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Probability Probability

27. A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(c) Calculate the probability that the card is (i) an ace (ii) black card.

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alok kumar singh

Contributor-Level 10

27. (a) Since there are 52 cards in the sample space,

n (S) = 52.

So, there are 52 sample points.

(b) In a deck of 52 cards there are 4 ace cards of which only one is of spades.

Hence, if A be an event of getting an ace of spades.

n (A) = 1

So, P (A) = $\frac{n (A)}{n (S)} = \frac{1}{52}$ .

(c) (i) Let B be an event of drawing an ace. As there are 4 ace cards we have,

n (B) = 4

So, P (B) = $\frac{n (B)}{n (S)} = \frac{4}{52} = \frac{1}{13}$ .

(ii) Let D be an event of drawing black cards. Since there are 26 black cards we have,

n (D) = 26.

So, P (D) = $\frac{n (D)}{n (S)} = \frac{26}{52} = \frac{1}{2}$

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Probability Probability

26. A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

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alok kumar singh

Contributor-Level 10

26. The sample space of throwing s dice is

S = {1, 2, 3, 4, 5, 6}, n (S) = 6.

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n (A)}{n (S)} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So, n (B) = 4

Therefore P (B) = $\frac{n (B)}{n (S)} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be event such that a number less than or equal to one will appear. Then,

C = {1}

So, n (C) = 1

? P (C) = $\frac{n (C)}{n (S)} = \frac{1}{6}$

(iv) Let D be event such that a number more than 6 appears. Then,

D =∅

So, n (D) = 0

? P (D) = $\frac{n (D)}{n (S)} = \frac{0}{6} = 0$

(v) Let E be event such that a number less than 6 appears. Then

E = {1, 2, 3, 4, 5}

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26. The sample space of throwing s dice is

S = {1, 2, 3, 4, 5, 6}, n (S) = 6.

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n (A)}{n (S)} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So, n (B) = 4

Therefore P (B) = $\frac{n (B)}{n (S)} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be event such that a number less than or equal to one will appear. Then,

C = {1}

So, n (C) = 1

? P (C) = $\frac{n (C)}{n (S)} = \frac{1}{6}$

(iv) Let D be event such that a number more than 6 appears. Then,

D =∅

So, n (D) = 0

? P (D) = $\frac{n (D)}{n (S)} = \frac{0}{6} = 0$

(v) Let E be event such that a number less than 6 appears. Then

E = {1, 2, 3, 4, 5}

So, n (E) = 5

? P (E) = $\frac{n (E)}{n (S} = \frac{5}{6}$

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Probability Probability

25. A coin is tossed twice, what is the probability that atleast one tail occurs?

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alok kumar singh

Contributor-Level 10

25. When a coin is tossed twice we have the sample space

S = {TT, TH, HT, HH}

So, n (S) = 4

Let A be the event of getting at least one tail.

Then, A = {TH, HT, TT}

So, n (A) = 3

Therefore, P (A)= Numver of outcome favorable to A/Total possible outcomes = $\frac{}{} = \frac{n (A)}{n (S)}$ .

$= \frac{3}{4}$

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Probability Probability

24. Which of the following can not be valid assignment of probabilities for outcomes

of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇,}

Assignment ω ₁ ω ₂ ω ₃ ω ₄ ω ₅ ω ₆ ω ₇

(a) 0.1 &nbs

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24. Which of the following can not be valid assignment of probabilities for outcomes

of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇,}

Assignment ω ₁ ω ₂ ω ₃ ω ₄ ω ₅ ω ₆ ω ₇

(a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6

(b)

(c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7

(d) – 0.1 0.2 0.3 0.4 – 0.2 0.1 0.3

(e) $\frac{\bar{1}}{\bar{14}}$ $\frac{\bar{2}}{\bar{14}}$ $\frac{\bar{3}}{\bar{14}}$ $\frac{\bar{4}}{\bar{14}}$ $\frac{\bar{5}}{\bar{14}}$ $\frac{\bar{6}}{\bar{14}}$ $\frac{\bar{15}}{\bar{14}}$

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alok kumar singh

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24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is 'one' the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

$= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = \frac{7}{7} = 1$ .

P (S) = 1

Hence, the given assignment of probability is valid.

(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7

= 2.8

i.e., P (S) > 1

As probability of the sample space S should always be '1'. The given assignment is invalid.

(d) Here P (W1) = –0.1 is negative.

As p

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24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is 'one' the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

$= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = \frac{7}{7} = 1$ .

P (S) = 1

Hence, the given assignment of probability is valid.

(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7

= 2.8

i.e., P (S) > 1

As probability of the sample space S should always be '1'. The given assignment is invalid.

(d) Here P (W1) = –0.1 is negative.

As probability of happening an event should always be positive and less than or equal to 1.

The given assignment is invalid.

(e) As P (W7) = $\frac{15}{14} = 1.07$

i.e. probability of happening an event is greater than one. The given assignment is invalid.

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Probability Probability

23. Refer to question 6 above, state true or false: (give reason for your answer)

(i) A and B are mutually exclusive

(ii) A and B are mutually exclusive and exhaustive

(iii) A = B′

(iv) A and C are mutually exclusive

(v) A and B′ are mutually exclusive.

(vi) A′, B′, C are mutually exclusive and exhaustive.

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alok kumar singh

Contributor-Level 10

23. (i) A ∩B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

A ∩B =∅

Hence, A and B are mutually exclusive.

The given statement is true.

(ii) A ∪B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1,

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23. (i) A ∩B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

A ∩B =∅

Hence, A and B are mutually exclusive.

The given statement is true.

(ii) A ∪B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

= { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

= S

Hence, A B = S and A∩ B =∅

A and B are mutually exclusive and mutually exhaustive.

So, the give statement is true.

