NCERT Solutions for Class 11 Math Chapter 16 - Probability

Answer:

Sample space = {HH, HT, TH, TT}, where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

(i) Sample space = {(1, 1), (1, 2),.......,(1, 6) ;
(2, 1), (2, 2),.........,(2, 6) ; (6, 1), (6, 2),.........., (6, 6)}.
(ii) No. of elements = 6 × 6 = 36.

Answer:

Let Q denote a 1 rupee coin, H a 2 rupee coin and R a 5 rupee coin.
The first coin may be any one of Q, H and R.
Corresponding to Q, the second draw may be H or R. Thus the two draws may be QH or QR.
Similarly corresponding to H, the second draw be Q or R. Thus the two draws may be HQ or HR.
Lastly the two draws may be RH or RQ.
Hence, the sample space = {QH, QR, HQ, HR, RH, RQ}.

Answer:

The number of accidents can be 0 or 1 or 2; and so on.
$\therefore$ Sample space = {0, 1, 2, ... }.

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Answer:

Sample space = {H, TH, TTH, TTTH, TTTTH,.......},
where H $\equiv$ Head and
T $\equiv$ Tail.

Answer:

Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}.
‘A or B’ = A $\mathrm{\upsilon }$ B = {1, 2, 3, 5}.

Answer:

Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}.
‘A and B’ = A $\cap$ B = {3, 5}.

Answer:

Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}.
‘A but not B’ = A – B = {2}.

Answer:

Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}.
‘not A’ = A′ = {1, 4, 6}.

Answer:

We have 36 elements in the sample space :
S = {(x, y) : x, y = 1, 2, 3, 4, 5, 6}.
$\therefore$ 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)}
[_ 1 + 1 = 2 (even) ; etc.]
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)}
[_ 1 + 2 = 3, which is a multiple of 3 ; etc.]
C = {(1, 1), (1, 2), (2, 1)} [_ 1 + 1 = 2 < 4 ; etc.]
D = {(6, 6)}. [ 6 + 6 = 12 > 11]
Now A $\cap$ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1),
(6, 6)} $\ne$ $\mathrm{\varphi }$
$\therefore$ A and B are not mutually exclusive events.
Similarly A $\cap$ C $\equiv$ $\mathrm{\varphi }$, A $\cap$ D $\equiv$ $\mathrm{\varphi }$, B $\cap$ C $\equiv$ $\mathrm{\varphi }$
and B $\cap$ D $\equiv$ $\mathrm{\varphi }$.
Thus the pairs (A, C), (A, D), (B, C) and (B, D) are not mutually exclusive
But C $\cap$ D = $\mathrm{\varphi }$ and as each C, D are mutually exclusive events.

Answer:

The sample space is :
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Here A= {TTT}, B = {HTT, THT, TTH} and C= {HHT, HTH, THH, HHH}.
(i) Now A$\mathrm{\upsilon }$B$\mathrm{\upsilon }$C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S.
$\therefore$ A, B and C are exhaustive events.
(ii) A $\cap$ B = $\mathrm{\varphi }$, B $\cap$ C = $\mathrm{\varphi }$ and A $\cap$ C = $\mathrm{\varphi }$
$\Rightarrow$ events are pair-wise disjoint
$\Rightarrow$ events are mutually exclusive.
Hence, A, B and C form a set of mutually exclusive and exhaustive events.

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Total number of discs = 4 + 3 + 2 = 9.
Let the events A, B and C be :
A : Disc drawn is red
B : Disc drawn is yellow
and C : Disc drawn is blue.

Answer:

Total number of discs = 4 + 3 + 2 = 9.
Let the events A, B and C be :
A : Disc drawn is red
B : Disc drawn is yellow
and C : Disc drawn is blue.

Answer:

Total number of discs = 4 + 3 + 2 = 9.
Let the events A, B and C be :
A : Disc drawn is red
B : Disc drawn is yellow
and C : Disc drawn is blue.

