Chapter 14: Probability

Each toss gives either Head (H) or Tail (T), giving $2 \times 2 \times 2 = 8$ total outcomes.

All heads $= \mathrm{HHH}$

Two heads, one tail $= \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}$

One head, two tails $= \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}$

All tails $= \mathrm{TTT}$

Thus, the sample space is:

Let $x$ represent the result of the first throw and $y$ represent the result of the second throw, where $x, y \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6 \times 6 = 36$.

The sample space is:

Each toss yields H or T, so there are $2 \times 2 \times 2 \times 2 = 2^4 = 16$ outcomes. Listing them by pattern:

All heads: HHHH

Three heads, one tail: HHHT, HHTH, HTHH, THHH

Two heads, two tails: HHTT, HTHT, HTTH, THHT, THTH, TTHH

One head, three tails: HTTT, THTT, TTHT, TTTH

All tails: TTTT

Thus, the sample space is:

The coin gives Head (H) or Tail (T). The die gives one of 6 faces: $1, 2, 3, 4, 5, 6$. Together, there are $2 \times 6 = 12$ possible outcomes.

Thus, the sample space is:

If the coin shows Head (H), the die is rolled, giving outcomes H1, H2, H3, H4, H5, H6. If the coin shows Tail (T), no die is rolled, so T is the only outcome in that case.

Thus, the sample space is:

Label the people: room X has boys $\mathrm{B}_1, \mathrm{B}_2$ and girls $\mathrm{G}_1, \mathrm{G}_2$; room Y has boy $\mathrm{B}_3$ and girls $\mathrm{G}_3, \mathrm{G}_4, \mathrm{G}_5$. First a room (X or Y) is chosen, then a person from that room is selected.

The sample space is:

The selected die can be Red (R), White (W), or Blue (B), and the uppermost face can show $1, 2, 3, 4, 5$, or $6$. Each colour-number pair forms one outcome.

The sample space is:

(i) The first child can be a boy (B) or a girl (G), followed by another boy or girl. Listing all ordered pairs:

(ii) A family with 2 children may have no girl, one girl, or two girls:

Denote the red ball by R and the white ball by W. When the first ball drawn is red, the second must be white. When the first is white, the second can be red or white (since there are 3 identical white balls).

If the first toss is Head (H), the coin is tossed again giving H or T. If the first toss is Tail (T), a die is rolled giving $1, 2, 3, 4, 5$, or $6$.

Each of the 3 bulbs is classified in 2 ways (D or N), giving a total of $2 \times 2 \times 2 = 8$ outcomes.

If the coin shows Tail (T): the experiment ends — outcome is just T.

If the coin shows Head (H): a die is rolled. If an odd number (1, 3, 5) appears, the experiment ends. If an even number (2, 4, 6) appears, the die is rolled again.

The first draw can yield any of the four numbers. For each first draw, the second draw yields any of the remaining three. This gives $4 \times 3 = 12$ ordered pairs.

If the die shows an even number (2, 4, or 6), the coin is tossed once giving H or T: outcomes are $2\mathrm{H}, 2\mathrm{T}, 4\mathrm{H}, 4\mathrm{T}, 6\mathrm{H}, 6\mathrm{T}$.

If the die shows an odd number (1, 3, or 5), the coin is tossed twice giving HH, HT, TH, or TT: outcomes are $1\mathrm{HH}, 1\mathrm{HT}, 1\mathrm{TH}, 1\mathrm{TT}, 3\mathrm{HH}, 3\mathrm{HT}, 3\mathrm{TH}, 3\mathrm{TT}, 5\mathrm{HH}, 5\mathrm{HT}, 5\mathrm{TH}, 5\mathrm{TT}$.

Label the red balls $\mathrm{R}_1, \mathrm{R}_2$ and the black balls $\mathrm{B}_1, \mathrm{B}_2, \mathrm{B}_3$. If the coin shows Tail (T), one of the five balls is drawn. If the coin shows Head (H), a die is rolled giving $1, 2, 3, 4, 5, 6$.

