Solution:

Complete the Following Statements

(i) Probability of an event E + Probability of the event ‘not E’ = 1.

(ii) The probability of an event that cannot happen is called 0.

(iii) The probability of an event that is certain to happen is called 1.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

Solution:

Equally Likely Outcomes in Experiments

(i) A driver attempts to start a car. The car starts or does not start.

Explanation: This experiment has two outcomes that may not be equally likely, as the car may have a higher chance of starting if it is in good condition.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Explanation: This experiment may not have equally likely outcomes, as a player's skill affects the likelihood of making the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Explanation: This experiment may have equally likely outcomes if the question is random and does not favor either answer.

(iv) A baby is born. It is a boy or a girl.

Explanation: This experiment is often considered to have equally likely outcomes, assuming equal probability for boys and girls at birth.

Solution:
Tossing a coin is considered fair because each outcome—heads or tails—has an equal probability of 50%. This randomness ensures that neither team has an advantage, allowing for an impartial decision to determine possession.

Solution:
Probability of an event :
\[0 \leq P(E) \leq 1\] (B) –1.5 cannot be the probability of an event, as probabilities must be between 0 and 1, both extreme values inclusive.

Solution:
The probability of "not E," often denoted as \( P(\text{not } E) \), can be calculated using the formula: \[ P(\text{not } E) = 1 - P(E) \] Given \( P(E) = 0.05 \): \[ P(\text{not } E) = 1 - 0.05 = 0.95 \] So, the probability of "not E" is 0.95.

Solution:
(i) The probability that Malini takes out an orange flavoured candy is 0, since the bag contains only lemon flavoured candies.

(ii) The probability that she takes out a lemon flavoured candy is 1, as that is the only type in the bag.

Solution:
The probability that the two students have the same birthday can be found by subtracting the probability of them not having the same birthday from 1: \[ P(\text{same birthday}) = 1 - P(\text{not same birthday}) = 1 - 0.992 = 0.008 \] So, the probability is 0.008.

Solution:
To find the probabilities, we first determine the total number of balls in the bag: Total balls = 3 (red) + 5 (black) = 8 balls.

(i) Probability of drawing a red ball: \[P(\text{red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \dfrac{3}{8}\] (ii) Probability of drawing a ball that is not red: \[P(\text{not red}) = 1 - P(\text{red}) = 1 - \dfrac{3}{8} = \dfrac{5}{8}\] So, the probabilities are:

(i) \( \dfrac{3}{8} \)
(ii) \( \dfrac{5}{8} \)

Solution:
To find the probabilities, we first calculate the total number of marbles: Total marbles = 5 (red) + 8 (white) + 4 (green) = 17. (i) Probability of drawing a red marble: \[ P(\text{red}) = \frac{5}{17} \] (ii) Probability of drawing a white marble: \[ P(\text{white}) = \frac{8}{17} \] (iii) Probability of not drawing a green marble: Total non-green marbles = 5 (red) + 8 (white) = 13. \[ P(\text{not green}) = \frac{13}{17} \]

Solution:
To find the probabilities, we first calculate the total number of coins in the piggy bank. Total number of coins: - 50p coins: 100 - Re. 1 coins: 50 - Rs. 2 coins: 20 - Rs. 5 coins: 10 Total coins = 100 + 50 + 20 + 10 = 180. (i) Probability of drawing a 50p coin: \[ P(\text{50p coin}) = \frac{100}{180} = \frac{5}{9} \] (ii) Probability of not drawing a Rs. 5 coin: Total coins that are not Rs. 5 = 100 (50p) + 50 (Re. 1) + 20 (Rs. 2) = 170. \[ P(\text{not Rs. 5 coin}) = \frac{170}{180} = \frac{17}{18} \]

Solution:
To find the probability that the fish taken out is a male fish, we first determine the total number of fish in the tank.

Total fish = 5 male fish + 8 female fish = 13 fish.

Now, the probability of drawing a male fish is given by the ratio of the number of male fish to the total number of fish:
\[ P(\text{male fish}) = \frac{\text{Number of male fish}}{\text{Total number of fish}} = \dfrac{5}{13} \]
So, the probability that the fish taken out is a male fish is \( \dfrac{5}{13} \).