Answer:

Total number of discs = 4 + 3 + 2 = 9.
Let the events A, B and C be :
A : Disc drawn is red
B : Disc drawn is yellow
and C : Disc drawn is blue.

Answer:

Total number of discs = 4 + 3 + 2 = 9.
Let the events A, B and C be :
A : Disc drawn is red
B : Disc drawn is yellow
and C : Disc drawn is blue.

Answer:

Let A and B be two events :
A : Anil will qualify the examination
B : Ashima will qualify the examination.
We have : P (A) = 0·05, P (B) = 0·10 and P (A $\cap$ B) = 0·02.
A′ is ‘not A’ and B′ is ‘not B’.
‘Both Anil and Ashima will not qualify the examination’ may be expressed as A′ $\cap$ B′.
Also A′ $\cap$ B′ = (A $\mathrm{\upsilon }$ B)′
[By De–Morgan’s Law]
Now P (A $\mathrm{\upsilon }$ B) = P (A) + P (B) – P (A $\cap$ B)
= 0·05 + 0·10 – 0·02 = 0·13
$\therefore$ P (A′ $\cap$ B′) = P (A $\mathrm{\upsilon }$ B)′ = 1 – 0·13 = 0·87.
Hence, the probability that both Anil and Ashima will not qualify the examination = 0·87.

Answer:

Let A and B be two events :
A : Anil will qualify the examination
B : Ashima will qualify the examination.
We have : P (A) = 0·05, P (B) = 0·10 and P (A $\cap$ B) = 0·02.
P (at least one of Anil and Ashima will not qualify)
= P (A $\cap$ B′ or A′ $\cap$ B or A′ $\cap$ B′)
But S = {A$\cap$B, A$\cap$B′, A′$\cap$B, A′$\cap$B′}
$\Rightarrow$P (A $\cap$ B) + P (A $\cap$ B′) + P (A′ $\cap$ B)
+ P (A′ $\cap$ B′) = 1
$\Rightarrow$ P (A $\cap$ B′) + P (A′ $\cap$ B) + P (A′ $\cap$ B′)
= 1 – P (A $\cap$ B)
= 1 – 0·02 = 0·98.
Hence, the probability that at least one of Anil and Ashima will not qualify = 0·98.

Answer:

Let A and B be two events :
A : Anil will qualify the examination
B : Ashima will qualify the examination.
We have : P (A) = 0·05, P (B) = 0·10 and P (A $\cap$ B) = 0·02.
P (only one of them will qualify)
= P (A $\cap$ B′ or A′ $\cap$ B)
= P (A $\cap$ B′) + P (A′ $\cap$ B)
= P (A) – P (A $\cap$ B) + P (B)
– P (A $\cap$ B)
= 0·10 – 0·02 + 0·05 – 0·02 = 0·11.
Hence, the probability that only one of Anil and Ashima will qualify = 0·11.

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Answer:

The number of arrangements in which Veena can visit four cities (A, B, C and D) = 4 ! = 24.
Thus 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}.
Let E be event when Veena visits ‘A before B’
Then E = {ABCD, ABDC, ACBD, ACDB, ADBC,
ADCB, CABD, CADB, CDAB, DABC, DACB, DCAB}

Answer:

The number of arrangements in which Veena can visit four cities (A, B, C and D) = 4 ! = 24.
Thus 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}.
Let F be the event when Veena visits A before B and B before C.
Then F = {ABCD, ABDC, ADBC, DABC}.

Answer:

The number of arrangements in which Veena can visit four cities (A, B, C and D) = 4 ! = 24.
Thus 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}.
Let G be the event when Veena visits A first and B last.
Then G = {ACDB, ADCB}

Answer:

The number of arrangements in which Veena can visit four cities (A, B, C and D) = 4 ! = 24.
Thus 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}.
Let H be the event when Veena visits A either first or second.
Then H = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, CABD, CADB, DABC, DACB}.

Answer:

The number of arrangements in which Veena can visit four cities (A, B, C and D) = 4 ! = 24.
Thus 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}.
Let I be the event when Veena visits A just before B.
Then I = {ABCD, ABDC, CABD, CDAB, DABC, DCAB}.

Answer:

Total number of possible hands = ⁵²C₇.

Answer:

Total number of possible hands = ⁵²C₇.

Answer:

Total number of possible hands = ⁵²C₇.

Answer:

Proof. Let B $\mathrm{\upsilon }$ C = D
Then P (A $\mathrm{\upsilon }$ B $\mathrm{\upsilon }$ C) = P (A $\mathrm{\upsilon }$ D) = P (A) + P
(D) – P (A $\cap$ D) ...(1) [Th. IV]
But A $\cap$ D = A $\cap$ (B $\mathrm{\upsilon }$ C)
= (A $\cap$ B) $\mathrm{\upsilon }$ (A $\cap$ C)
$\therefore$ P (A $\cap$ D) = P [(A $\cap$ B) $\mathrm{\upsilon }$ (A $\cap$ C)]
= P (A $\cap$ B) + P (A $\cap$ C) – P
[(A $\cap$ B) $\cap$ (A $\cap$ C)] [Th. VI]
= P (A $\cap$ B + P (A $\cap$ C) – P
(A $\cap$ B $\cap$ C) ...(2)
[$\because$(A $\cap$ B) $\cap$ (A $\cap$ C) = A $\cap$ B $\cap$ C]
and P (D) = P (B $\mathrm{\upsilon }$ C) = P (B) + P (C) – P (B $\cap$ C)
...(3)
From (1), using (2) and (3), we have :
P (A $\mathrm{\upsilon }$ B $\mathrm{\upsilon }$ C) = P (A) + P (B) + P (C) – P
(B $\cap$ C) – [P (A $\cap$ B) + P (A $\cap$ C) – P (A $\cap$ B $\cap$ C)]
= P (A) + P (B) + P (C) – P
(A $\cap$ B) – P (B $\cap$ C) – P (A $\cap$ C) + P (A $\cap$ B $\cap$ C).
Cor. If A, B, C are mutually exclusive events, then P (A $\cap$ B) = P (B $\cap$ C) = P (A $\cap$ C)
= P (A $\cap$ B $\cap$ C) = 0.
$\therefore$ P (A $\mathrm{\upsilon }$ B $\mathrm{\upsilon }$ C) = P (A) + P (B) + P (C).

Answer:

No. of elements in sample space = ⁵P₃
= 5 × 4 × 3
= 60.

Answer:

No. of elements in sample space = ⁵P₃
= 5 × 4 × 3
= 60.

Answer:

When a coin is tossed three times, then the possible outcomes of the experiment are :
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT,
where H $\equiv$ Head and T $\equiv$ Tail.
$\therefore$ S, the sample sapce
={HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

Answer:

When a die is thrown two times.
$\therefore$ S, the sample space={(1, 1), (1, 2), ........., (1, 6) ; (2, 1), (2, 2), ...... (2, 6) ;
................................. ; (6, 1), (6, 2), ........... (6, 6)}.

Answer:

When a coin is tossed four times.
$\therefore$ S, the sample space
= {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH,
THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT},
where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

When a coin is tossed and a die is thrown.
$\therefore$ S, the sample space
= {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6},
where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

When a coin is tossed and a die is rolled only in case a head is shown on the coin.
$\therefore$ S, the sample space
={H1, H2, H3, H4, H5, H6, T},
where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

Answer:

Let R, W and B denote the red, white and blue die respectively.
$\therefore$ S, the sample space
= {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4,
W5, W6, B1, B2, B3, B4, B5, B6}.

Answer:

S, the required sample space
= {BB, BG, GB, GG},
where B $\equiv$ Boy and G $\equiv$ Girl.

Answer:

S, the required sample space
= {0, 1, 2}.

Answer:

The box contains 1 red and 3 identical white balls.
$\therefore$ S, the reqd. sample space
= {RW, WR, WW}.
where R $\equiv$ Red ball and W $\equiv$ White ball.

Answer:

S, the reqd. sample space.
= {HH, HT, T1, T2, T3, T4, T5, T6},
where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

S, the reqd. sample space
= {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN},
where D $\equiv$ Defective bulb and N $\equiv$ Non-defective bulb.

Answer:

S, the reqd. sample space.
= {T, H1, H3, H5, H21, H22, H23, H24,
H25, H26, H41, H42, H43, H44, H45, H46,
H61, H62, H63, H64, H65, H66},
where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

S, the sample space
= {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)}.

Answer:

S, the reqd. sample space is :
{1HH, 1TH, 1HT, 1TT, 2H, 2T, 3HH, 3TH, 3HT, 3TT,
4H, 4T, 5HH, 5TH, 5HT, 5TT, 6H, 6T},
where H $\equiv$ Head and T $\equiv$ Tail.

Answer:

Answer:

S, the reqd. sample space
= {6, (1, 6) (2, 6), (3, 6), (4, 6), (5, 6),
(1, 1, 6), (1, 2, 6), (1, 3, 6), ....... (1, 5, 6),
(2, 1, 6), (2, 2, 6), ..........., (2, 5, 6), ..........
(5, 1, 6), (5, 2, 6), .............}.

Answer:

Here sample space,
S = {1, 2, 3, 4, 5, 6}.
E (die shows 4) = {4}.
F (die shows an even number} = {2, 4, 6}.
Here H $\cap$ F = {4} $\Rightarrow$ E $\cap$ F $\ne$ $\mathrm{\varphi }$.
Hence E and F are not mutually exclusive.

Answer:

(a) (i) A = {1, 2, 3, 4, 5, 6}
(ii) B=$\mathrm{\varphi }$
(iii) C = {3, 6}
(iv) D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}.
(b) (i)A $\mathrm{\upsilon }$ B = {1, 2, 3, 4, 5, 6} $\mathrm{\upsilon }$ $\mathrm{\varphi }$
= {1, 2, 3, 4, 5, 6}
(ii) A $\cap$ B = {1, 2, 3, 4, 5, 6} $\cap$ $\mathrm{\varphi }$ = $\mathrm{\varphi }$
(iii) E $\mathrm{\upsilon }$ F = {6} $\mathrm{\upsilon }$ {3, 4, 5, 6}
= {3, 4, 5, 6}
(iv) D $\cap$ E = {1, 2, 3} $\cap$ {6} = $\mathrm{\varphi }$
(v) A – C = {1, 2, 3, 4, 5, 6} – {3, 6}
= {1, 2, 4, 5}
(vi) D – E = {1, 2, 3} – {6} = {1, 2, 3}
(vii) F′ = {1, 2, 3, 4, 5, 6} –
{3, 4, 5, 6} = {1, 2}
(viii) E $\cap$ F′ = {6} $\cap$ {1, 2} = $\mathrm{\varphi }$.

Answer:

A = {(3, 6), (6, 3), (4, 5), (5, 4), (4, 6),
(6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2),
(6, 2), (2, 1), (2, 3), (2, 4), (2, 5),
(2, 6)},
C = {(3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}.
Now A $\cap$ B = $\mathrm{\varphi }$, B $\cap$ C = $\mathrm{\varphi }$
and A $\cap$ C = {(3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}.
Hence A, B and B, C are mutually exclusive.

Answer:

A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}.
A and B, B and C, A and C, C and D are mutually exclusive.

Answer:

A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}.
A and C are simple events.

Answer:

A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}.
B and D are compound events.

Answer:

‘‘Getting at least two heads’’ and ‘‘getting at least two tails.’’