Let A denote any outcome other than 6 (i.e., 1, 2, 3, 4, or 5) and B denote the outcome 6. Since 6 can appear on the first, second, third, or any subsequent throw, the sample space is infinite:

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

Since $\mathrm{E} \cap \mathrm{F} = \{4\} \cap \{2, 4, 6\} = \{4\} \neq \phi$, the events E and F are not mutually exclusive.

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

(i) Every number on a die is less than 7: $A = \{1,2,3,4,5,6\} = S$

(ii) No number on a die exceeds 7: $B = \phi$

(iii) Multiples of 3: $C = \{3, 6\}$

(iv) Numbers less than 4: $D = \{1, 2, 3\}$

(v) Even numbers greater than 4: $E = \{6\}$

(vi) Numbers $\geq 3$: $F = \{3, 4, 5, 6\}$

Now computing the required set operations:

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

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

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

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

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

$\mathrm{A} – \mathrm{C} = \{1,2,3,4,5,6\} – \{3,6\} = \{1,2,4,5\}$

$\mathrm{D} – \mathrm{E} = \{1,2,3\} – \{6\} = \{1,2,3\}$

$\mathrm{E} \cap \mathrm{F}’ = \{6\} \cap \{1,2\} = \phi$

$\mathrm{F}’ = \mathrm{S} – \{3,4,5,6\} = \{1,2\}$

The sample space has $6 \times 6 = 36$ elements. Identify each event:

Checking intersections:

$\mathrm{A} \cap \mathrm{B} = \phi$ $\Rightarrow$ A and B are mutually exclusive.

$\mathrm{B} \cap \mathrm{C} = \phi$ $\Rightarrow$ B and C are mutually exclusive.

$\mathrm{A} \cap \mathrm{C} = \{(3,6),(6,3),(4,5),(5,4),(6,6)\} \neq \phi$ $\Rightarrow$ A and C are not mutually exclusive.

The sample space is $\mathrm{S} = \{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$.

(i) Mutually exclusive pairs:

$\mathrm{A} \cap \mathrm{B} = \phi$ — A and B are mutually exclusive.

$\mathrm{A} \cap \mathrm{C} = \phi$ — A and C are mutually exclusive.

$\mathrm{A} \cap \mathrm{D} = \{\mathrm{HHH}\} \neq \phi$ — A and D are NOT mutually exclusive.

$\mathrm{B} \cap \mathrm{C} = \phi$ — B and C are mutually exclusive.

$\mathrm{B} \cap \mathrm{D} = \{\mathrm{HHT}, \mathrm{HTH}\} \neq \phi$ — B and D are NOT mutually exclusive.

$\mathrm{C} \cap \mathrm{D} = \phi$ — C and D are mutually exclusive.

Hence, A and B; A and C; B and C; C and D are mutually exclusive pairs.

(ii) Simple events (containing exactly one sample point): A and C.

(iii) Compound events (containing more than one sample point): B and D.

The sample space is $\mathrm{S} = \{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$.

(i) Let A: exactly one head, B: exactly one tail. Then $A = \{\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}\}$, $B = \{\mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}\}$. Since $\mathrm{A} \cap \mathrm{B} = \phi$, they are mutually exclusive.

(ii) Let A: at most one head, B: exactly two heads, C: exactly three heads. Then $A = \{\mathrm{TTT}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}\}$, $B = \{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$, $C = \{\mathrm{HHH}\}$. Since $A \cap B = \phi$, $A \cap C = \phi$, $B \cap C = \phi$, and $A \cup B \cup C = S$, they are mutually exclusive and exhaustive.

(iii) Let A: at most one head, B: at least one head. Then $A \cap B = \{\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}\} \neq \phi$, so they are not mutually exclusive.

(iv) Let A: exactly one head, B: exactly two heads. Then $A \cap B = \phi$ (mutually exclusive), but $A \cup B \neq S$ (not exhaustive).

(v) Let A: exactly one head, B: exactly two heads, C: exactly three heads. Then $A \cap B = A \cap C = B \cap C = \phi$ (mutually exclusive), but $A \cup B \cup C \neq S$ (not exhaustive, since TTT is missing).

