00097665625
So, the probability of getting 10 heads is:

P(x) = ⁿC_r p^r (1 - p)^{n - r}
= 66 × 0.00097665625 × (1 – 0.5)^(12-10)
= 0.0644593125 × (0.5)²
= 0.016114828125

The probability of getting 10 heads = 0.0161...

Example 5: Suppose that each time you take a free throw shot, you have a 35% chance of making it. If you take 25 shots, what is the probability of making exactly 15 of them?

Solution:

Given,

• n = 25
• r = 15
• p = 0.35
• q = 0.65

Compute
C_25,15 (0.35)¹⁵ (0.65)¹⁰ = 0.165

There is a 16.5% chance of making exactly 15 shots.

Example 6: There is a total of 5 people in the room, what is the possibility that someone in the room shares His / Her birthday with at least someone else?

Solution:

P(s) = p(At least someone shares with someone else)                        

P(d) = p(No one share their birthday everyone has a different birthday)

p(s) + p(d) = 1 or 100%
p(s) =100% - p(d)

There are 5 people in the room, the possibility that no one shares his/her birthday

= (365 × 364 × 363 × 362 × 361) ⁄ (365)⁵
= (365! ⁄ (365 - 5)!) ⁄ 365⁵
= (365! ⁄ 360!) ⁄ 365⁵
= 0.9728

p(d) = 0,9728 or 97.28%

p(s) = 100% - p(d)

= 100% - 97.28% or 1 - 0.9728
= 2.72%  ≈ 0.0272

Practice Questions

Q1: Find the Probability Distribution of the the Number of Heads when two coins are tossed Simultaneously.

Q2: What is the Probability Distribution of the of number of Kings when three cards are drawn at random.

Q3: A die is thrown twice. Find the probability of getting a number of sixes.

Q4: A coin is thrown until a tail appears or the the head appears three times continuously. Find the probability distribution of tosses.