Class 11 Maths NCERT Solutions Chapter 14 Ex 14.1 – Probability

The sample space is S=\{1,2,3,4,5,6\}.

E=\{4\} and F=\{2,4,6\}

E\cap F=\{4\}\cap\{2,4,6\}=\{4\}

Since E\cap F=\{4\}\ne\phi, the outcome 4 belongs to both events, so they can occur together.

The sample space for throwing a die is S=\{1,2,3,4,5,6\}.

A=\{1,2,3,4,5,6\} — every number on the die is less than 7, so A=S.

B=\phi — no number on the die is greater than 7.

C=\{3,6\} — the multiples of 3 on the die.

D=\{1,2,3\} — numbers less than 4.

E=\{6\} — the only even number greater than 4 on the die.

F=\{3,4,5,6\} — numbers not less than 3, i.e. numbers ≥ 3.

A\cup B=\{1,2,3,4,5,6\}\cup\phi=\{1,2,3,4,5,6\}

A\cap B=\{1,2,3,4,5,6\}\cap\phi=\phi

B\cup C=\phi\cup\{3,6\}=\{3,6\}

E\cap F=\{6\}\cap\{3,4,5,6\}=\{6\}

D\cap E=\{1,2,3\}\cap\{6\}=\phi

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

D-E=\{1,2,3\}-\{6\}=\{1,2,3\}

F'=S-F=\{1,2,3,4,5,6\}-\{3,4,5,6\}=\{1,2\}

E\cap F'=\{6\}\cap\{1,2\}=\phi

The sample space is S=\{(x,y):x,y=1,2,\ldots,6\}, with 36 equally likely outcomes.

A=\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)\} — pairs summing to 9, 10, 11 or 12.

B=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2)\} — every pair where a 2 appears on either die (11 outcomes).

C=\{(3,6),(4,5),(5,4),(6,3),(6,6)\} — sums of 9 or 12 that are also ≥ 7 (sums 7, 8, 10, 11 are not multiples of 3, so only 9 and 12 qualify).

A\cap B=\phi — no outcome in A has a 2 on either die, since every sum in A is 9 or more, needing both dice showing at least 3. So A and B are mutually exclusive.

A\cap C=\{(3,6),(4,5),(5,4),(6,3),(6,6)\}\ne\phi — every outcome in C also lies in A (all these sums exceed 8, except that (3,6) sums to 9 > 8). So A and C are not mutually exclusive.

B\cap C=\phi — no outcome in C has a 2 on either die, since a 2 paired with anything gives a sum too small or not a multiple of 3 in the required range. So B and C are mutually exclusive.

The sample space is S=\{\text{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}\} (8 outcomes).

A=\{\text{HHH}\}, B=\{\text{HHT,HTH,THH}\}, C=\{\text{TTT}\}, D=\{\text{HHH,HHT,HTH,HTT}\}

Checking every pair: A\cap B=\phi, A\cap C=\phi, B\cap C=\phi, C\cap D=\phi. But A\cap D=\{\text{HHH}\}\ne\phi and B\cap D=\{\text{HHT,HTH}\}\ne\phi.

So the mutually exclusive pairs are: A and B; A and C; B and C; C and D.

A simple event has exactly one sample point. A=\{\text{HHH}\} and C=\{\text{TTT}\} each have one outcome, so A and C are simple events.

A compound event has more than one sample point. B has 3 outcomes and D has 4 outcomes, so B and D are compound events.

The sample space is S=\{\text{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}\} (8 outcomes). These answers are illustrative — any events satisfying the stated conditions are acceptable.

Let A=\{\text{HHH}\} ('three heads') and B=\{\text{TTT}\} ('three tails'). A\cap B=\phi, so A and B are mutually exclusive.

Let A=\{\text{TTT}\} ('no head'), B=\{\text{HTT,THT,TTH}\} ('exactly one head'), C=\{\text{HHT,HTH,THH,HHH}\} ('at least two heads'). These are pairwise disjoint, and A\cup B\cup C=S, so they are mutually exclusive and exhaustive.

Let A = 'at least one head' (every outcome except TTT) and B = 'at least one tail' (every outcome except HHH). Both HHT and other mixed outcomes lie in A\cap B, so A\cap B\ne\phi — not mutually exclusive.

Let A=\{\text{HHT,HTH,THH}\} ('exactly two heads') and B=\{\text{HTT,THT,TTH}\} ('exactly two tails'). A\cap B=\phi, but A\cup B has only 6 outcomes and misses HHH and TTT, so A\cup B\ne S — mutually exclusive but not exhaustive.

Let A=\{\text{HTT,THT,TTH}\} ('exactly one head'), B=\{\text{HHT,HTH,THH}\} ('exactly one tail'), C=\{\text{HHH}\} ('three heads'). These are pairwise disjoint, but their union has only 7 outcomes, missing TTT — mutually exclusive but not exhaustive.

The sample space is S=\{(x,y):x,y=1,2,\ldots,6\}, 36 outcomes, where x is the number on the first die.

A = all 18 pairs with x\in\{2,4,6\}; B = all 18 pairs with x\in\{1,3,5\}; C = pairs with x+y\le5.

A′ is 'not an even number on the first die', which is exactly the same set as B: the first die shows an odd number. So A'=B (18 outcomes).

'Not B' is 'not an odd number on the first die', which is exactly A: the first die shows an even number. So B'=A (18 outcomes).

Every outcome has the first die either even or odd, so A\cup B=S, all 36 outcomes.

No outcome can have the first die both even and odd, so A\cap B=\phi.

A-C is the set of pairs with an even first die AND a sum greater than 5. This removes from A the pairs (2,1), (2,2), (2,3), (4,1) — the only pairs in A with sum ≤ 5 — leaving 14 outcomes.

B\cup C combines all 18 odd-first-die outcomes with the (even-first-die) outcomes in C not already counted — this works out to 22 outcomes in total.

B\cap C is pairs with an odd first die and sum ≤ 5: \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\} — 6 outcomes.

Since B'=A, this is A\cap A\cap C'=A\cap C', i.e. 'even first die AND sum greater than 5' — the same 14 outcomes found in part (v).

No outcome can have the first die both even and odd, so A\cap B=\phi — A and B are mutually exclusive.

From (i), A\cap B=\phi (mutually exclusive). Also every outcome's first die is either even or odd, so A\cup B=S (exhaustive). Both conditions hold.

B′ is 'not an odd number on the first die', which means the first die is even — exactly the definition of A. So A=B'.

A and C share outcomes such as (2,1), (2,2), (2,3) and (4,1) — an even first die with a sum ≤ 5 is entirely possible. So A\cap C\ne\phi, and A, C are not mutually exclusive.

Since B'=A (from part (iii)), A\cap B'=A\cap A=A\ne\phi. So A and B′ share every outcome of A — they are not mutually exclusive.

A'\cup B'\cup C does equal S (since A'=B and B'=A, so A'\cup B'=A\cup B=S already, and adding C changes nothing) — so this part is exhaustive. But A'\cap C=B\cap C=\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}\ne\phi (from Q6 part (vii)), so A′ and C overlap. Since they are not pairwise disjoint, A′, B′, C are not mutually exclusive, even though they are exhaustive.