Probability
Terminology
Deterministic Experiment: is the one, which gives a certain definite result. e.g. acid is added to a base.
Random Experiment: is the one, which gives one or more results under identical conditions. E.g. a coin is
tossed.
The set of all possible outcomes of a random experiment is known as the sample space and every outcome is a sample point.
Probability of an Event:
If there are n-elementary events associated with a random experiment and m of them are favorable to an
event A, then the probability of happening of A is denoted by P (A) and is defined as the ratio m/n.
Probability of an event occurring =Number of allpossible outcomes
..............................
Number of favourable outcomes
Thus, P (A) =M/N
Clearly, 0 ≤ m ≤ n, therefore 0 ≤ M/N ≤ 1, so that 0 ≤ P (A) ≤ 1
TIP:
a. Probability always
lies between 0 and 1.
b. Probability of sure
event is 1.
c. Probability is 0 for
impossible event.
Complement of an event:
Since the number of cases in which the event A will not happen is n - m, therefore if A denotes not happening
of A, then the probability P (A) of not happening of A is given by :
P(a)= n-m/n = 1-m/n = 1- p(a) or p(a)+ p(a)=1
Ex.1 What is the probability both the two dices thrown show a number multiple of 3?
Sol. There are only two numbers divisible by 3 (3, 6) in a dice.
Favorable chances = 2
Hence probability of a dice showing number divisible by 3 =2/6= 1/3
Probability that both the dices show number divisible by 3 & 6 = 1/3*1/3= 1/9
The odds in favour of occurrence of the event A are defined by m : n – m i.e., P (A): P (A) and the odds against the occurrence of A are defined by n-m: m, i.e., P (A): P (A).
Ex.2 A spinner has 4 equal sectors colored yellow, blue, green, and red. What are the chances of
landing on blue after spinning the spinner?
Sol. This problem asked us to find the probability that the spinner will
land on blue. Let's look at some definitions and examples relating
to the problem above.
Impossible event and certain event:
An impossible event has no chance of occurring. If event A is impossible, then P(A) = 0
A certain event is certain to occur. If event A is certain, then P(A) = 1.
Mutually Exclusive Events:
Two or more events are said to be mutually exclusive these events cannot occur simultaneously. Two or
more events are said to be compatible if they can occur simultaneously.
Two events (A and B) are mutually exclusive if the intersection of two events is null or they have no common
element i.e. A ∩ B = φ .
e.g. In drawing a card from a deck of 52 cards:
A: The event that it is a red.
B: The event that it is a black.
C: The event that it is a king.
In the above case events A and B are mutually exclusive but the events B and C are not mutually exclusive or
disjoint since they may have common outcomes.
Sum Rule:
If E and F are two mutually exclusive events, then the probability that either event E or event F will occur in a
single trial is given by :P(E or F) or P (E ∪ F) = P(E) + P(F)
If the event are not mutually exclusive, then
P(E ∪ F) = P(E) + P(F) – P(E and F together )
Note: P (neither E nor F) = 1 – P(E or F).
Ex.3 The probability of two events A and B are 0.25 and 0.50 respectively. The probability of their
simultaneous occurrence is 0.1. Find the probability that neither A nor B occurs.
Sol. We have P(A) = 0.25, P(B) = 0.50 and P(A ∩ B) = 0.14.
Required probability = P (A ∩ B) = P (A ∪ B) [∴ (A ∪ B)= A ∩ B]
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – [0.25 + 0.50 – 0.14] = 0.39.
Independent/ Dependent events:
Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring.
Two events, A and B, are dependent if the fact that A occurs affects the probability of B occurring.
Some other examples of independent events are:
• Tossing a coin and landing on heads, and rolling a die and getting a 5.
• Choosing a card from a deck of cards and getting a three, replacing it,
and choosing a second card and getting an ace.
• Rolling a die and getting a 4, and then rolling a second die and getting a 1.
Multiplication Rule
When two events, A and B, are independent, the probability of both occurring is:
P(A and B) = P(A ∩ B) = P(A) × P(B)
Ex.4 A coin is tossed and a single 6-sided dice is rolled. Find the probability of getting a head on the
coin and a 3 on the dice.
Sol. P(head) = 1/2
P(3) = 1/6
P(head and 3) = 1/2*1/6= 1/12
Binomial Distribution: If n trials are performed under the same condition and probability of success in
each trial is p and q = 1 – p then the probability of exactly r successes in n trials is:P(r) = n^Cr p^r q^n – r
Ex.5 A dice is tossed 5 times. What is the probability that 5 shows up exactly thrice?
