Probability

(i) For any event E and its complement $\bar{E}$, we know that $P(E) + P(\bar{E}) = 1$.

Answer: 1

(ii) An event that cannot happen has no favourable outcomes, so its probability = $\dfrac{0}{n} = 0$.

Answer: 0; such an event is called an impossible event.

(iii) An event that is certain to happen has all outcomes favourable, so its probability = $\dfrac{n}{n} = 1$.

Answer: 1; such an event is called a sure (or certain) event.

(iv) Since all elementary events together cover the entire sample space, their probabilities add up to:

Answer: 1

(v) Probability is always a number between 0 and 1 (inclusive).

Answer: greater than or equal to 0 and less than or equal to 1.

Concept: Outcomes are equally likely if each outcome has the same chance of occurring.

(i) A car starting or not starting depends on the condition of the car, fuel, battery, etc. These two outcomes are not equally likely.

(ii) Whether a player shoots or misses depends on the player's skill and practice. These two outcomes are not equally likely.

(iii) A true-false question has exactly two options — right or wrong. Since there is no prior knowledge assumed, each option is equally likely. These outcomes are equally likely.

(iv) A baby being born a boy or a girl are two outcomes that are (approximately) equally likely in nature. These outcomes are equally likely.

Given: A coin is tossed to decide which team gets the ball.

Reason: When a fair coin is tossed, there are only two possible outcomes — Head (H) and Tail (T). Both outcomes are equally likely, each with probability $\dfrac{1}{2}$.

Since neither team has an advantage over the other (each has an equal chance of winning the toss), tossing a coin is considered a fair and unbiased method of making the decision.

Concept: The probability of any event E must satisfy $0 \leq P(E) \leq 1$.

Checking each option:
- (A) $\dfrac{2}{3} \approx 0.667$ — lies between 0 and 1. ✓ Valid.
- (B) $-1.5$ — is negative, which is less than 0. ✗ Not valid.
- (C) $15\% = 0.15$ — lies between 0 and 1. ✓ Valid.
- (D) $0.7$ — lies between 0 and 1. ✓ Valid.

Answer: (B) $-1.5$ cannot be the probability of an event.

Given: $P(E) = 0.05$

Formula: $P(E) + P(\bar{E}) = 1$

$$P(\bar{E}) = 1 - P(E) = 1 - 0.05 = 0.95$$

The probability of 'not E' is $0.95$.

Given: The bag contains only lemon flavoured candies.

(i) Probability of taking out an orange flavoured candy:

Since there are no orange flavoured candies in the bag, this is an impossible event.
$$P(\text{orange flavoured candy}) = 0$$

(ii) Probability of taking out a lemon flavoured candy:

Since all candies are lemon flavoured, this is a certain event.
$$P(\text{lemon flavoured candy}) = 1$$

Given: $P(\text{2 students do NOT have the same birthday}) = 0.992$

Formula: $P(E) + P(\bar{E}) = 1$

Let $E$ = event that 2 students have the same birthday.

Then $\bar{E}$ = event that 2 students do not have the same birthday.

$$P(E) = 1 - P(\bar{E}) = 1 - 0.992 = 0.008$$

The probability that the 2 students have the same birthday is $0.008$.

Given: 3 red balls + 5 black balls = 8 balls total.

Total number of possible outcomes = 8.

(i) Probability of drawing a red ball:

Number of red balls = 3
$$P(\text{red}) = \frac{3}{8}$$

(ii) Probability of drawing a ball that is not red:

Number of non-red (black) balls = 5
$$P(\text{not red}) = \frac{5}{8}$$

Verification: $\dfrac{3}{8} + \dfrac{5}{8} = 1$ ✓

Given: 5 red + 8 white + 4 green = 17 marbles total.

Total number of possible outcomes = 17.