Answer:

‘‘Getting no heads’’, ‘‘getting exactly one head’’ and ‘‘getting at least two heads’’.

Answer:

‘‘Getting at most two tails’’ and ‘‘getting exactly two tails’’.

Answer:

‘‘Getting exactly one head’’ and ‘‘getting exactly two heads’’.

Answer:

‘‘Getting exactly one tail’’, ‘‘getting exactly two tails’’ and ‘‘getting exactly three tails’’.

Answer:

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)}.
A′ = Getting an odd number on the first die = B.

Answer:

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)}.
not B = Getting an even number on the first die = A.

Answer:

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)}.
A or B = A $\mathrm{\upsilon }$ 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), (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)}
= S.

Answer:

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)}.
A and B = A $\cap$ B = $\mathrm{\varphi }$.

Answer:

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)}.
A but not 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)}.

Answer:

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)}.
B or C = B $\mathrm{\upsilon }$ C = {(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)}.

Answer:

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)}.
B $\cap$ C = {(1, 1), (1, 2), (1, 3), (1, 4),
(3, 1), (3, 2)}.

Answer:

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)}.
A $\cap$ B′ $\cap$ C′.
Here 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′ = A.
and C′ = {(1, 5), (2, 4), (3, 3), (4, 2),
(5, 1), (1, 6), (2, 5), (3, 4)
(4, 3), (5, 2), (6, 1), (2, 6),
(3, 5), (4, 4), (5, 3), (6, 2),
(3, 6), (4, 5), (5, 4), (6, 3),
(4, 6), (5, 5), (6, 4), (5, 6),
(6, 5), (6, 6)}.
$\therefore$A $\cap$ B′ $\cap$ 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)}.

Answer:

True [Even and odd numbers cannot come on first die simultaneously]

Answer:

True [Above reason along with the first die can have either even or odd number]

Answer:

True [Not B means B has even number on first die, which is same as A]

Answer:

False [Sum can be ≤ 5 when first die shows even number, for ex. (2, 1) ; etc.]

Answer:

False.

Answer:

False.

Answer:

First Condition. Each p ($\mathrm{\omega }$) is positive and less than 1.
Second Condition. Sum of probabilities
= 0·1 + 0·01 + 0·05 + 0·03 + 0·01 + 0·2 + 0·6 = 1
$\therefore$ Assignment is valid.

Answer:

Valid [Reasons as in part (a)]

Answer:

First Condition. Each p ($\mathrm{\omega }$) is positive and less than 1.
Second Condition. Sum of probabilities.
= 0·1 + 0·2 + 0·3 + 0·4 + 0·5 + 0·6 + 0·7 > 1
$\therefore$ Assignment is valid.

Answer:

First Condition. Each p ($\mathrm{\omega }$) is not possible.
$\therefore$ Assignment is valid.

Answer:

First Condition. Each assignment is not less than 1.
$\therefore$ Assignment is valid.

Answer:

Here Sample space,
S = {HH, HT, TH, TT},
where H $\equiv$ Head and T $\equiv$ Tail.
Let A be the event of getting at least one tail.
$\therefore$ A = {HT, TH, TT}.

Answer:

Here n the number of exhaustive cases = 6.

Answer:

Here n the number of exhaustive cases = 6.

Answer:

Here n the number of exhaustive cases = 6.

Answer:

Here n the number of exhaustive cases = 6.

Answer:

Here n the number of exhaustive cases = 6.

Answer:

Since there are 52 cards,
$\therefore$there are 52 points in the sample space.

Answer:

Answer:

Answer:

No. of exhaustive cases = 2 × 6 = 12 No. of favourable cases = 1 [(1, 2)]

Answer:

No. of exhaustive cases = 2 × 6 = 12 No. of favourable cases = 1 [(6, 6)]

Answer:

Any one person can be selected out of 10 (6W + 4M)
$\therefore$ No. of exhaustive cases = 10.
One woman can be selected out of 6.
$\therefore$ No. of favourable cases = 6

Answer:

Here S = {HHHH, HHHT, HHTH, HTHH,
THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH,
HTTT, THTT, TTHT, TTTH, TTTT}

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Sample space, S = {HHH, HHT,
HTH, THH, HTT, THT, TTH, TTH}.

