Students can go through AP Inter 1st Year Maths Notes 14th Lesson Probability will help students in revising the entire concepts quickly.
Probability Class 11 Notes AP Inter 1st Year Maths 14th Lesson
→ Event: A subset of the sample space
→ Impossible event: The empty set
→ Sure event: The whole sample space
→ Complementary event or ‘not event’: The set A’ or S – A
→ Event A or B: The set A ∪ B
→ Event A and B: The set A ∩ B
→ Event A and not B: The set A – B
→ Mutually exclusive event: A and B are mutually exclusive if A ∩ B = φ
→ Exhaustive and mutually exclusive events: Events E1, E2,…, En are mutually exclusive and exhaustive if E1 ∪ E2 ∪ …. ∪ En = S and Ei ∩ Ej = φ ∀ i ≠ j
→ Probability: Number P(ωi) associated with sample point ωi such that
(i) 0 ≤ P (ωi) ≤ 1
(ii) ΣP(ωi) for all ωi ∈ S = 1
(iii) P(A) = ΣP(ωi) for all ωi ∈ A.
The number P(ωi) is called the probability of the outcome ωi.
→ Equally likely outcomes: All outcomes with equal probability.
→ Probability of an event: For a finite sample space with equally likely outcomes,
Probability of an event P(A) = n(A)/n(S), where n(A) = number of elements in the set A, n(S) = number of elements in the set S.
→ If A and B are any two events, then P(A or B) = P(A) + P(B) – P(A and B) equivalently,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
→ If A and B are mutually exclusive, then P(A or B) = P(A) + P(B)
→ If A is any event, then P(not A) = 1 – P(A)
Example Problems
Question 1.
Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events
(i) A or B
(ii) A and B
(iii) A but not B
(iv) ‘not A.’
Solution:
Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5}, and B = {1, 3, 5}
Obviously
(i) ‘A or B’ = A ∪ B = {1, 2, 3, 5}
(ii) ‘A and B’ = A ∩ B = {3, 5}
(iii) ‘A but not B’ = A – B = {2}
(iv) ‘not A’ = A’ = {1, 4, 6}
Question 2.
Two dice are thrown, and the sum of the numbers that come up on the dice is noted. Let us consider the following events associated with this experiment
A: ‘the sum is even’,
B: ‘the sum is a multiple of 3’,
C: ‘the sum is less than 4’ and
D: ‘the sum is greater than 11’.
Which pairs of these events are mutually exclusive?
Solution:
There are 36 elements in the sample space S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}.
Then A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
B = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)}
C = {(1, 1), (2, 1), (1, 2)}
and D = {(6, 6)}
We find that A ∩ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)} ≠ φ
Therefore, A and B are not mutually exclusive events.
Similarly, A ∩ C ≠ φ, A ∩ D ≠ φ, B ∩ C ≠ φ, and B ∩ D ≠ φ.
Thus, the pairs of events, (A, C), (A, D), (B, C), (B, D) are not mutually exclusive events.
Also, C ∩ D = φ, and so C and D are mutually exclusive events.
Question 3.
A coin is tossed three times. Consider the following events.
A: ‘No head appears’,
B: ‘Exactly one head appears’ and
C: ‘At least two heads appear’.
Do they form a set of mutually exclusive and exhaustive events?
Solution:
The sample space of the experiment is S = {HHH, HHT, HTH, THH, HIT, THT, TTH, TTT}
and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH}
Now A ∪ B ∪ C = {TTT, HIT, THT, TTH, HHT, HTH, THH, HHH} = S
Therefore, A, B, and C are exhaustive events.
Also, A ∩ B = φ, A ∩ C = φ and B ∩ C = φ
Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive.
Hence, A, B, and C form a set of mutually exclusive and exhaustive events.
Question 4.
Let a sample space be S = {ω1, ω2, …., ω6}. Which of the following assignments of probabilities to each outcome are valid?
Solution:
(a) Condition (i): Each of the numbers p(ω1) is positive and less than one.
Condition (ii): Sum of probabilities = \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\) = 1
Therefore, the assignment is valid.
(b) Condition (i): Each of the numbers p(ω1) is either 0 or 1.
Condition (ii): Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1
Therefore, the assignment is valid.
(c) Condition (i): Two of the probabilities p(ω5) and p(ω6) are negative; the assignment is not valid.
(d) Since p(ω6) = \(\frac {3}{2}\) > 1, the assignment is not valid.
(e) Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1, the assignment is not valid.
Question 5.
One card is drawn from a well-shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be
(i) a diamond
(ii) not an ace
(iii) a black card (i.e., a club or a spade)
(iv) not a diamond
(v) not a black card
Solution:
When a card is drawn from a well-shuffled deck of 52 cards, the number of possible outcomes is 52.
(i) Let A be the event ‘the card drawn is a diamond.’
Clearly, the number of elements in set A is 13.
Therefore, P(A) = \(\frac{13}{52}=\frac{1}{4}\)
i.e., probability of a diamond card = \(\frac {1}{4}\)
(ii) We assume that the event ‘Card drawn is an ace’ is B.
Therefore, ‘Card drawn is not an ace’ should be B.
We know that P(B’) = 1 – P(B)
= 1 – \(\frac {4}{52}\)
= 1 – \(\frac {1}{13}\)
= \(\frac {12}{13}\)
(iii) Let C denote the event ‘card drawn is a black card.’
Therefore, the number of elements in the set C = 26
ie P(C) = \(\frac{26}{52}=\frac{1}{2}\)
Thus, probability of a black card = \(\frac {1}{2}\)
(iv) We assumed in (i) above that A is the event ‘card drawn is a diamond’, so the event ‘card drawn is not a diamond’ may be denoted as A’ or ‘not A.’
Now P(not A) = 1 – P(A)
= 1 – \(\frac {1}{4}\)
= \(\frac {3}{4}\)
(v) The event ‘card drawn is not a black card’ may be denoted as C’ or ‘not C’.
We know that P(not C) = 1 – P(C)
= 1 – \(\frac {1}{2}\)
= \(\frac {1}{2}\)
Therefore, probability of not a black card = 1 – \(\frac {1}{2}\) = \(\frac {1}{2}\)
Question 6.
A bag contains 9 discs, of which 4 are red, 3 are blue, and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be
(i) red,
(ii) yellow,
(iii) blue,
(iv) not blue,
(v) either red or blue.
Solution:
There are 9 discs in all, so the total number of possible outcomes is 9.
Let the events A, B, C be defined as
A: ‘The disc drawn is red.’
B: ‘The disc drawn is yellow.’
C: ‘The disc drawn is blue’.
(i) The number of red discs = 4,
i.e., n(A) = 4
Hence P(A) = \(\frac {4}{9}\)
(ii) The number of yellow discs = 2
i.e., n(B) = 2
Therefore, P(B) = \(\frac {2}{9}\)
(iii) The number of blue discs = 3,
i.e., n(C) = 3
Therefore, P(C) = \(\frac{3}{9}=\frac{1}{3}\)
(iv) Clearly, the event ‘not blue’ is ‘not C’.
We know that P(not C) = 1 – P(C)
Therefore, P(not C) = 1 – \(\frac {1}{3}\) = \(\frac {2}{3}\)
(v) The event ‘either red or blue’ may be described by the set ‘A or C.’
Since A and C are mutually exclusive events, we have
P(A or C) = P(A ∪ C) = P(A) + P(C)
= \(\frac{4}{9}+\frac{1}{3}\)
= \(\frac {7}{9}\)
Question 7.
Two students, Anil and Ashima, appeared in an examination. The probability that Anil will qualify for the examination is 0.05, and that Ashima will qualify for the examination is 0.10. The probability that both will qualify for the examination is 0.02. Find the probability that
(a) Neither Anil nor Ashima will qualify for the examination.
(b) At least one of them will not qualify for the examination and
(c) Only one of them will qualify for the examination.
Solution:
Let E and F denote the events that Anil and Ashima will qualify for the examination, respectively.
Given that P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02. Then
(a) The event ‘both Anil and Ashima will not qualify for the examination’ may be expressed as E’ ∩ F’.
Since E’ is ‘not E’, i.e., Anil will not qualify for the examination, and F’ is ‘not F’,
i.e., Ashima will not qualify for the examination.
Also E’ ∩ F’ = (E ∪ F)’ (by Demorgan’s Law)
Now P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
or P(E ∪ F) = 0.05 + 0.10 – 0.02 = 0.13
Therefore P(E’ ∩ F’) = P(E ∪ F)’
= 1 – P(E ∪ F)
= 1 – 0.13
= 0.87
(b) P(at least one of them will not qualify) = 1 – P(both of them will qualify)
= 1 – 0.02
= 0.98
(c) The event that only one of them will qualify for the examination is the same as the event either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify)
i.e., E ∩ F’ or E’ ∩ F, where E ∩ F’ and E’ ∩ F are mutually exclusive.