(i) Probability of getting a red marble:
$$P(\text{red}) = \frac{5}{17}$$

(ii) Probability of getting a white marble:
$$P(\text{white}) = \frac{8}{17}$$

(iii) Probability of getting a marble that is not green:

Number of non-green marbles = 5 + 8 = 13
$$P(\text{not green}) = \frac{13}{17}$$

Given:
- 50p coins = 100
- ₹1 coins = 50
- ₹2 coins = 20
- ₹5 coins = 10

Total number of coins = $100 + 50 + 20 + 10 = 180$

(i) Probability that the coin is a 50p coin:
$$P(\text{50p coin}) = \frac{100}{180} = \frac{5}{9}$$

(ii) Probability that the coin is not a ₹5 coin:

Number of coins that are not ₹5 = $180 - 10 = 170$
$$P(\text{not a ₹5 coin}) = \frac{170}{180} = \frac{17}{18}$$

Given: 5 male fish + 8 female fish = 13 fish total.

Total number of possible outcomes = 13.

Number of favourable outcomes (male fish) = 5.

$$P(\text{male fish}) = \frac{5}{13}$$

The probability that the fish taken out is a male fish is $\dfrac{5}{13}$.

Given: The spinner has 8 equally likely outcomes: {1, 2, 3, 4, 5, 6, 7, 8}.

Total number of outcomes = 8.

(i) Probability of pointing at 8:

Favourable outcomes = {8} → 1 outcome
$$P(8) = \frac{1}{8}$$

(ii) Probability of pointing at an odd number:

Odd numbers in {1,2,3,4,5,6,7,8} = {1, 3, 5, 7} → 4 outcomes
$$P(\text{odd}) = \frac{4}{8} = \frac{1}{2}$$

(iii) Probability of pointing at a number greater than 2:

Numbers greater than 2 = {3, 4, 5, 6, 7, 8} → 6 outcomes
$$P(\text{greater than 2}) = \frac{6}{8} = \frac{3}{4}$$

(iv) Probability of pointing at a number less than 9:

All numbers 1 through 8 are less than 9 → 8 outcomes
$$P(\text{less than 9}) = \frac{8}{8} = 1$$

This is a certain event, as all outcomes satisfy the condition.

Given: A die is thrown once. Possible outcomes = {1, 2, 3, 4, 5, 6}.

Total number of outcomes = 6.

(i) Probability of getting a prime number:

Prime numbers in {1,2,3,4,5,6} = {2, 3, 5} → 3 outcomes
$$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$$

(ii) Probability of getting a number lying between 2 and 6:

Numbers strictly between 2 and 6 = {3, 4, 5} → 3 outcomes
$$P(\text{between 2 and 6}) = \frac{3}{6} = \frac{1}{2}$$

(iii) Probability of getting an odd number:

Odd numbers = {1, 3, 5} → 3 outcomes
$$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$

Given: A well-shuffled deck of 52 cards. Total outcomes = 52.

(i) A king of red colour:

There are 2 red kings (King of Hearts and King of Diamonds).
$$P(\text{red king}) = \frac{2}{52} = \frac{1}{26}$$

(ii) A face card:

Face cards = Jack, Queen, King of each of 4 suits = $3 \times 4 = 12$ cards.
$$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$$

(iii) A red face card:

Red face cards = Face cards of Hearts and Diamonds = $3 + 3 = 6$ cards.
$$P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$$

(iv) The jack of hearts:

There is only 1 jack of hearts.
$$P(\text{jack of hearts}) = \frac{1}{52}$$

(v) A spade:

There are 13 spade cards in a deck.
$$P(\text{spade}) = \frac{13}{52} = \frac{1}{4}$$

(vi) The queen of diamonds:

There is only 1 queen of diamonds.
$$P(\text{queen of diamonds}) = \frac{1}{52}$$

Given: 5 cards: Ten, Jack, Queen, King, Ace (all of diamonds).

(i) Probability that the card is the queen:

Total outcomes = 5; Favourable outcomes = 1 (Queen)
$$P(\text{queen}) = \frac{1}{5}$$

(ii) The queen is drawn and put aside. Remaining cards = {Ten, Jack, King, Ace} → 4 cards.

(a) Probability that the second card is an ace:

Favourable outcomes = 1 (Ace)
$$P(\text{ace}) = \frac{1}{4}$$

(b) Probability that the second card is a queen:

The queen has already been removed, so there are 0 queens left.
$$P(\text{queen}) = \frac{0}{4} = 0$$

This is an impossible event.

Given:
- Defective pens = 12
- Good pens = 132
- Total pens = $12 + 132 = 144$

Total number of outcomes = 144.

Number of favourable outcomes (good pen) = 132.

$$P(\text{good pen}) = \frac{132}{144} = \frac{11}{12}$$

The probability that the pen taken out is a good one is $\dfrac{11}{12}$.

(i) Probability that the drawn bulb is defective:

Given: Total bulbs = 20; Defective bulbs = 4.

$$P(\text{defective}) = \frac{4}{20} = \frac{1}{5}$$

(ii) After a non-defective bulb is drawn and not replaced:

Remaining total bulbs = $20 - 1 = 19$

Remaining defective bulbs = 4 (unchanged, since a non-defective was removed)

Remaining non-defective bulbs = $19 - 4 = 15$

$$P(\text{not defective}) = \frac{15}{19}$$

The probability that the second bulb drawn is not defective is $\dfrac{15}{19}$.

Given: Discs numbered 1 to 90. Total outcomes = 90.

(i) Probability of a two-digit number:

Two-digit numbers from 1 to 90: 10, 11, 12, …, 90 → $90 - 10 + 1 = 81$ numbers.
$$P(\text{two-digit}) = \frac{81}{90} = \frac{9}{10}$$

(ii) Probability of a perfect square number:

Perfect squares from 1 to 90: $1, 4, 9, 16, 25, 36, 49, 64, 81$ → 9 numbers.
$$P(\text{perfect square}) = \frac{9}{90} = \frac{1}{10}$$

(iii) Probability of a number divisible by 5:

Numbers divisible by 5 from 1 to 90: $5, 10, 15, \ldots, 90$ → $\dfrac{90}{5} = 18$ numbers.
$$P(\text{divisible by 5}) = \frac{18}{90} = \frac{1}{5}$$

Given: The six faces of the die show: A, B, C, D, E, A.

Total outcomes = 6.

(i) Probability of getting A:

The letter A appears on 2 faces.
$$P(A) = \frac{2}{6} = \frac{1}{3}$$

(ii) Probability of getting D:

The letter D appears on 1 face.
$$P(D) = \frac{1}{6}$$

Given:
- Rectangular region: length = 3 m, width = 2 m (assumed standard dimensions from the figure).
- Circle with diameter = 1 m, so radius $r = \dfrac{1}{2}$ m.

Area of the rectangle:
$$A_{\text{rect}} = 3 \times 2 = 6 \text{ m}^2$$

Area of the circle:
$$A_{\text{circle}} = \pi r^2 = \pi \times \left(\frac{1}{2}\right)^2 = \frac{\pi}{4} \text{ m}^2$$

Probability that the die lands inside the circle:
$$P = \frac{A_{\text{circle}}}{A_{\text{rect}}} = \frac{\pi/4}{6} = \frac{\pi}{24}$$

The required probability is $\dfrac{\pi}{24}$.

Given:
- Total pens = 144
- Defective pens = 20
- Good pens = $144 - 20 = 124$

(i) Probability that she will buy it (pen is good):
$$P(\text{good}) = \frac{124}{144} = \frac{31}{36}$$

(ii) Probability that she will not buy it (pen is defective):
$$P(\text{defective}) = \frac{20}{144} = \frac{5}{36}$$

Verification: $\dfrac{31}{36} + \dfrac{5}{36} = 1$ ✓

Given: Two dice are thrown simultaneously. Total outcomes = $6 \times 6 = 36$.

