1.0Types of Probability
Conditional Probability
Conditional probability assesses the likelihood of an event occurring given that another event has already happened. It is denoted as P(A|B), where A is the event of interest and B is the condition under which event A occurs. The formula for conditional probability is:
P(A∣B)=P(B)P(A and B)
This formula states that the probability of event A given event B is equal to the probability of both events A and B occurring divided by the probability of event B occurring. Conditional probability is widely used in various fields, including statistics, machine learning, and decision-making processes.
Empirical Probability
Empirical probability, also known as experimental probability, is derived from observations or experiments. It is calculated by conducting trials or experiments and observing the frequency of a specific event occurring. The empirical probability of an event is the ratio of the number of times the event occurs to the total number of trials conducted. As the number of trials increases, the empirical probability tends to approach the theoretical probability, providing a practical estimate of likelihood based on real-world data.
2.0Bayes Theorem
Bayes' theorem, also known as Conditional Probability, calculates the likelihood of an event given the occurrence of other events. It aids in determining the probability of an event based on certain conditions. The probability is determined considering all possible outcomes. The theorem's formula is expressed as
P(A∣B)=P(B)P(B∣A)⋅P(A)
where P(A|B) signifies the probability of event A given event B.
where P(B|A) represents the probability of event B given event A.
P(A) and P(B) denote the likelihood of occurrence of events A and B, respectively.
3.0Law of Total Probability
In an experiment with n outcomes, the total probability of all these outcomes combined always equals 1.
P(A1) + P(A2) + P(A3) + … + P(An) = 1
Important Notes on Probability: Probability quantifies the likelihood of an event occurring, typically expressed as a fraction between 0 and 1. Events are subsets of the sample space, such as {head, tail} for coin flips and {1, 2, 3, 4, 5, 6} for dice rolls.
4.0Independent Events Probability
Independent events in probability refer to events where the occurrence of one event does not affect the probability of the other event occurring. Mathematically, two events A and B are independent if and only if:
P (A ∩ B) = P (A)⋅P (B)
This equation states that the probability of both events A and B occurring is equal to the product of their individual probabilities. In other words, knowing that one event has occurred does not provide any information about the likelihood of the other event occurring.
5.0Multiplication Theorem of Probability
The Multiplication Theorem of Probability, also known as the Product Rule, is a fundamental principle in probability theory used to calculate the probability of the intersection of two or more events.
For two events A and B, the theorem states:
P (A ∩ B) = P(A). P(B|A)
where:
P (A ∩ B) is the probability that both events A and B occur,
P (A) is the probability of event A,
P (B|A) is the conditional probability of event B given that event A has occurred.
This formula allows us to find the probability of the joint occurrence of two events by multiplying the probability of the first event by the conditional probability of the second event given the first event has occurred.
The Multiplication Theorem can be extended to more than two events by repeatedly applying the product rule. It is a fundamental tool in probability calculations and is widely used in various fields such as statistics, finance, and engineering.
Also Read : Probability and Statistics
6.0Probability Solved Examples
Example 1: Two coins (a one-rupee coin and a two-rupee coin) are tossed once. Find a sample space.
Solution:
Clearly the coins are distinguishable in the sense that we can speak of the first coin and the second coin. Since either coin can turn up Head (H) or Tail (T), the possible outcomes may be Heads on both coins = (H, H) = HH
Head on first coin and Tail on the other = (H, T) = HT
Tail on first coin and Head on the other = (T, H) = TH
Tail on both coins = (T, T) = TT
Thus, the sample space is S = {HH, HT, TH, TT}
Example 2: When a coin is tossed twice. If head appears in the second throw, then a dice is thrown. Write down the sample space of the experiment.
Solution:
When a coin is tossed two times then possible outcomes are {(TT), (HT), (TH), (HH)}
If the head appears in the second throw, then the dice is thrown.
∴ All possible outcomes of the experiment are-
S = {(TT), (HT), (TH1), (TH2), (TH3), (TH4), (TH5), (TH6), (HH1), (HH2), (HH3), (HH4), (HH5), (HH6)}
Example 3: A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls; if it shows tail, we throw a die. Describe the sample space of this experiment.
Solution:
Let us denote blue balls by B1, B2, B3 and the white balls by W1, W2, W3, W4. Then a sample space of the experiment is
S = {HB1, HB2, HB3, HW1, HW2, HW3, HW4, T1, T2, T3, T4, T5, T6}.
Here HBi means head on the coin and ball Bi is drawn, HWi means head on the coin and ball Wi is drawn. Similarly, Ti means tail on the coin and the number i on the die.
Example 4: Two natural numbers are randomly selected from the set of first 20 natural numbers. Find the probability that-
(A) Their sum is odd
(B) sum is even
(C) The selected pair is twin prime.
Solution:
(A) S = {1, 2, 3, ......, 19, 20}; n(S)=20C2
N(A)=110C1⋅10C1=100⇒P(A)=190100=1910
(sum odd ⇒one odd and one even)
(B) n (B) =10C2 +10C2=2. 10C2 = 90 ⇒ P(B)= 19090=199
(sum even ⇒ both odd or both even)
(C) n (C) = {(3, 5), (5, 7), (11, 13), (17, 19)} ⇒ P (C) =1904=952
Example 5: Roll a fair die twice. Let A be the event that the sum of the two rolls equals six and let B be the event that the same number comes up twice. What is P (A/B)?
(A) 1/6 (B) 5/36 (C) 1/5 (D) none
Solution:
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
B = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
P(BA)=P(B)P(A∩B)=n(B)n(A∩B)
=61
Example 6: In a class, 30% of the students failed in Physics, 25% failed in Mathematics and 15% failed in both Physics and Mathematics. If a student is selected at random failed in Mathematics, find the probability that he failed in Physics also.
Solution:
Let A be the event "failed in Physics" and B be the event "failed in Mathematics". We want to find
P(BA).
It is given that P(A)=10030 and P(B)= 10025
Also,
P(A∩B)=10015
Therefore
P. (BA)=P(B)P(A∩B)=25/10015/100=2515=53
Example 7: Let P(A) = 21, P(B) = 31, then find P (A ∪ B) if
- A & B are mutually exclusive
- A & B are independent
Solution:
Let P (A)=21 P (B) = 31
then find P (A ∪ B) if
(i) A & B are mutually exclusive
∴ P (A ∩ B) = 0
P (A ∪ B) = 21+31=65
(ii) A & B are independent
P (A ∩ B) = 21⋅31=61
P (A ∪ B) = 21+31−61=65−61=64=32
7.0Probability Practice Problems
- Find the sample space associated with the experiment of rolling a pair of dice (plural of die) once. Find the number of elements of the sample space.
- The probability of drawing a white ball from a bag containing 3 black balls and 4 white balls, is
(A) 74 (B) 73 (C) 71 (D) None of these
- If P (A ∪ B) = P (A ∩ B) for any two events A and B, then
(A) P (A) = P (B) (B) P (A) > P (B) (C) P (A) < P (B) (D) none of these
- A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?
- Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that the first two cards are kings and the third card drawn is an ace?
Also Read: