Class 11 Maths NCERT Solutions Chapter 14 Ex 14.2 – Probability

A valid assignment needs every probability \ge0, and all the probabilities must add up to exactly 1.

(a) Every value is positive, and the sum is 0.1+0.01+0.05+0.03+0.01+0.2+0.6=1. Valid.

(b) Every value is positive, and the sum is 7\times\dfrac{1}{7}=1. Valid.

(c) Every value is positive, but the sum is 0.1+0.2+0.3+0.4+0.5+0.6+0.7=2.8\ne1. Not valid.

(d) Two of the values (−0.1 and −0.2) are negative. Not valid.

(e) Every value is positive, but the sum is \dfrac{1+2+3+4+5+6+15}{14}=\dfrac{36}{14}\ne1. Not valid.

The sample space is S=\{\text{HH,HT,TH,TT}\}, 4 equally likely outcomes.

'At least one tail' excludes only HH, so the favourable outcomes are \{\text{HT,TH,TT}\}, 3 outcomes.

P(\text{at least one tail})=\dfrac{3}{4}

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

(i) Prime numbers on a die: \{2,3,5\}, 3 outcomes. P=\dfrac{3}{6}=\dfrac{1}{2}

(ii) Numbers \ge3: \{3,4,5,6\}, 4 outcomes. P=\dfrac{4}{6}=\dfrac{2}{3}

(iii) Numbers \le1: \{1\}, 1 outcome. P=\dfrac{1}{6}

(iv) Numbers >6: no such outcome on a die. P=\dfrac{0}{6}=0

(v) Numbers <6: \{1,2,3,4,5\}, 5 outcomes. P=\dfrac{5}{6}

(a) Each of the 52 cards is a possible outcome, so the sample space has 52 points.

(b) There is exactly 1 ace of spades in the deck. P(\text{ace of spades})=\dfrac{1}{52}

(c)(i) There are 4 aces in the deck. P(\text{ace})=\dfrac{4}{52}=\dfrac{1}{13}

(c)(ii) There are 26 black cards (clubs and spades). P(\text{black card})=\dfrac{26}{52}=\dfrac{1}{2}

The coin shows 1 or 6, and the die shows 1 to 6, so the sample space has 2\times6=12 equally likely outcomes.

(i) A sum of 3 needs coin = 1 and die = 2 (since coin = 6 would need die = −3, impossible). Only (1,2) works — 1 outcome. P=\dfrac{1}{12}

(ii) A sum of 12 needs coin = 6 and die = 6 (since coin = 1 would need die = 11, impossible). Only (6,6) works — 1 outcome. P=\dfrac{1}{12}

There are 4+6=10 council members in total, and one is selected at random.

P(\text{woman})=\dfrac{6}{10}=\dfrac{3}{5}

The sample space of tossing a coin four times has 2^4=16 equally likely outcomes. The amount won depends only on the number of heads, k, out of 4 tosses: amount =k(1)-(4-k)(1.5)=2.5k-6, and the number of outcomes with exactly k heads is \binom{4}{k}.

These 5 amounts and probabilities together account for all 16 outcomes (1+4+6+4+1=16), confirming the total probability is 1.

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

(i) 3 heads: \{\text{HHH}\}, 1 outcome. P=\dfrac{1}{8}

(ii) Exactly 2 heads: \{\text{HHT,HTH,THH}\}, 3 outcomes. P=\dfrac{3}{8}

(iii) At least 2 heads (2 or 3 heads): 3+1=4 outcomes. P=\dfrac{4}{8}=\dfrac{1}{2}

(iv) At most 2 heads (0, 1 or 2 heads): all except HHH, 7 outcomes. P=\dfrac{7}{8}

(v) No head: \{\text{TTT}\}, 1 outcome. P=\dfrac{1}{8}

(vi) 3 tails: \{\text{TTT}\}, 1 outcome. P=\dfrac{1}{8}

(vii) Exactly 2 tails: \{\text{HTT,THT,TTH}\}, 3 outcomes. P=\dfrac{3}{8}

(viii) No tail: \{\text{HHH}\}, 1 outcome. P=\dfrac{1}{8}

(ix) At most 2 tails (0, 1 or 2 tails): all except TTT, 7 outcomes. P=\dfrac{7}{8}

Using the complement rule P(\text{not }A)=1-P(A):

P(\text{not }A)=1-\dfrac{2}{11}=\dfrac{9}{11}

The word 'ASSASSINATION' has 13 letters: A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, T appears 1 time, and O appears 1 time (3+4+2+2+1+1=13).

