Probability - Exercise 14.1

(i) 1. By definition of complementary events, $P(E) + P(\bar{E}) = 1$.
(ii) 0; impossible event. An event with zero favorable outcomes has probability 0.
(iii) 1; sure or certain event. An event containing all sample outcomes is certain to happen.
(iv) 1. The total probability across all mutually exclusive elementary outcomes always sums to 1.
(v) 0 and 1. For any event $E$, $0 \le P(E) \le 1$.

(i) Not equally likely. A car's starting depends on its mechanical health, battery, and fuel. Starting and not starting are not equally probable.
(ii) Not equally likely. Shooting success depends heavily on the player's training, shooting distance, and skill.
(iii) Equally likely. In a true/false trial with no prior bias, guessing leads to two equally probable outcomes: right or wrong ($P = \frac{1}{2}$).
(iv) Equally likely. Biologically and statistically, a newborn is equally likely to be a boy or a girl ($P = \frac{1}{2}$).

When an unbiased coin is tossed, there are exactly two possible outcomes: Head ($H$) and Tail ($T$).
These two outcomes are mutually exclusive and equally likely: $$P(H) = P(T) = \frac{1}{2}$$ The outcome of any individual toss is entirely random and impossible to predict beforehand. Hence, tossing a coin is completely fair and unbiased.

For any random event $E$, the probability must satisfy: $$0 \le P(E) \le 1$$ Evaluating the given options:
(A) $\frac{2}{3} \approx 0.67$ (Valid)
(B) $-1.5 < 0$ (Invalid; probability cannot be negative)
(C) $15\% = 0.15$ (Valid)
(D) $0.7$ (Valid)

The probability of the complement event $\bar{E}$ ('not $E$') is given by: $$P(\bar{E}) = 1 - P(E)$$ $$P(\text{not } E) = 1 - 0.08 = 0.92$$

(i) The bag contains 0 orange candies. Drawing an orange candy is an impossible event: $$P(\text{Orange Candy}) = \frac{0}{\text{Total}} = 0$$ (ii) Every candy in the bag is lemon flavoured. Drawing a lemon candy is a sure event: $$P(\text{Lemon Candy}) = \frac{\text{Total}}{\text{Total}} = 1$$

Let $E$ be the event that 2 students have the same birthday.
Then $\bar{E}$ is the event that they do not have the same birthday.
Given $P(\bar{E}) = 0.992$: $$P(E) = 1 - P(\bar{E}) = 1 - 0.992 = 0.008$$

Total number of balls $= 4 + 6 = 10$.
(i) Number of red balls $= 4$: $$P(\text{Red}) = \frac{4}{10} = \frac{2}{5} = 0.4$$ (ii) Probability of not red: $$P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - 0.4 = 0.6 = \frac{3}{5}$$

Total marbles $= 6 + 7 + 5 = 18$.
(i) Red marbles $= 6 \implies P(\text{Red}) = \frac{6}{18} = \frac{1}{3}$.
(ii) White marbles $= 7 \implies P(\text{White}) = \frac{7}{18}$.
(iii) Not green means either red or white $= 6 + 7 = 13$: $$P(\text{Not Green}) = \frac{13}{18}$$

Total number of coins $= 100 + 50 + 20 + 10 = 180$.
(i) Favourable outcomes for $50\text{p}$ coin $= 100$: $$P(50\text{p}) = \frac{100}{180} = \frac{5}{9}$$ (ii) Number of $₹5$ coins $= 10$. Therefore, coins that are not $₹5 = 180 - 10 = 170$: $$P(\text{Not } ₹5) = \frac{170}{180} = \frac{17}{18}$$

Number of male fish $= 6$.
Number of female fish $= 9$.
Total number of fish in tank $= 6 + 9 = 15$.
$$P(\text{Male Fish}) = \frac{\text{Number of Male Fish}}{\text{Total Fish}} = \frac{6}{15} = \frac{2}{5} = 0.4$$