Write out the sets A, B, C from the 36-element sample space:

(i) $A’ =$ not A = getting an odd number on the first die $\Rightarrow A’ = B$

(ii) not B = getting an even number on the first die $\Rightarrow B’ = A$

(iii) $A$ or $B = A \cup B = S$

(iv) $A$ and $B = A \cap B = \phi$

(v) A but not C $= A – C = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

(vi) $B$ or $C = B \cup C = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(2,1),(2,2),(2,3),(4,1)\}$

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

(viii) Since $B’ = A$, we have $A \cap B’ \cap C’ = A \cap A \cap C’ = A \cap C’ = A – C = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

(i) True, since $\mathrm{A} \cap \mathrm{B} = \phi$.

(ii) True, since $\mathrm{A} \cap \mathrm{B} = \phi$ and $\mathrm{A} \cup \mathrm{B} = \mathrm{S}$.

(iii) True (from Q.6 (ii), $B’ = A$).

(iv) False, since $\mathrm{A} \cap \mathrm{C} = \{(2,1),(2,2),(2,3),(4,1)\} \neq \phi$.

(v) False: since $B’ = A$, we have $A \cap B’ = A \cap A = A \neq \phi$.

(vi) False: since $A’ = B$, we have $A’ \cap B’ = B \cap A = \phi$, but $B’ \cap C = A \cap C \neq \phi$. Although $A’ \cup B’ \cup C = B \cup A \cup C = S \cup C = S$, they are not all mutually exclusive.

A valid probability assignment requires: (i) each $P(\omega_i) \geq 0$, and (ii) $\sum P(\omega_i) = 1$.

(a) Valid. All values are positive and $0.1+0.01+0.05+0.03+0.01+0.2+0.6 = 1$. ✓

(b) Valid. Each $P(\omega_i) = \dfrac{1}{7} > 0$ and $\dfrac{1}{7} \times 7 = 1$. ✓

(c) Not valid. $\sum P(\omega_i) = 0.1+0.2+0.3+0.4+0.5+0.6+0.7 = 2.8 \neq 1$. ✗

(d) Not valid. $P(\omega_1) = -0.1 < 0$. ✗

(e) Not valid. $P(\omega_7) = \dfrac{15}{14} > 1$. ✗

The sample space is $\mathrm{S} = \{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$, so $n(\mathrm{S}) = 4$.

Let E: at least one tail occurs. Then $\mathrm{E} = \{\mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$, so $n(\mathrm{E}) = 3$.

The sample space is $\mathrm{S} = \{1,2,3,4,5,6\}$, so $n(\mathrm{S}) = 6$.

(i) Prime numbers on a die: $\mathrm{E}_1 = \{2,3,5\}$, $n(\mathrm{E}_1) = 3$. $\mathrm{P}(\mathrm{E}_1) = \dfrac{3}{6} = \dfrac{1}{2}$

(ii) Numbers $\geq 3$: $\mathrm{E}_2 = \{3,4,5,6\}$, $n(\mathrm{E}_2) = 4$. $\mathrm{P}(\mathrm{E}_2) = \dfrac{4}{6} = \dfrac{2}{3}$

(iii) Numbers $\leq 1$: $\mathrm{E}_3 = \{1\}$, $n(\mathrm{E}_3) = 1$. $\mathrm{P}(\mathrm{E}_3) = \dfrac{1}{6}$

(iv) No die face exceeds 6: $\mathrm{E}_4 = \phi$ is an impossible event. $\mathrm{P}(\mathrm{E}_4) = 0$

(v) Numbers less than 6: $\mathrm{E}_5 = \{1,2,3,4,5\}$, $n(\mathrm{E}_5) = 5$. $\mathrm{P}(\mathrm{E}_5) = \dfrac{5}{6}$

(a) $n(\mathrm{S}) = 52$

(b) There is only one ace of spades: $n(\mathrm{E}) = 1$

(c)(i) There are 4 aces in the pack: $n(\mathrm{F}) = 4$

(c)(ii) There are 26 black cards (13 clubs + 13 spades): $n(\mathrm{G}) = 26$

The coin shows either 1 or 6 and the die shows 1–6, giving $n(\mathrm{S}) = 2 \times 6 = 12$ equally likely outcomes:

