NCERT Solutions · Class 11 Mathematics Probability

Axioms of probability. An assignment \(\displaystyle P(\omega_i) \) to the seven outcomes of \(\displaystyle S \) is valid only if it satisfies both conditions\[\text{(1) } 0 \leq P(\omega_i) \leq 1 \ \text{ for every } i, \qquad \text{(2) } \sum_{i=1}^{7} P(\omega_i) = 1 \]Test each row.(a) All seven numbers lie between $\displaystyle 0$ and $\displaystyle 1$, and \[0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1 \] Both conditions hold — valid.(b) Each entry is \(\displaystyle \tfrac{1}{7} \in [0,1] \), and \(\displaystyle 7 \times \tfrac{1}{7} = 1 \) — valid.(c) Each entry lies in \(\displaystyle [0,1] \), but \[0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 \neq 1 \] Condition ($\displaystyle 2$) fails — not valid.(d) The sum is \(\displaystyle -0.1 + 0.2 + 0.3 + 0.4 - 0.2 + 0.1 + 0.3 = 1 \), but \(\displaystyle P(\omega_1) = -0.1 \) and \(\displaystyle P(\omega_5) = -0.2 \) are negative. A probability can never be negative, so condition ($\displaystyle 1$) fails — not valid.(e) Here \(\displaystyle P(\omega_7) = \tfrac{15}{14} > 1 \), which condition ($\displaystyle 1$) forbids; also \[\frac{1+2+3+4+5+6+15}{14} = \frac{36}{14} = \frac{18}{7} \neq 1 \] so condition ($\displaystyle 2$) fails too — not valid.Assignments (c), (d) and (e) are not valid: (c) because the probabilities add to $\displaystyle 2.8$ instead of $\displaystyle 1$, (d) because two entries are negative, and (e) because \(\displaystyle \frac{15}{14} > 1 \) and the total is \(\displaystyle \frac{18}{7} \). Assignments (a) and (b) are valid.

Equally likely outcomes. Tossing a coin twice, \[S = \{HH,\; HT,\; TH,\; TT\}, \qquad n(S) = 4 \] and all four outcomes are equally likely, so \(\displaystyle P(E) = \dfrac{n(E)}{n(S)} \).Let E be the event "at least one tail occurs" — that is, one tail or two tails: \[E = \{HT,\; TH,\; TT\}, \qquad n(E) = 3 \]\[P(E) = \frac{3}{4} \](Check by the complement: the only outcome with no tail is \(\displaystyle HH \), so \(\displaystyle P(E) = 1 - \tfrac{1}{4} = \tfrac{3}{4} \).)\(\displaystyle P(\text{at least one tail}) = \dfrac{3}{4} \).

Equally likely outcomes. For one throw of a die, \(\displaystyle S = \{1,2,3,4,5,6\} \) with \(\displaystyle n(S) = 6 \), and for any event E, \[P(E) = \frac{n(E)}{n(S)} = \frac{n(E)}{6} \](i) A prime number will appear. The primes among $\displaystyle 1$–$\displaystyle 6$ are $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 5$, so \(\displaystyle E = \{2,3,5\} \): \[P = \frac{3}{6} = \frac{1}{2} \](ii) A number greater than or equal to 3. \(\displaystyle E = \{3,4,5,6\} \): \[P = \frac{4}{6} = \frac{2}{3} \](iii) A number less than or equal to one. \(\displaystyle E = \{1\} \): \[P = \frac{1}{6} \](iv) A number more than 6. No face exceeds $\displaystyle 6$, so \(\displaystyle E = \phi \) — an impossible event: \[P = \frac{0}{6} = 0 \](v) A number less than 6. \(\displaystyle E = \{1,2,3,4,5\} \): \[P = \frac{5}{6} \](i) \(\displaystyle \dfrac{1}{2} \) (ii) \(\displaystyle \dfrac{2}{3} \) (iii) \(\displaystyle \dfrac{1}{6} \) (iv) \(\displaystyle 0 \) (v) \(\displaystyle \dfrac{5}{6} \).