Total possible outcomes $= 8$ (numbers $\{1, 2, 3, 4, 5, 6, 7, 8\}$).
(i) Number 5 appears once $\implies P(5) = \frac{1}{8}$.
(ii) Even numbers: $\{2, 4, 6, 8\}$ (4 numbers) $\implies P(\text{Even}) = \frac{4}{8} = \frac{1}{2}$.
(iii) Numbers greater than 3: $\{4, 5, 6, 7, 8\}$ (5 numbers) $\implies P(>3) = \frac{5}{8}$.
(iv) Numbers less than 9: all 8 numbers $\implies P(<9) = \frac{8}{8} = 1$.

Sample space $S = \{1, 2, 3, 4, 5, 6\}$, Total outcomes $= 6$.
(i) Prime numbers in $S = \{2, 3, 5\}$ (3 outcomes): $$P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$$ (ii) Numbers strictly between 2 and 5 $= \{3, 4\}$ (2 outcomes): $$P(\text{Between 2 and 5}) = \frac{2}{6} = \frac{1}{3}$$ (iii) Odd numbers in $S = \{1, 3, 5\}$ (3 outcomes): $$P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$$

Total cards $n(S) = 52$.
(i) Black queens: Queen of Spades and Queen of Clubs (2 cards): $$P(\text{Black Queen}) = \frac{2}{52} = \frac{1}{26}$$ (ii) Face cards: 4 Jacks, 4 Queens, 4 Kings (12 cards): $$P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}$$ (iii) Red face cards: Hearts and Diamonds (6 cards: 2 Jacks, 2 Queens, 2 Kings): $$P(\text{Red Face Card}) = \frac{6}{52} = \frac{3}{26}$$ (iv) Jack of Spades: Exactly 1 card: $$P(\text{Jack of Spades}) = \frac{1}{52}$$

(i) Total cards $= 5$ (10, J, Q, K, A of diamonds). Number of queens $= 1$: $$P(\text{Queen}) = \frac{1}{5}$$ (ii) The queen is drawn and placed aside. Remaining cards $= 5 - 1 = 4$ (Ten, Jack, King, Ace).
(a) Number of aces $= 1$: $$P(\text{Ace}) = \frac{1}{4}$$ (b) Number of queens in the remaining 4 cards $= 0$: $$P(\text{Queen}) = \frac{0}{4} = 0$$

Defective pens $= 10$
Good pens $= 140$
Total number of pens in the lot $= 10 + 140 = 150$
$$P(\text{Good Pen}) = \frac{\text{Number of Good Pens}}{\text{Total Pens}} = \frac{140}{150} = \frac{14}{15}$$

(i) Total bulbs $= 50$. Defective bulbs $= 5$: $$P(\text{Defective}) = \frac{5}{50} = \frac{1}{10} = 0.1$$ (ii) A non-defective bulb is drawn and not replaced.
New total number of bulbs $= 50 - 1 = 49$.
Initial non-defective bulbs $= 50 - 5 = 45$.
Remaining non-defective bulbs $= 45 - 1 = 44$.
$$P(\text{Not Defective}) = \frac{44}{49}$$

Total cards in box $= 100$.
(i) Two-digit numbers run from $10$ to $99$ inclusive: $$\text{Count} = 99 - 10 + 1 = 90$$ $$P(\text{Two-Digit}) = \frac{90}{100} = \frac{9}{10} = 0.9$$ (ii) Perfect squares from $1$ to $100$: $\{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$ (10 numbers): $$P(\text{Perfect Square}) = \frac{10}{100} = \frac{1}{10} = 0.1$$ (iii) Numbers divisible by 7: $7 \times 1, 7 \times 2, \dots, 7 \times 14 = 98$ (14 numbers): $$P(\text{Divisible by 7}) = \frac{14}{100} = \frac{7}{50} = 0.14$$

Total faces $= 6$.
(i) The letter 'P' appears on 2 faces: $$P(\text{P}) = \frac{2}{6} = \frac{1}{3}$$ (ii) The letter 'T' appears on 1 face: $$P(\text{T}) = \frac{1}{6}$$