(i) Sum $= 3$: only $(1,2)$ qualifies. $\mathrm{P}(\mathrm{E}) = \dfrac{1}{12}$

(ii) Sum $= 12$: only $(6,6)$ qualifies. $\mathrm{P}(\mathrm{F}) = \dfrac{1}{12}$

There are $4 + 6 = 10$ council members in total. The number of ways to select one woman is ${}^6C_1 = 6$ and total ways to select any member is ${}^{10}C_1 = 10$.

The sample space has $2^4 = 16$ equally likely outcomes. The net amount depends on the number of heads and tails:

(i) 4 heads, 0 tails: wins ₹4

(ii) 3 heads, 1 tail: wins ₹3, loses ₹1.50 → net ₹1.5

(iii) 2 heads, 2 tails: wins ₹2, loses ₹3 → net $-$₹1

(iv) 1 head, 3 tails: wins ₹1, loses ₹4.50 → net $-$₹3.50

(v) 0 heads, 4 tails: loses ₹6

The sample space is $\mathrm{S} = \{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$, so $n(\mathrm{S}) = 8$.

(i) $\mathrm{E}_1 = \{\mathrm{HHH}\}$, $\mathrm{P}(\mathrm{E}_1) = \dfrac{1}{8}$

(ii) $\mathrm{E}_2 = \{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$, $\mathrm{P}(\mathrm{E}_2) = \dfrac{3}{8}$

(iii) At least 2 heads: $\mathrm{E}_3 = \{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HHH}\}$, $\mathrm{P}(\mathrm{E}_3) = \dfrac{4}{8} = \dfrac{1}{2}$

(iv) At most 2 heads: $\mathrm{E}_4 = \{\mathrm{TTT}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$, $\mathrm{P}(\mathrm{E}_4) = \dfrac{7}{8}$

(v) No head (all tails): $\mathrm{E}_5 = \{\mathrm{TTT}\}$, $\mathrm{P}(\mathrm{E}_5) = \dfrac{1}{8}$

(vi) $\mathrm{E}_6 = \{\mathrm{TTT}\}$, $\mathrm{P}(\mathrm{E}_6) = \dfrac{1}{8}$

(vii) Exactly two tails: $\mathrm{E}_7 = \{\mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}\}$, $\mathrm{P}(\mathrm{E}_7) = \dfrac{3}{8}$

(viii) No tail (all heads): $\mathrm{E}_8 = \{\mathrm{HHH}\}$, $\mathrm{P}(\mathrm{E}_8) = \dfrac{1}{8}$

(ix) At most two tails: $\mathrm{E}_9 = \{\mathrm{HHH}, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}, \mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}\}$, $\mathrm{P}(\mathrm{E}_9) = \dfrac{7}{8}$

Using the complement rule:

ASSASSINATION has 13 letters: 6 vowels (A, A, A, I, I, O) and 7 consonants (S, S, S, S, N, N, T).

(i) $\mathrm{P}(\text{vowel}) = \dfrac{6}{13}$

(ii) $\mathrm{P}(\text{consonant}) = \dfrac{7}{13}$

Six numbers can be chosen from 1 to 20 in ${}^{20}C_6$ ways, so $n(\mathrm{S}) = {}^{20}C_6$. There is only one winning combination, so $n(\mathrm{E}) = 1$.

(i) Since $A \cap B \subset A$, we must have $\mathrm{P}(A \cap B) \leq \mathrm{P}(A)$. Here $\mathrm{P}(A \cap B) = 0.6 > 0.5 = \mathrm{P}(A)$, which is a contradiction. So the probabilities are not consistently defined.

(ii) Using $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) – \mathrm{P}(A \cap B)$:

Since $0 \leq 0.1 \leq 1$, this is valid. The probabilities are consistently defined.