Therefore, P(only one of them will qualify) = P(E ∩ F’ or E’ ∩ F)
= P(E ∩ F’) + P(E’ ∩ F)
= P(E) – P(E ∩ F) + P(F) – P(E ∩ F)
= 0.05 – 0.02 + 0.10 – 0.02
= 0.11
Question 8.
A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(a) No man?
(b) One man?
(c) Two men?
Solution:
The total number of persons = 2 + 2 = 4
Out of these four persons, two can be selected in 4C2 ways.
(a) No men in the committee of two means there will be two women in the committee.
Out of two women, two can be selected in 2C2 = 1 way.
Therefore P(no man) = \(\frac{{ }^2 C_2}{{ }^4 C_2}=\frac{1 \times 2 \times 1}{4 \times 3}=\frac{1}{6}\)
(b) One man in the committee means that there is one woman.
One man out of 2 can be selected in 2C1 ways, and one woman out of 2 can be selected in 2C1 ways.
Together, they can be selected in 2C1 × 2C1 ways.
Therefore P(One man) = \(\frac{{ }^2 C_1 \times{ }^2 C_1}{{ }^4 C_2}=\frac{2 \times 2}{2 \times 3}=\frac{2}{3}\)
(c) Two men can be selected in 2C2 ways.
Hence P(Two men) = \(\frac{{ }^2 C_2}{{ }^4 C_2}=\frac{1}{{ }^4 C_2}=\frac{1}{6}\)
Question 9.
On her vacations, Veena visits four cities (A, B, C, and D) in a random order. What is the probability that she visits?
(i) A before B?
(ii) A before B and B before C?
(iii) A first and B last?
(iv) Which A is either first or second?
(v) A just before B?
Solution:
The number of arrangements (orders) in which Veena can visit four cities, A, B, C, or D, is 4! i.e., 24.
Therefore, n(S) = 24
Since the number of elements in the sample space of the experiment is 24, all of these outcomes are considered to be equally likely.
A sample space for the experiment is S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CBAD, CDAB, CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}
(i) Let the event ‘she visits A before B’ be denoted by E.
Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB, ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}
Thus P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{12}{24}=\frac{1}{2}\)
(ii) Let the event ‘Veena visits A before B and D before C’ be denoted by F.
Here, F = {ABCD, DABC, ABDC, ADBC}
Therefore, P(F) = \(\frac{\mathrm{n}(\mathrm{~F})}{\mathrm{n}(\mathrm{~S})}=\frac{4}{24}=\frac{1}{6}\)
Students are advised to find the probability in case of (iii), (iv), and (v).
Question 10.
Find the probability that when a hand of 7 cards is drawn from a well-shuffled deck of 52 cards, it contains
(i) all Kings
(ii) 3 Kings
(iii) at least 3 Kings
Solution:
Total number of possible hands = 52C7
(i) Number of hands with 4 Kings = 4C4 × 48C3
(the other 3 cards must be chosen from the rest 48 cards)
Hence P(a hand will have 4 Kings) = \(\frac{{ }^4 C_4 \times{ }^{48} C_3}{{ }^{52} C_7}=\frac{1}{7735}\)
(ii) Number of hands with 3 Kings and 4 non-King cards = 4C3 × 48C4
Therefore P(3 Kings) = \(\frac{{ }^4 C_3 \times{ }^{48} C_4}{{ }^{52} C_7}=\frac{9}{1547}\)
(iii) P(at least 3 Kings) = P(3 Kings or 4 Kings)
= P(3 Kings) + P(4 Kings)
= \(\frac{9}{1547}+\frac{1}{7735}\)
= \(\frac {46}{7735}\)
Question 11.
If A, B, C are three events associated with a random experiment, prove that P(A ∪ B ∪ C) = P(A) + P (B) + P (C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
Solution:
Consider E = B ∪ C so that
P(A ∪ B ∪ C) = P(A ∪ E) = P (A) + P (E) – P (A ∩ E) ……….(1)
Now P(E) = P(B ∪ C) = P(B) + P(C) – P(B ∩ C)
Also, A ∩ E = A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) [Using the distribution property of intersection of sets over the union]
Thus P(A ∩ E) = P(A ∩ B) + P(A ∩ C) – P[(A ∩ B) ∩ (A ∩ C)]
P(A ∩ E) = P(A ∩ B) + P(A ∩ C) – P[A ∩ B ∩ C] ……….(3)
Using (2) and (3) in (1), we get
P[A ∪ B ∪ C] = P(A) + P(B) + P(C) – P(B ∩ C) – P(A ∩ B) – P(A ∩ C) + P(A ∩ B ∩ C)
Question 12.