(iii) A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B = S – B = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} – { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, (4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

= { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B' = A

Hence, the given relation is true.

(iv) A ∩C = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, (1)}

= { (2, 1), (2, 2), (2, 3), (4, 1)}≠∅

So, A∩ C≠∅

i.e., A and C are not mutually exclusive.

Hence, the given statement is false.

(v) A∩ B' = A∩ A = A≠∅ { B' = A}

Hence A∩B'≠∅ i.e., A and B are not mutually exclusive.

So, the given statement is false.

(vi) As A' = B

B = A

So, A'∩B = B∩A =∅

A'∩ C = B∩C = { (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}=∅

B'∩C = A∩C = { (2, 1), (2, 2), (2, 3), (4, 1)}=∅

Hence, A', B' and C are not mutually exclusive.

The given statement is false.

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Probability Probability

22. Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5.

Describe the events

(i) A′ (ii) Not B (iii) A or B

(iv) A and B (v) A but not C

(vi) B or C (vii) B and C (viii) A ∩ B′ ∩ C′

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alok kumar singh

Contributor-Level 10

22. The sample space of the experiment is

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Now, A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(3, 1), (3, 2), (3, 3), (3, 4

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22. The sample space of the experiment is

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Now, A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

C = { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

(i) A? = S – A = Sample space – getting even number on 1st dice = getting odd number on 1st dice = B

(ii) B' = S – B = Sample space – getting odd number on 1st dice = getting even number on 1st dice = A

(iii) A or B = A∪B = (getting even on 1st die)∪ (getting odd o 1st dice) = Sample space = S

(iv) A and B = A∩ B = (getting even on 1st dice)∩ (getting odd on 1st dice) =∅

(v) A but not C = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} – { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

A – C = { (2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(vi) B or C = B∪C = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}∪ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

= { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(vii) B and C = B∩ C = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} = { (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2), (4, 1)}

(viii) A∩B ∩C = A ∩A ∩C (as B' = A)

= A ∩C' (A ∩A = A)

So, C' = S – C = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} – { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

C' = { (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

So, A ∩B ∩ C' = (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

{ (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, (3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

= { (2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

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Probability Probability

21. Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive.

(iii) Two events, which are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive.

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alok kumar singh

Contributor-Level 10

21. The sample space of the experiment is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

(i) Let A and B be two events such that

A: only head occurs

A = {HHH}

B: only tail occurs

B = {TTT}

And A ∩B = {HHH}∩ {TTT} =∅

(ii) Let A, B and C be two events such that

A: at most one head (i.e., the event in which we get maximum one head; one head or no head at all) occurs

So, A = {HTT, THT, TTH, TTT}

B: exactly two head occurs

So, B = {HHT, HTH, THH}

C: all are heads

So, C = {HHH}

Hence, A ∩ B = {HTT, THT, TTH, TTT} ∩ {HHT, HTH, THH} =∅

A ∩C = {HTT, THT, TTH, TTT} ∩ {HHH} =∅

B∩ C = {HHT, HTH, THH}∩ {HHH} =∅

And A∪ B ∪C = {HTT, T

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Probability Probability

20. Three coins are tossed once. Let A denote the event 'three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails

show and D denote the event 'a head shows on the first coin”. Which events are

(i) Mutually exclusive ? (ii) Simple? (iii) Compound?

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alok kumar singh

Contributor-Level 10

20. The sample space of the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Now, A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

(i) A∩ B = {HHH}∩ {HHT, HTH, THH} =∅

A ∩ C = {HHH}∩ {TTT} =∅

A∩ D = {HHH}∩ {HHH, HTH, HTT} = {HHH}

B ∩ C = {HHT, HTH, THH}∩ {TTT} =∅

B∩ D = {HHT, HTH, THH}∩ {HHH, HHT, HTH, HTT} = {HHT, HTH}

C∩ D = {TTT}∩ {HHH, HHT, HTH, HTT} =∅

Hence, (A, B), (A, C), (B, C) and (C, D) are pairs of mutually exclusive events.

(ii) Simple event are those which has only one sample point. So, the simple event are A and C.

(iii) Event having more than one sample point are called com

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Probability Probability

19. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

A: the sum is greater than 8, B: 2 occurs on either die

C: the sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?

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alok kumar singh

Contributor-Level 10

19. The sample space of the experiment is

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

And A = { (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

B = { (1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}

C = { (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

Now, A∩

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Probability Probability

18. A die is thrown. Describe the following events:

(i) A: a number less than 7

(ii) B: a number greater than 7

(iii) C: a multiple of 3 (iv) D: a number less than 4

(v) E: an even number greater than 4 (vi) F: a number not less than 3

Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′

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alok kumar singh

Contributor-Level 10

18. The sample space of the experiment is

S = {1, 2, 3, 4, 5, 6}

(i) A = {1, 2, 3, 4, 5, 6}

(ii) B =∅

(iii) C = {3, 6}

(iv) D = {1, 2, 3}

(v) E = {6}

(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6}∩ {∅}= ∅h

= {1, 2, 3, 4, 5, 6}

A ∩B = {1, 2, 3, 4, 5, 6}∩ {∅}

B ∪C = ∅∪ {3, 6) = {3, 6}

E∩ F = {6}∩ {3, 4, 5, 6} = {6}

D∩ E = {1, 2, 3}∩ {6} =∅

A – C = {1, 2, 3, 4, 5, 6} – {3, 6} = {1, 2, 4, 5}

D – E = {1, 2, 3} – {6} = {1, 2, 3}

E ∩F' = E∩ (S – F) = {6}∩ [ {1, 2, 3, 4, 5, 6} – {3, 4, 5, 6}]

= {6} ∩ {1, 2}

=∅

F' = S – F = {1, 2, 3, 4,

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