Sol. Here the 'random experiment' consists in tossing a die 5 times and observing the number '5' assuccess, then
p = Probability of getting '5' with single dice = 1/6 so that q = 1 – 1/6=5/6.
Since the value of p is constant for each dice and the trials are independent, using formula for Binomial
probability law, the probability of r successes is given by :
P(r) = P(X = r) = 5^Cr (1/6)^r (5/6)^5-r ; r = 0, 1, 2,….5.
Required probability that 5 shows up exactly thrice is given by:
P(X = 3) = P(3) = 5^c3(1/6)^r (5/6)^5-3 = 10*1/216*25/36=125/3888.
Ex.6 An unbiased dice is thrown. What is the probability of getting
(i) an even number;
(ii) a multiple of 3;
(iii) an even number or a multiple of 3;
(iv) an even number & a multiple of 3?
Sol. In a single throw of an unbiased dice, we can get any one of the outcomes 1, 2, 3, 4, 5, 6. So,
exhaustive number of cases = 6.
(i) An even number is obtained if we obtain any one of 2, 4, 6 as an outcome. So, favorable number of cases = 3. Thus, required probability = 3/6= 1/2.
(ii) A multiple of 3 is obtained if we obtain any one of 3, 6 as an outcome. So, favorable number of cases = 2. Thus, required probability = 2/6= 1/3.
(iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6. So,favorable number of cases = 4. Thus, required probability =
4/6= 2/3.
(iv) An even number and a multiple of 3 is obtained if we get 6 as an outcome. So, favorable number of cases = 1. Thus, required probability = 1/6.
Ex.7 What is the chance of drawing an ace from a deck of cards?
Sol. There being 52 cards in a deck of cards, the total number of possible outcomes of drawing a card is
52. Since there are 4 aces in a pack of cards, the total number of favorable outcomes is 4. The chance
or probability of drawing an ace from the given pack is the ratio of total number of favorable outcomes
to the total number of possible outcomes.
Which in this case is: 4/52 = 1/13.
Ex.8 One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be
drawn. Find the probability that the card drawn is:
(i) an ace, (ii) red,
(iii) either red or king, (iv) red and a king
Sol. Out of 52 cards, one card can be drawn in 52^c1 ways.
(i) There are four aces in a pack of 52 cards, out of which one can be drawn in 4^C1 Therefore Favourable number of cases = 4^C1 = 4.
Terminology
Deterministic Experiment: is the one, which gives a certain definite result. e.g. acid is added to a base.
Random Experiment: is the one, which gives one or more results under identical conditions. E.g. a coin is
tossed.
The set of all possible outcomes of a random experiment is known as the sample space and every outcome is a sample point.
Probability of an Event:
If there are n-elementary events associated with a random experiment and m of them are favorable to an
event A, then the probability of happening of A is denoted by P (A) and is defined as the ratio m/n.
Probability of an event occurring =Number of allpossible outcomes
..............................
Number of favourable outcomes
Thus, P (A) =M/N
Clearly, 0 ≤ m ≤ n, therefore 0 ≤ M/N ≤ 1, so that 0 ≤ P (A) ≤ 1
TIP:
a. Probability always
lies between 0 and 1.
b. Probability of sure
event is 1.
c. Probability is 0 for
impossible event.
Complement of an event:
Since the number of cases in which the event A will not happen is n - m, therefore if A denotes not happening
of A, then the probability P (A) of not happening of A is given by :
P(a)= n-m/n = 1-m/n = 1- p(a) or p(a)+ p(a)=1
Ex.1 What is the probability both the two dices thrown show a number multiple of 3?
Sol. There are only two numbers divisible by 3 (3, 6) in a dice.
Favorable chances = 2
Hence probability of a dice showing number divisible by 3 =2/6= 1/3
Probability that both the dices show number divisible by 3 & 6 = 1/3*1/3= 1/9
The odds in favour of occurrence of the event A are defined by m : n – m i.e., P (A): P (A) and the odds against the occurrence of A are defined by n-m: m, i.e., P (A): P (A).
Ex.2 A spinner has 4 equal sectors colored yellow, blue, green, and red. What are the chances of
landing on blue after spinning the spinner?
Sol. This problem asked us to find the probability that the spinner will
land on blue. Let's look at some definitions and examples relating
to the problem above.
Impossible event and certain event:
An impossible event has no chance of occurring. If event A is impossible, then P(A) = 0
A certain event is certain to occur. If event A is certain, then P(A) = 1.
Mutually Exclusive Events:
Two or more events are said to be mutually exclusive these events cannot occur simultaneously. Two or
more events are said to be compatible if they can occur simultaneously.