(i) Completing the table:

We list the number of ways each sum can occur:

- Sum = 2: {(1,1)} → 1 way → $P = \dfrac{1}{36}$
- Sum = 3: {(1,2),(2,1)} → 2 ways → $P = \dfrac{2}{36} = \dfrac{1}{18}$
- Sum = 4: {(1,3),(2,2),(3,1)} → 3 ways → $P = \dfrac{3}{36} = \dfrac{1}{12}$
- Sum = 5: {(1,4),(2,3),(3,2),(4,1)} → 4 ways → $P = \dfrac{4}{36} = \dfrac{1}{9}$
- Sum = 6: {(1,5),(2,4),(3,3),(4,2),(5,1)} → 5 ways → $P = \dfrac{5}{36}$
- Sum = 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} → 6 ways → $P = \dfrac{6}{36} = \dfrac{1}{6}$
- Sum = 8: {(2,6),(3,5),(4,4),(5,3),(6,2)} → 5 ways → $P = \dfrac{5}{36}$
- Sum = 9: {(3,6),(4,5),(5,4),(6,3)} → 4 ways → $P = \dfrac{4}{36} = \dfrac{1}{9}$
- Sum = 10: {(4,6),(5,5),(6,4)} → 3 ways → $P = \dfrac{3}{36} = \dfrac{1}{12}$
- Sum = 11: {(5,6),(6,5)} → 2 ways → $P = \dfrac{2}{36} = \dfrac{1}{18}$
- Sum = 12: {(6,6)} → 1 way → $P = \dfrac{1}{36}$

Completed Table:

| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ |

(ii) The student's argument is not correct.

The 11 outcomes (sums 2 through 12) are not equally likely. For example, sum = 7 can occur in 6 ways while sum = 2 can occur in only 1 way. Equally likely outcomes means each outcome has the same chance, which is not the case here. The correct probabilities are as computed in the table above.

Given: A coin is tossed 3 times.

Total possible outcomes: $2^3 = 8$

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Outcomes where Hanif wins (all same): {HHH, TTT} → 2 outcomes

$$P(\text{win}) = \frac{2}{8} = \frac{1}{4}$$

Probability that Hanif loses:
$$P(\text{lose}) = 1 - P(\text{win}) = 1 - \frac{1}{4} = \frac{3}{4}$$

The probability that Hanif will lose the game is $\dfrac{3}{4}$.

Given: A die is thrown twice. Total outcomes = $6 \times 6 = 36$.

(i) Probability that 5 will not come up either time:

On each throw, outcomes other than 5 = {1, 2, 3, 4, 6} → 5 outcomes.

Number of outcomes where 5 does not appear at all = $5 \times 5 = 25$.

$$P(\text{5 not at all}) = \frac{25}{36}$$

(ii) Probability that 5 will come up at least once:

Using the complement:
$$P(\text{5 at least once}) = 1 - P(\text{5 not at all}) = 1 - \frac{25}{36} = \frac{11}{36}$$

Verification by direct count: Outcomes with at least one 5:
- 5 on first die: (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) → 6
- 5 on second die: (1,5),(2,5),(3,5),(4,5),(6,5) → 5 (excluding (5,5) already counted)
- Total = 11 outcomes ✓

$$P(\text{5 at least once}) = \frac{11}{36}$$

(i) This argument is NOT correct.

When two coins are tossed, the correct sample space is:
{HH, HT, TH, TT} → 4 equally likely outcomes, not 3.

The outcome 'one head and one tail' can occur in 2 ways (HT and TH), so it is not equally likely with the other two outcomes.

Correct probabilities:
- $P(\text{two heads}) = \dfrac{1}{4}$
- $P(\text{two tails}) = \dfrac{1}{4}$
- $P(\text{one of each}) = \dfrac{2}{4} = \dfrac{1}{2}$

Since the three outcomes are not equally likely, assigning probability $\dfrac{1}{3}$ to each is incorrect.

(ii) This argument is CORRECT.

When a die is thrown, the sample space = {1, 2, 3, 4, 5, 6} — 6 equally likely outcomes.

Odd numbers = {1, 3, 5} → 3 outcomes; Even numbers = {2, 4, 6} → 3 outcomes.

Both groups have the same number of equally likely outcomes, so:
$$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$

The argument is correct.