In a relay race, there are five teams: A, B, C, D, and E.
(a) What is the probability that A, B, and C finish first, second, and third, respectively?
(b) What is the probability that A, B, and C are the first three to finish (in any order)?
(Assume that all finishing orders are equally likely)
Solution:
If we consider the sample space consisting of all finishing orders in the first three places,
we will have 5P3 = \(\frac{5!}{(5-3)!}\)
= 5 × 4 × 3
= 60
Sample points, each with a probability of \(\frac {1}{60}\)
(a) A, B, and C finish first, second, and third, respectively.
There is only one finishing order for this, i.e., ABC.
Thus P(A, B and C finish first, second and third respectively) = \(\frac {1}{60}\)
(b) A, B, and C are the first three finishers.
There will be 3! arrangements for A, B, and C.
Therefore, the sample points corresponding to this event will be 3! in number.
So P(A, B and C are first three to finish) = \(\frac{3!}{60}=\frac{6}{60}=\frac{1}{10}\)
Multiple Choice Questions
Question 1.
Let A and B be two events such that P(A) = 0.8, P(B) = 0.7. Then P(A ∩ B)
(1) ≤ 0.6
(2) ≥ 0.5
(3) ≥ 0.7
(4) ≤ 0.4
Answer:
(2) ≥ 0.5
Question 2.
The probability that a leap year selected at random will contain 53 Sundays is
(1) \(\frac {1}{7}\)
(2) \(\frac {2}{7}\)
(3) \(\frac {6}{7}\)
(4) \(\frac {5}{7}\)
Answer:
(2) \(\frac {2}{7}\)
Question 3.
If two dice are thrown, then the probability of getting the same number on both faces is
(1) \(\frac {2}{6}\)
(2) \(\frac {3}{6}\)
(3) \(\frac {5}{6}\)
(4) \(\frac {1}{6}\)
Answer:
(4) \(\frac {1}{6}\)
Question 4.
If two dice are thrown, then the probability of getting a total score of seven is
(1) \(\frac {1}{6}\)
(2) \(\frac {2}{6}\)
(3) \(\frac {3}{6}\)
(4) \(\frac {5}{6}\)
Answer:
(1) \(\frac {1}{6}\)
Question 5.
If A and B are two events of a sample space with P(A ∪ B) = P(A) + P(B), then they are
(1) Mutually exclusive
(2) Exhaustive
(3) 1 and 2
(4) None
Answer:
(1) Mutually exclusive
Question 6.
An integer is picked from 1 to 20 (both inclusive). Then the probability that it is a prime
(1) \(\frac {3}{20}\)
(2) \(\frac {3}{5}\)
(3) \(\frac {2}{5}\)
(4) \(\frac {1}{5}\)
Answer:
(3) \(\frac {2}{5}\)
Question 7.
If P(A) = 0.5, P(B) = 0.3. (Given that A and B are mutually exclusive) Then (A’ ∩ B’) = _____
(1) 0.6
(2) 0.5
(3) 0.7
(4) 0.2
Answer:
(4) 0.2
Question 8.
One card is drawn at random from a well-shuffled deck of 52 cards. Then the probability that the card is not red.
(1) \(\frac {2}{5}\)
(2) \(\frac {1}{5}\)
(3) \(\frac {1}{2}\)
(4) \(\frac {1}{3}\)
Answer:
(3) \(\frac {1}{2}\)
Question 9.
Which of the following is false?
(1) P(E) = 0 ⇔ E is an impossible event
(2) 0 ≤ P(E) < 1
(3) P(E) = 1 ⇔ E is a certain event
(4) P(E) + P(\(\bar{E}\)) = 1
Answer:
(2) 0 ≤ P(E) < 1
Question 10.
Which of the following statements is not correct? P is a probability function of the sample space S.
(1) P(E) > 0 ∀ E ∈ P(S)
(2) P(S) = 1
(3) P(φ) = 0
(4) P(E1 ∩ E2) = P(E1) + P(E2) where E1, E2 are mutually exclusive
Answer:
(4) P(E1 ∩ E2) = P(E1) + P(E2) where E1, E2 are mutually exclusive