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 A1 ∩ B1 = ∅
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 = 4C2
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,