Two events (A and B) are mutually exclusive if the intersection of two events is null or they have no common
element i.e. A ∩ B = φ .
e.g. In drawing a card from a deck of 52 cards:
A: The event that it is a red.
B: The event that it is a black.
C: The event that it is a king.
In the above case events A and B are mutually exclusive but the events B and C are not mutually exclusive or
disjoint since they may have common outcomes.
Sum Rule:
If E and F are two mutually exclusive events, then the probability that either event E or event F will occur in a
single trial is given by :P(E or F) or P (E ∪ F) = P(E) + P(F)
If the event are not mutually exclusive, then
P(E ∪ F) = P(E) + P(F) – P(E and F together )
Note: P (neither E nor F) = 1 – P(E or F).
Ex.3 The probability of two events A and B are 0.25 and 0.50 respectively. The probability of their
simultaneous occurrence is 0.1. Find the probability that neither A nor B occurs.
Sol. We have P(A) = 0.25, P(B) = 0.50 and P(A ∩ B) = 0.14.
Required probability = P (A ∩ B) = P (A ∪ B) [∴ (A ∪ B)= A ∩ B]
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – [0.25 + 0.50 – 0.14] = 0.39.
Independent/ Dependent events:
Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring.
Two events, A and B, are dependent if the fact that A occurs affects the probability of B occurring.
Some other examples of independent events are:
• Tossing a coin and landing on heads, and rolling a die and getting a 5.
• Choosing a card from a deck of cards and getting a three, replacing it,
and choosing a second card and getting an ace.
• Rolling a die and getting a 4, and then rolling a second die and getting a 1.
Multiplication Rule
When two events, A and B, are independent, the probability of both occurring is:
P(A and B) = P(A ∩ B) = P(A) × P(B)
Ex.4 A coin is tossed and a single 6-sided dice is rolled. Find the probability of getting a head on the
coin and a 3 on the dice.
Sol. P(head) = 1/2
P(3) = 1/6
P(head and 3) = 1/2*1/6= 1/12
Binomial Distribution: If n trials are performed under the same condition and probability of success in
each trial is p and q = 1 – p then the probability of exactly r successes in n trials is:P(r) = n^Cr p^r q^n – r
Ex.5 A dice is tossed 5 times. What is the probability that 5 shows up exactly thrice?
Sol. Here the 'random experiment' consists in tossing a die 5 times and observing the number '5' assuccess, then
p = Probability of getting '5' with single dice = 1/6 so that q = 1 – 1/6=5/6.
Since the value of p is constant for each dice and the trials are independent, using formula for Binomial
probability law, the probability of r successes is given by :
P(r) = P(X = r) = 5^Cr (1/6)^r (5/6)^5-r ; r = 0, 1, 2,….5.
Required probability that 5 shows up exactly thrice is given by:
P(X = 3) = P(3) = 5^c3(1/6)^r (5/6)^5-3 = 10*1/216*25/36=125/3888.
Ex.6 An unbiased dice is thrown. What is the probability of getting
(i) an even number;
(ii) a multiple of 3;
(iii) an even number or a multiple of 3;
(iv) an even number & a multiple of 3?
Sol. In a single throw of an unbiased dice, we can get any one of the outcomes 1, 2, 3, 4, 5, 6. So,
exhaustive number of cases = 6.
(i) An even number is obtained if we obtain any one of 2, 4, 6 as an outcome. So, favorable number of cases = 3. Thus, required probability = 3/6= 1/2.
(ii) A multiple of 3 is obtained if we obtain any one of 3, 6 as an outcome. So, favorable number of cases = 2. Thus, required probability = 2/6= 1/3.
(iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6. So,favorable number of cases = 4. Thus, required probability =
4/6= 2/3.
(iv) An even number and a multiple of 3 is obtained if we get 6 as an outcome. So, favorable number of cases = 1. Thus, required probability = 1/6.
Ex.7 What is the chance of drawing an ace from a deck of cards?
Sol. There being 52 cards in a deck of cards, the total number of possible outcomes of drawing a card is
52. Since there are 4 aces in a pack of cards, the total number of favorable outcomes is 4. The chance
or probability of drawing an ace from the given pack is the ratio of total number of favorable outcomes
to the total number of possible outcomes.
Which in this case is: 4/52 = 1/13.
Ex.8 One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be
drawn. Find the probability that the card drawn is:
(i) an ace, (ii) red,
(iii) either red or king, (iv) red and a king
Sol. Out of 52 cards, one card can be drawn in 52^c1 ways.
(i) There are four aces in a pack of 52 cards, out of which one can be drawn in 4^C1 Therefore Favourable number of cases = 4^C1 = 4.
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