Equally likely outcomes. One card is drawn from a well-shuffled pack, so every card is equally likely and \(\displaystyle P(E) = \dfrac{n(E)}{n(S)} \).(a) Sample space. The experiment can end in any one of the $\displaystyle 52$ cards, so \[n(S) = 52 \](b) Ace of spades. There is exactly one such card: \[P = \frac{1}{52} \](c)(i) An ace. The pack has four aces, one in each suit: \[P = \frac{4}{52} = \frac{1}{13} \](c)(ii) A black card. The black suits are clubs and spades, \(\displaystyle 13 + 13 = 26 \) cards: \[P = \frac{26}{52} = \frac{1}{2} \](a) $\displaystyle 52$ sample points; (b) \(\displaystyle \dfrac{1}{52} \); (c)(i) \(\displaystyle \dfrac{1}{13} \), (ii) \(\displaystyle \dfrac{1}{2} \).

Equally likely outcomes. The coin shows $\displaystyle 1$ or $\displaystyle 6$, and the die shows $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$ or 6. Writing an outcome as (coin, die), \[S = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\;(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \] so \(\displaystyle n(S) = 2 \times 6 = 12 \), and since both the coin and the die are fair, all $\displaystyle 12$ outcomes are equally likely.(i) Sum = 3. With the coin showing $\displaystyle 1$ we need the die to show $\displaystyle 2$; with the coin showing $\displaystyle 6$ the sum is already at least 7. So the event is \(\displaystyle \{(1,2)\} \): \[P = \frac{1}{12} \](ii) Sum = 12. With the coin showing $\displaystyle 1$ the largest sum is \(\displaystyle 1 + 6 = 7 \), so the coin must show $\displaystyle 6$ and then the die must show 6. The event is \(\displaystyle \{(6,6)\} \): \[P = \frac{1}{12} \](i) \(\displaystyle \dfrac{1}{12} \) (ii) \(\displaystyle \dfrac{1}{12} \).

Equally likely outcomes. The council has \(\displaystyle 4 + 6 = 10 \) members, and one is chosen at random, so each member is equally likely: \[n(S) = 10 \]Let W be the event "the member selected is a woman". There are $\displaystyle 6$ women, so \(\displaystyle n(W) = 6 \) and\[P(W) = \frac{n(W)}{n(S)} = \frac{6}{10} = \frac{3}{5} = 0.6 \]The probability that a woman is selected is \(\displaystyle \dfrac{3}{5} \).

Sample space of four tosses, then the money each outcome pays. Four tosses of a fair coin give \[n(S) = 2^4 = 16 \] equally likely outcomes. The amount depends only on how many heads occur, so let \(\displaystyle k \) be the number of heads; then \(\displaystyle 4 - k \) tails occur and the amount won is\[\text{Amount} = 1 \times k - 1.50 \times (4-k) = k - 6 + 1.5k = 2.5k - 6 \quad \text{(in rupees)} \]The number of outcomes with exactly \(\displaystyle k \) heads is \(\displaystyle \binom{4}{k} \), so \(\displaystyle P = \dbinom{4}{k}\big/16 \).

Heads \(\displaystyle k \)	Amount \(\displaystyle 2.5k - 6 \)	Number of outcomes \(\displaystyle \binom{4}{k} \)	Probability
$\displaystyle 4$	\(\displaystyle +\,\text{Rs } 4.00 \)	$\displaystyle 1$ (HHHH)	\(\displaystyle \dfrac{1}{16} \)
$\displaystyle 3$	\(\displaystyle +\,\text{Rs } 1.50 \)	$\displaystyle 4$	\(\displaystyle \dfrac{4}{16} = \dfrac{1}{4} \)
$\displaystyle 2$	\(\displaystyle -\,\text{Rs } 1.00 \)	$\displaystyle 6$	\(\displaystyle \dfrac{6}{16} = \dfrac{3}{8} \)
$\displaystyle 1$	\(\displaystyle -\,\text{Rs } 3.50 \)	$\displaystyle 4$	\(\displaystyle \dfrac{4}{16} = \dfrac{1}{4} \)
$\displaystyle 0$	\(\displaystyle -\,\text{Rs } 6.00 \)	$\displaystyle 1$ (TTTT)	\(\displaystyle \dfrac{1}{16} \)

(A minus sign means a loss.) The counts add to \(\displaystyle 1+4+6+4+1 = 16 \) and the probabilities to \(\displaystyle \tfrac{1}{16}+\tfrac{1}{4}+\tfrac{3}{8}+\tfrac{1}{4}+\tfrac{1}{16} = 1 \), as they must.Five different amounts are possible: Rs $\displaystyle 4.00$, Rs $\displaystyle 1.50$, \(\displaystyle -\)Rs $\displaystyle 1.00$, \(\displaystyle -\)Rs $\displaystyle 3.50$ and \(\displaystyle -\)Rs $\displaystyle 6.00$, with probabilities \(\displaystyle \dfrac{1}{16},\; \dfrac{1}{4},\; \dfrac{3}{8},\; \dfrac{1}{4},\; \dfrac{1}{16} \) respectively.