Using the addition formula: $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) – \mathrm{P}(A \cap B)$

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

(ii)

(iii)

Since A and B are mutually exclusive, $\mathrm{P}(A \cap B) = 0$. Applying the addition rule:

(i)

(ii) By De Morgan’s Law:

Applying De Morgan’s Law step by step:

Since $\mathrm{P}(E \cap F) \neq 0$, E and F are not mutually exclusive.

(i) $\mathrm{P}(\text{not A}) = 1 – \mathrm{P}(A) = 1 – 0.42 = 0.58$

(ii) $\mathrm{P}(\text{not B}) = 1 – \mathrm{P}(B) = 1 – 0.48 = 0.52$

(iii)

Let M: student studies Mathematics, B: student studies Biology. The given probabilities are:

Let $\mathrm{E}_1$: passes first exam, $\mathrm{E}_2$: passes second exam. Given $\mathrm{P}(\mathrm{E}_1) = 0.8$, $\mathrm{P}(\mathrm{E}_2) = 0.7$, $\mathrm{P}(\mathrm{E}_1 \cup \mathrm{E}_2) = 0.95$.

Applying the addition formula:

$\therefore$ The probability of passing both examinations is $\mathbf{0.55}$.

Let E: passes English, H: passes Hindi. Given $\mathrm{P}(E \cap H) = 0.5$, $\mathrm{P}(E’ \cap H’) = 0.1$, $\mathrm{P}(E) = 0.75$.

By De Morgan’s Law: $\mathrm{P}(E’ \cap H’) = \mathrm{P}(E \cup H)’ = 0.1$, so

Applying the addition formula:

Let E: opted for NCC, F: opted for NSS. The given probabilities are:

(i)

(ii)

(iii)

Total marbles $= 10 + 20 + 30 = 60$. Any 5 can be drawn in ${}^{60}C_5$ ways.

(i) All 5 blue: choose 5 from 20 blue marbles in ${}^{20}C_5$ ways.

(ii) It is easier to use the complement. There are $10 + 20 = 30$ non-green marbles.

The total number of ways to draw 4 cards from 52 is ${}^{52}C_4$. To get 3 diamonds from 13 and 1 spade from 13:

The six faces are marked $1, 1, 2, 2, 2, 3$.

(i) Three faces show 2: $\mathrm{P}(2) = \dfrac{3}{6} = \dfrac{1}{2}$

(ii) Two faces show 1 and one shows 3: $\mathrm{P}(1 \text{ or } 3) = \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

(iii) $\mathrm{P}(\text{not } 3) = 1 – \mathrm{P}(3) = 1 – \dfrac{1}{6} = \dfrac{5}{6}$

Total tickets $= 10{,}000$; prize tickets $= 10$; non-prize tickets $= 9{,}990$.

(a) $\mathrm{P}(\text{no prize with 1 ticket}) = \dfrac{9990}{10000} = \dfrac{999}{1000}$

(b) Two tickets from 10,000 in ${}^{10000}C_2$ ways; two non-prize tickets from 9,990 in ${}^{9990}C_2$ ways:

(c) Ten tickets from 10,000 in ${}^{10000}C_{10}$ ways; ten non-prize tickets from 9,990 in ${}^{9990}C_{10}$ ways:

Two sections (A with 40, B with 60) are formed from 100 students in ${}^{100}C_{40}$ ways.

(a) Both in section A: need 38 more from the remaining 98 students (${}^{98}C_{38}$ ways). Both in section B: need 40 from the remaining 98 (${}^{98}C_{40}$ ways).

(b)

Alternatively, by direct counting: one of the two enters section A in ${}^2C_1 = 2$ ways, and 39 more are chosen from the remaining 98 in ${}^{98}C_{39}$ ways:

Three letters $\mathrm{L}_1, \mathrm{L}_2, \mathrm{L}_3$ can be placed in envelopes $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ in ${}^3P_3 = 3! = 6$ ways. Listing all 6 arrangements, exactly 2 of them (arrangements (iv) and (v)) have no letter in its proper envelope.

Using the complement:

(i)

(ii) By De Morgan’s Law:

(iii)

(iv)