Solution:
In this game of chance, the total number of possible outcomes when the arrow is spun is 8 (the numbers 1 through 8).

(i) Probability that it will point at 8:
\[ P(8) = \dfrac{1}{8} \]
(ii) Probability that it will point at an odd number (1, 3, 5, 7):
There are 4 odd numbers.
\[ P(\text{odd number}) = \dfrac{4}{8} = \frac{1}{2} \]
(iii) Probability that it will point at a number greater than 2 (3, 4, 5, 6, 7, 8):
There are 6 numbers greater than 2.
\[ P(\text{greater than 2}) = \dfrac{6}{8} = \dfrac{3}{4} \]
(iv) Probability that it will point at a number less than 9:
All numbers (1, 2, 3, 4, 5, 6, 7, 8) are less than 9, so there are 8 favorable outcomes.
\[ P(\text{less than 9}) = \dfrac{8}{8} = 1 \]
To summarize:
(i) \( \dfrac{1}{8} \)
(ii) \( \dfrac{1}{2} \)
(iii) \( \dfrac{3}{4} \)
(iv) \( 1 \)

Solution:
When a die is thrown, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. This gives us a total of 6 outcomes. (i) **Probability of getting a prime number (2, 3, 5):** The prime numbers on a die are 2, 3, and 5, which gives us 3 favorable outcomes. \[ P(\text{prime number}) = \frac{3}{6} = \frac{1}{2} \] (ii) **Probability of getting a number lying between 2 and 6 (3, 4, 5):** The numbers that lie between 2 and 6 are 3, 4, and 5, giving us 3 favorable outcomes. \[ P(\text{between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \] (iii) **Probability of getting an odd number (1, 3, 5):** The odd numbers on a die are 1, 3, and 5, giving us 3 favorable outcomes. \[ P(\text{odd number}) = \frac{3}{6} = \frac{1}{2} \] To summarize: - (i) \( \frac{1}{2} \) - (ii) \( \frac{1}{2} \) - (iii) \( \frac{1}{2} \)

Solution:
In a standard deck of 52 cards, we can calculate the probabilities for each scenario:

(i) Probability of getting a king of red color (2 cards: King of Hearts, King of Diamonds): \[ P(\text{red king}) = \dfrac{2}{52} = \dfrac{1}{26} \]
(ii) Probability of getting a face card (12 cards: Kings, Queens, Jacks of all suits): \[ P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13} \]
(iii) Probability of getting a red face card (6 cards: King, Queen, Jack of Hearts and Diamonds): \[ P(\text{red face card}) = \dfrac{6}{52} = \dfrac{3}{26} \]
(iv) Probability of getting the jack of hearts (1 card): \[ P(\text{jack of hearts}) = \dfrac{1}{52} \]
(v) Probability of getting a spade (13 cards): \[ P(\text{spade}) = \dfrac{13}{52} = \dfrac{1}{4}\]
(vi) Probability of getting the queen of diamonds (1 card): \[ P(\text{queen of diamonds}) = \dfrac{1}{52}\]
To summarize:
(i) \( \dfrac{1}{26} \)
(ii) \( \dfrac{3}{13} \)
(iii) \( \dfrac{3}{26} \)
(iv) \( \dfrac{1}{52} \)
(v) \( \dfrac{1}{4} \)
(vi) \( \dfrac{1}{52} \)

Solution:
In this scenario, there are 5 cards: the ten, jack, queen, king, and ace of diamonds. (i) **Probability that the card is the queen:** There is 1 queen among the 5 cards. \[ P(\text{queen}) = \frac{1}{5} \] (ii) If the queen is drawn and put aside, there are now 4 cards remaining (ten, jack, king, and ace). (a) **Probability that the second card picked up is an ace:** There is 1 ace among the 4 remaining cards. \[ P(\text{ace}) = \frac{1}{4} \] (b) **Probability that the second card picked up is a queen:** There are no queens left since the queen was put aside. \[ P(\text{queen}) = 0 \] To summarize: - (i) \( \frac{1}{5} \) - (ii)(a) \( \frac{1}{4} \) - (ii)(b) \( 0 \)

Solution:
To determine the probability that a pen taken out at random is a good one, we first need to find the total number of pens. Total pens = Number of good pens + Number of defective pens Total pens = 132 good pens + 12 defective pens = 144 pens. Now, the probability of picking a good pen is given by the ratio of the number of good pens to the total number of pens: \[ P(\text{good pen}) = \frac{\text{Number of good pens}}{\text{Total number of pens}} = \frac{132}{144} = \frac{11}{12}. \] Thus, the probability that the pen taken out is a good one is \( \frac{11}{12} \).