Equally likely outcomes. Tossing three coins once, \[S = \{HHH,\; HHT,\; HTH,\; THH,\; HTT,\; THT,\; TTH,\; TTT\}, \qquad n(S) = 8 \] and every outcome is equally likely, so \(\displaystyle P(E) = \dfrac{n(E)}{8} \).(i) $\displaystyle 3$ heads — \(\displaystyle \{HHH\} \): \(\displaystyle P = \dfrac{1}{8} \).(ii) $\displaystyle 2$ heads (exactly two) — \(\displaystyle \{HHT, HTH, THH\} \): \(\displaystyle P = \dfrac{3}{8} \).(iii) At least $\displaystyle 2$ heads — two heads or three heads: \(\displaystyle \{HHT, HTH, THH, HHH\} \): \(\displaystyle P = \dfrac{4}{8} = \dfrac{1}{2} \).(iv) At most $\displaystyle 2$ heads — $\displaystyle 0$, $\displaystyle 1$ or $\displaystyle 2$ heads, i.e. every outcome except \(\displaystyle HHH \): \[P = 1 - P(3 \text{ heads}) = 1 - \frac{1}{8} = \frac{7}{8} \](v) No head — \(\displaystyle \{TTT\} \): \(\displaystyle P = \dfrac{1}{8} \).(vi) $\displaystyle 3$ tails — the same event as "no head", \(\displaystyle \{TTT\} \): \(\displaystyle P = \dfrac{1}{8} \).(vii) Exactly two tails — \(\displaystyle \{TTH, THT, HTT\} \): \(\displaystyle P = \dfrac{3}{8} \).(viii) No tail — \(\displaystyle \{HHH\} \): \(\displaystyle P = \dfrac{1}{8} \).(ix) At most two tails — every outcome except \(\displaystyle TTT \): \[P = 1 - P(3 \text{ tails}) = 1 - \frac{1}{8} = \frac{7}{8} \](i) \(\displaystyle \dfrac{1}{8} \) (ii) \(\displaystyle \dfrac{3}{8} \) (iii) \(\displaystyle \dfrac{1}{2} \) (iv) \(\displaystyle \dfrac{7}{8} \) (v) \(\displaystyle \dfrac{1}{8} \) (vi) \(\displaystyle \dfrac{1}{8} \) (vii) \(\displaystyle \dfrac{3}{8} \) (viii) \(\displaystyle \dfrac{1}{8} \) (ix) \(\displaystyle \dfrac{7}{8} \).

Complement rule. For any event A, the event "not A" is the complement \(\displaystyle A' \), and since A and \(\displaystyle A' \) are mutually exclusive and exhaustive, \[P(A) + P(A') = 1 \qquad \Longrightarrow \qquad P(A') = 1 - P(A) \]Here \(\displaystyle P(A) = \dfrac{2}{11} \), so\[P(\text{not } A) = 1 - \frac{2}{11} = \frac{11 - 2}{11} = \frac{9}{11} \]\(\displaystyle P(\text{not } A) = \dfrac{9}{11} \).

Equally likely outcomes. Count the letters of ASSASSINATION, treating the $\displaystyle 13$ printed letters as the sample points (each equally likely to be picked):\[\text{A} \to 3, \quad \text{S} \to 4, \quad \text{I} \to 2, \quad \text{N} \to 2, \quad \text{T} \to 1, \quad \text{O} \to 1 \]and \(\displaystyle 3 + 4 + 2 + 2 + 1 + 1 = 13 = n(S) \).(i) A vowel. The vowels present are A, I and O: \[n(\text{vowels}) = 3 + 2 + 1 = 6 \qquad \Longrightarrow \qquad P = \frac{6}{13} \](ii) A consonant. The consonants are S, N and T: \[n(\text{consonants}) = 4 + 2 + 1 = 7 \qquad \Longrightarrow \qquad P = \frac{7}{13} \](As a check, \(\displaystyle \tfrac{6}{13} + \tfrac{7}{13} = 1 \), since a letter is either a vowel or a consonant.)(i) \(\displaystyle \dfrac{6}{13} \) (ii) \(\displaystyle \dfrac{7}{13} \).