Total area of rectangular region: $$\text{Area of Rectangle} = 4 \times 3 = 12\text{ m}^2$$ Diameter of circle $= 1.4\text{ m} \implies$ Radius $r = 0.7\text{ m}$.
$$\text{Area of Circle} = \pi r^2 = \frac{22}{7} \times (0.7)^2 = \frac{22}{7} \times 0.49 = 22 \times 0.07 = 1.54\text{ m}^2$$ $$P(\text{Inside Circle}) = \frac{\text{Area of Circle}}{\text{Total Area}} = \frac{1.54}{12} = \frac{154}{1200} = \frac{77}{600} \approx 0.128$$

Total pens $= 200$.
Defective pens $= 25$.
Good pens $= 200 - 25 = 175$.
(i) Rohit buys the pen if it is good: $$P(\text{Buys}) = \frac{175}{200} = \frac{7}{8} = 0.875$$ (ii) Rohit does not buy if it is defective: $$P(\text{Does not buy}) = \frac{25}{200} = \frac{1}{8} = 0.125$$

Total possible outcomes when two dice are rolled $= 6 \times 6 = 36$.
(i) Outcomes giving a sum of 10: $$\{(4, 6), (5, 5), (6, 4)\} \implies 3 \text{ outcomes}$$ $$P(\text{Sum } = 10) = \frac{3}{36} = \frac{1}{12}$$ (ii) Outcomes giving a sum $\le 5$: $$\text{Sum 2: } (1, 1)$$ $$\text{Sum 3: } (1, 2), (2, 1)$$ $$\text{Sum 4: } (1, 3), (2, 2), (3, 1)$$ $$\text{Sum 5: } (1, 4), (2, 3), (3, 2), (4, 1)$$ Total favourable outcomes $= 1 + 2 + 3 + 4 = 10$ outcomes.
$$P(\text{Sum } \le 5) = \frac{10}{36} = \frac{5}{18}$$

Total possible outcomes $= 2^3 = 8$: $$S = \{\text{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}\}$$ Winning outcomes (all same) $= \{\text{HHH, TTT}\} \implies 2 \text{ outcomes}$.
Losing outcomes (not all same) $= 8 - 2 = 6$ outcomes: $$P(\text{Lose}) = \frac{6}{8} = \frac{3}{4} = 0.75$$

Throwing a die twice has $6 \times 6 = 36$ possible outcomes.
Outcomes where 4 comes up at least once: $$\text{4 on 1st die: } (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) \quad (6 \text{ pairs})$$ $$\text{4 on 2nd die: } (1, 4), (2, 4), (3, 4), (5, 4), (6, 4) \quad (5 \text{ pairs})$$ Total outcomes with 4 at least once $= 6 + 5 = 11$.
(ii) $P(\text{4 at least once}) = \frac{11}{36}$.
(i) By complement rule, 4 does not come up either time: $$P(\text{4 not either time}) = 1 - P(\text{4 at least once}) = 1 - \frac{11}{36} = \frac{25}{36}$$

(i) Incorrect. When two coins are tossed, the sample space is: $$S = \{\text{HH, HT, TH, TT}\} \implies 4 \text{ equally likely outcomes}$$ "Two heads" $= \{\text{HH}\}$ has $P = \frac{1}{4}$.
"Two tails" $= \{\text{TT}\}$ has $P = \frac{1}{4}$.
"One of each" $= \{\text{HT, TH}\}$ has $P = \frac{2}{4} = \frac{1}{2}$.
The outcomes are not equally likely, so the probability is not $\frac{1}{3}$.

(ii) Correct. When a fair die is rolled, the six elementary outcomes $\{1, 2, 3, 4, 5, 6\}$ are equally likely.
Odd numbers $= \{1, 3, 5\}$ (3 outcomes).
Even numbers $= \{2, 4, 6\}$ (3 outcomes).
Since both sets have 3 equally likely outcomes: $$P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$$