Solution:
(i) To find the probability that a bulb drawn is defective, we use the number of defective bulbs and the total number of bulbs. Total bulbs = 20 Defective bulbs = 4 \[ P(\text{defective}) = \frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}} = \frac{4}{20} = \frac{1}{5}. \] (ii) If the first bulb drawn is not defective, then there are now 19 bulbs left (16 good and 3 defective). Now, the probability that the next bulb drawn is not defective is: \[ P(\text{not defective}) = \frac{\text{Number of good bulbs}}{\text{Total remaining bulbs}} = \frac{16}{19}. \] To summarize: - (i) \( \frac{1}{5} \) - (ii) \( \frac{16}{19} \)

Solution:
To find the probabilities, we start with the total number of discs, which is 90. (i) **Probability of drawing a two-digit number:** The two-digit numbers are from 10 to 90 (inclusive). Count of two-digit numbers = 90 - 10 + 1 = 81. \[ P(\text{two-digit number}) = \frac{81}{90} = \frac{9}{10}. \] (ii) **Probability of drawing a perfect square number:** The perfect squares between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81 (which are \(1^2\) to \(9^2\)). Count of perfect squares = 9. \[ P(\text{perfect square}) = \frac{9}{90} = \frac{1}{10}. \] (iii) **Probability of drawing a number divisible by 5:** The numbers divisible by 5 from 1 to 90 are: 5, 10, 15, ..., 90. These form an arithmetic sequence where \(a = 5\), \(d = 5\), and the last term is 90. To find the count: \[ n = \frac{90 - 5}{5} + 1 = 18. \] \[ P(\text{divisible by 5}) = \frac{18}{90} = \frac{1}{5}. \] To summarize: - (i) \( \frac{9}{10} \) - (ii) \( \frac{1}{10} \) - (iii) \( \frac{1}{5} \)

Solution:
The die has the following faces: A, B, C, D, E, A. This gives us a total of 6 faces, with A appearing twice.

(i) Probability of getting A:
There are 2 faces showing A.
\[P(A) = \dfrac{\text{Number of A faces}}{\text{Total faces}} = \dfrac{2}{6} = \dfrac{1}{3}\]
(ii) Probability of getting D:
There is 1 face showing D.
\[P(D) = \dfrac{\text{Number of D faces}}{\text{Total faces}} = \dfrac{1}{6}\]
To summarize:
(i) \( \dfrac{1}{3} \)
(ii) \( \dfrac{1}{6} \)

Solution:
To determine the probability that a die lands inside a circle with a diameter of 1 meter, we first need to calculate the area of the circle and the area of the rectangular region in which the die can land. 1. **Area of the circle:** The radius \( r \) of the circle is half of the diameter: \( r = \frac{1}{2} \) m. The area \( A \) of the circle is given by the formula: \[ A_{\text{circle}} = \pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4} \, \text{m}^2. \] 2. **Area of the rectangular region:** If the rectangular region has a width and height both greater than or equal to 1 m (common assumptions), let's assume it is a square with a side length of 1 m for simplicity: \[ A_{\text{rectangle}} = 1 \times 1 = 1 \, \text{m}^2. \] 3. **Probability of landing inside the circle:** The probability \( P \) is the ratio of the area of the circle to the area of the rectangle: \[ P = \frac{A_{\text{circle}}}{A_{\text{rectangle}}} = \frac{\frac{\pi}{4}}{1} = \frac{\pi}{4} \approx 0.785. \] Thus, the probability that the die will land inside the circle is approximately \( \frac{\pi}{4} \) or about 0.785.