(i) Vowels are A, I, O: 3+2+1=6 letters out of 13. P(\text{vowel})=\dfrac{6}{13}

(ii) Consonants are S, N, T: 4+2+1=7 letters out of 13. P(\text{consonant})=\dfrac{7}{13}

Since order doesn't matter, the total number of ways to choose 6 different numbers from 1 to 20 is \binom{20}{6}.

\binom{20}{6}=\dfrac{20!}{6!\,14!}=38760

Only one of these combinations matches the committee's fixed six numbers, so there is exactly 1 favourable outcome.

P(\text{winning})=\dfrac{1}{38760}

(i) Since A\cap B\subseteq A, it must always be true that P(A\cap B)\le P(A). Here P(A\cap B)=0.6, but P(A)=0.5<0.6. This is impossible, so these values are not consistently defined.

(ii) Using the addition rule, P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.8=0.1. Since 0\le0.1\le\min(0.5,0.4)=0.4, and P(A\cup B)=0.8\ge\max(0.5,0.4)=0.5, all conditions are satisfied, so these values are consistently defined.

In every row, use the addition rule P(A\cup B)=P(A)+P(B)-P(A\cap B) to solve for the missing value.

(i) P(A\cup B)=\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{15}=\dfrac{5}{15}+\dfrac{3}{15}-\dfrac{1}{15}=\dfrac{7}{15}

(ii) 0.6=0.35+P(B)-0.25\ \Rightarrow\ P(B)=0.6-0.35+0.25=0.5

(iii) 0.7=0.5+0.35-P(A\cap B)\ \Rightarrow\ P(A\cap B)=0.5+0.35-0.7=0.15

Since A and B are mutually exclusive, P(A\cap B)=0, so the addition rule simplifies to P(A\cup B)=P(A)+P(B).

P(A\text{ or }B)=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}

(i) Using the addition rule:

P(E\text{ or }F)=P(E)+P(F)-P(E\text{ and }F)=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{2}{8}+\dfrac{4}{8}-\dfrac{1}{8}=\dfrac{5}{8}

(ii) 'Not E and not F' is E'\cap F'=(E\cup F)' by De Morgan's law, so:

P(\text{not }E\text{ and not }F)=1-P(E\text{ or }F)=1-\dfrac{5}{8}=\dfrac{3}{8}

'Not E or not F' is E'\cup F'=(E\cap F)' by De Morgan's law, so:

P((E\cap F)')=0.25\ \Rightarrow\ P(E\cap F)=1-0.25=0.75

Since P(E\cap F)=0.75\ne0, the events E and F do share outcomes.

(i) P(\text{not }A)=1-P(A)=1-0.42=0.58

(ii) P(\text{not }B)=1-P(B)=1-0.48=0.52

(iii) Using the addition rule:

P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)=0.42+0.48-0.16=0.74

Let M: 'studies Mathematics' and B: 'studies Biology'. Given P(M)=0.4, P(B)=0.3, P(M\cap B)=0.1.

Using the addition rule:

P(M\text{ or }B)=P(M)+P(B)-P(M\cap B)=0.4+0.3-0.1=0.6

Let A: 'passes the first exam' and B: 'passes the second exam'. Given P(A)=0.8, P(B)=0.7, and 'at least one' means P(A\cup B)=0.95.

Using the addition rule, rearranged for P(A\cap B):

P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.8+0.7-0.95=0.55

Let E: 'passes English' and H: 'passes Hindi'. Given P(E\cap H)=0.5, P(\text{neither})=P((E\cup H)')=0.1, and P(E)=0.75.

Since 'passing neither' is the complement of 'passing at least one':

P(E\cup H)=1-0.1=0.9

Using the addition rule, rearranged for P(H):

P(H)=P(E\cup H)-P(E)+P(E\cap H)=0.9-0.75+0.5=0.65

Let N: 'opted for NCC' and S: 'opted for NSS'. Then P(N)=\dfrac{30}{60}, P(S)=\dfrac{32}{60}, P(N\cap S)=\dfrac{24}{60}.

(i) Using the addition rule:

P(N\cup S)=\dfrac{30}{60}+\dfrac{32}{60}-\dfrac{24}{60}=\dfrac{38}{60}=\dfrac{19}{30}

(ii) 'Neither' is the complement of 'NCC or NSS':

P(\text{neither})=1-P(N\cup S)=1-\dfrac{19}{30}=\dfrac{11}{30}

(iii) 'NSS but not NCC' is S-N=S\cap N':

P(S-N)=P(S)-P(N\cap S)=\dfrac{32}{60}-\dfrac{24}{60}=\dfrac{8}{60}=\dfrac{2}{15}