**Chapter 33: ****Probability**** – Exercise 33.2**

**Probability**** – Exercise 33.2 – Q.1**

Since a coin is tossed, so the total nos of elementary events is

S = {H, 7}

⟹ n(s) = 2

Also, the total no, of events

= {H}, {T}, {H, T}, {7, H}

= 4

**Probability**** – Exercise 33.2 – Q.2**

Since we are tossing two coins so, the all events associated with random experiment are

{HH), {HT}, {TH}, {TT}, {HH, HT}, {HH,TH}, {HH ,TT}, {HT ,TH}, {HT ,TT}, {TH,TT}{HH,HT,TH}, {HH, HT, TT}, {HH,TH,TT), {HT, TH, TT}, {HH, HT, TH, TT},

Total = 15

From above the elementary events are {HH}, {HT}, {TH}, {TT}

Total elementary event = 4

**Probability**** – Exercise 33.2 – Q.3**

A - Getting three heads = {HHH} =1

B - Getting two heads and one tail = {HHT, THH, HTH} = 3

C - Getting three tails = {TTT) = 1

D - Getting a head on the first coin = {HHH, HHT, HTH, HTT} = 4

i) Which pairs of events are mutually exclusive?

We know that A and 8 are said to be mutually exclusive if A ∩ B = ∅

a) A and B b) A and C c) B and C d) C and D are mutually exclusive

ii) Which events are elementary events?

A and C are elementary events.

iii) Which events are compound events?

Clearly B and D are union of three events and 4 events respectively.

∴ B and D are compound events.

**Probability**** – Exercise 33.2 – Q.4**

Since a die was thrown. So elementary events are

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

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

ii) B = Getting a number greater than 7.

B = ∅ [ ∵ A die has 1, 2, 3, 4, 5, 6 members only]

iii) C = Getting a multiple of 3.

C = {3, 6}

iv) D - Getting a number less than 4,

D = (1, 2, 3)

v) E = Getting an even number greater than 4.

E = {6}

vi) F = Getting a number not less than 3.

F = {3, 4, 5, 6}

Also, A ∪ B = {1, 2, 3, 4, 5, 6}

A ∩ B = {∅}

B ∩ C = {∅}

E ∩ F = {6}

D ∩ F = {3}

= 1 - F = {1, 2}

**Probability**** – Exercise 33.2 – Q.5**

Sample space associated with given event is

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

(i) A = {HTT, THT, TTH}, B = {HHT, THE, HTH}

A and B are mutually exclusive events

(ii) A = {HHH, TTT}, B = {HHT, THH, HTH} and

C = {HTT, THT, TTH}

Above events are exhaustive and mutually exclusive events.

Because A ∩ B = B ∩ C = C ∩ A = ∅ and A ∪ B ∪ C = S

(iii) A = {HHH, HHT, THH, HTH}

B = {HHT, THH, HTH, HTT, THT, TTH, TTT}

A and B are not mutually exclusive because A ∩ B = ∅

(iv) A = {HHH, HHT, THH}, B = {THT, TTH, TTT}

A and B are mutually exclusive but not exhaustive A ∩ B = 0 and A ∪ B ≠ S

**Probability**** – Exercise 33.2 – Q.6**

(i) A = both numbers are odd = {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5),(5,1), (5,3), (5,5)}

(ii) B = both numbers are even

= {(2,2), (2, 4), (2,6), (4,2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

(iii) C = Sum of numbers is less than 6

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

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

A ∩ B = ∅

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

A ∩ C = {(1, 1), (1, 3), (3, 1)}

B ∩ C = ∅

**Probability**** – Exercise 33.2 – Q.7**

A = Getting an even number on the first die.

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 = Getting an odd number on the first die.

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 = Getting at most 5 as sum of the numbers on the two dice.

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

D = Getting the sum of the numbers on the dice > 5 but < 10.

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

F = Getting an odd number on one of the dice.

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

Its clear that A and B are mutually exclusive events and A ∩ B = ∅

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

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

A ∩ E = {(4, 6), (6, 4), (6, 5), (6, 6)}

A ∪ F = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (5, 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 ∩ F = {(2, 1), (2,3), (2,5), (4,1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}

(ii)

a) T A ∩ B = ∅

b) T A ∩ B = ∅ and A ∪ B = S

c) F A ∩ C ≠ ∅

d) F A ∩ B = ∅ and A ∪ B ≠ S

e) T C ∩ D = D ∩ E = C ∩ E = Φ and C ∪ D ∪ E = S

f) T A^{1} ∩ B^{1} = ∅

g) F A ∩ F ≠ ∅

**Probability**** – Exercise 33.2 – Q.8**

We have four slips of paper with numbers 1, 2, 3 & 4.

A person draws two slips without replacement.

∴ Number of elementary events = ^{4}C_{2}

A = The number on the first slip is larger than the one on the second slip

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

B = The number on the second slip is greater than 2

∴ B = {(1,3), (2,3) , (1,4), (2, 4), (3, 4), (4,3)}

C = The sum of the numbers on the two slips is 6 or 7

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

and,

D = The number on the second slips is twice that on the first slip

D = {(1, 2), (2, 4)}

and, A and D form a pair of mutually exclusive events as A ∩ B = ∅

**Probability**** – Exercise 33.2 – Q.9**

(i) Sample space for picking up a card from a set of 52 cards is set of 52 cards itself

(ii) For an event of chosen card be black faced card, event is a set of jack, king, queen of spades and clubs,