Solution:
In the lot of 144 ball pens, there are 20 defective pens and \( 144 - 20 = 124 \) good pens. (i) Probability that Nuri will buy the pen: Since she will buy a pen only if it is good, the probability is given by the ratio of good pens to the total number of pens: \[ P(\text{buys}) = \frac{\text{Number of good pens}}{\text{Total number of pens}} = \frac{124}{144} = \frac{31}{36}. \] (ii) Probability that she will not buy the pen: This occurs if the pen is defective. The probability is given by the ratio of defective pens to the total number of pens: \[ P(\text{does not buy}) = \frac{\text{Number of defective pens}}{\text{Total number of pens}} = \frac{20}{144} = \frac{5}{36}. \] To summarize: - (i) \( \frac{31}{36} \) - (ii) \( \frac{5}{36} \)

Solution:
To determine the probability that Hanif loses the game, we first need to find the total number of possible outcomes when tossing a coin 3 times.

1. Total outcomes:
Each toss has 2 possible results (heads or tails). Therefore, for 3 tosses: \[\text{Total outcomes} = 2^3 = 8.\]
2. Winning outcomes:
Hanif wins if all tosses are the same, which can happen in two ways:
- All heads (HHH)
- All tails (TTT)

Thus, there are 2 winning outcomes.

3. Losing outcomes:
The number of outcomes where Hanif loses is the total outcomes minus the winning outcomes: \[ \text{Losing outcomes} = \text{Total outcomes} - \text{Winning outcomes} = 8 - 2 = 6.\]
4. Probability of losing:
Now, we can calculate the probability that Hanif loses: \[ P(\text{loses}) = \frac{\text{Losing outcomes}}{\text{Total outcomes}} = \frac{6}{8} = \frac{3}{4}. \]
Therefore, the probability that Hanif will lose the game is \( \frac{3}{4} \).

Solution:
When a die is thrown twice, there are a total of \(6 \times 6 = 36\) possible outcomes.


(i) Probability that 5 will not come up either time:
To find this probability, we first determine the number of outcomes where 5 does not appear. Each die has 5 outcomes that are not 5 (1, 2, 3, 4, 6).

The number of outcomes where 5 does not appear in two throws is:
\[5 \times 5 = 25.\]
Now, the probability is:
\[P(\text{not 5 either time}) = \dfrac{25}{36}\]

(ii) Probability that 5 will come up at least once:
The probability of getting at least one 5 can be found using the complement rule: \[P(\text{at least one 5}) = 1 - P(\text{not 5 either time}).\]
Using the result from (i):
\[P(\text{at least one 5}) = 1 - \dfrac{25}{36} = \dfrac{11}{36}.\]

To summarize:
(i) \( \dfrac{25}{36} \)
(ii) \( \dfrac{11}{36} \)

Solution:
Let's evaluate each argument:

(i) Argument: If two coins are tossed simultaneously, there are three possible outcomes — two heads, two tails, or one of each. Therefore, for each of these outcomes, the probability is \(\dfrac{1}{3}\).

Evaluation: This argument is not correct. When tossing two coins, the possible outcomes are:
- HH (two heads)
- HT (one head, one tail)
- TH (one head, one tail)
- TT (two tails)
Thus, there are actually 4 outcomes: HH, HT, TH, and TT. The correct probabilities for each outcome are: - \(P(\text{two heads}) = \dfrac{1}{4}\)
- \(P(\text{two tails}) = \dfrac{1}{4}\)
- \(P(\text{one of each}) = \dfrac{2}{4} = \dfrac{1}{2}\)

So, the probabilities do not equal \(\dfrac{1}{3}\).
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(ii) Argument: If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is \(\dfrac{1}{2}\).

Evaluation: This argument is correct. A standard die has six faces showing the numbers 1, 2, 3, 4, 5, and 6. The odd numbers are 1, 3, and 5 (3 outcomes), while the even numbers are 2, 4, and 6 (3 outcomes). Since there are 3 odd numbers and 3 even numbers, the probabilities are:
- \(P(\text{odd number}) = \dfrac{3}{6} = \dfrac{1}{2}\)
- \(P(\text{even number}) = \dfrac{3}{6} = \dfrac{1}{2}\)

Therefore, the argument is correct.