Answer:

Answer:

The word ‘ASSASSINATION’ has 13 letters in which 6 are vowels (viz. AAAIIO) and 7 consonants (viz. SSSSNNT)
$\therefore$ n (S) = 13.
No. of vowels = 6.

Answer:

The word ‘ASSASSINATION’ has 13 letters in which 6 are vowels (viz. AAAIIO) and 7 consonants (viz. SSSSNNT)
$\therefore$ n (S) = 13.
No. of consonants = 7.

Answer:

Answer:

P (A $\mathrm{\upsilon }$ B) = P (A) + P (B) – P (A $\cap$ B)
= 0·5 + 0·7 – 0·06 = 1·20 – 0·06
= 1·14 > 1.
Hence the probabilities are not consistently defined.

Answer:

P (A $\mathrm{\upsilon }$ B) = P (A) + P (B) – P (A $\cap$ B)
$\Rightarrow$ 0·8 = 0·5 + 0·4 – P (A $\cap$ B)
$\Rightarrow$ P (A $\cap$ B) = 0·9 – 0·8 = 0·1.
Hence the probabilities are consistently defined.

Answer:

Answer:

P (A $\mathrm{\upsilon }$ B) = P (A) + P (B) – P (A $\cap$ B)
$\Rightarrow$ 0·6 = 0·35 + P (B) – 0·25
$\Rightarrow$ P (B) = 0·6 – 0.1 = 0·5.

Answer:

P (A $\mathrm{\upsilon }$ B) = P (A) + P (B) – P (A $\cap$ B)
$\Rightarrow$ 0·7 = 0·5 + 0·35 – P (A $\cap$ B)
$\Rightarrow$ P (A $\cap$ B) = 0·35 + 0·5 – 0·7
= 0·35 – 0·2 = 0·15.

Answer:

Answer:

Answer:

Answer:

Answer:

P (not A) = 1 – P (A) = 1 – 0·42 = 0·58.

Answer:

P (not B) = 1 – P (B) = 1 – 0·48 = 0·52.

Answer:

P (A or B) = P (A) + P (B) – P (A $\cap$ B)
= 0·42 + 0·48 – 0·16
= 0·90 – 0·16 = 0·74.

Answer:

Answer:

Let A and B be two examinations.
Here P (A) = 0·8, P (B) = 0·7 and (P $\mathrm{\upsilon }$ B) = 0·95.
$\therefore$ P (A $\mathrm{\upsilon }$ B) = P (A) + P (B) – P (A $\cap$ B)
$\Rightarrow$ 0·95 = 0·8 + 0·7 – P (A $\cap$ B)
$\Rightarrow$ P(A $\cap$ B) = 1·5 – 0·95 = 0·55.
Hence P (Pasing both exam.) = 0·55.

Answer:

Let E and H be English and Hindi examination respectively.
Here P (E $\cap$ H) = 0·5, 1 – P (E $\mathrm{\upsilon }$ H) = 0·1
$\Rightarrow$ P (E $\mathrm{\upsilon }$ H) = 0·9
and P (E) = 0·75.
NowP (E $\mathrm{\upsilon }$ H) = P (E) + P (H) – P (E $\cap$ H)
$\Rightarrow$ 0·9 = 0·75 + P (H) – 0·5
Hence P (H) = 1·4 – 0·75 = 0·65.

Answer:

Let A and B be two events when a student opts for NCC and NSS respectively.

Answer:

Let A and B be two events when a student opts for NCC and NSS respectively.

Answer:

Let A and B be two events when a student opts for NCC and NSS respectively.