ProbabilityClass 10 Mathematics NCERT Solutions

(i) Probability of an event E + Probability of the event 'not E ' = 1. Reason: An event E and its complement 'not E' (denoted as $\overline{E}$) together cover all possible outcomes of an experiment. Therefore, $P(E) + P(\overline{E}) = 1$.

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event. Reason: An impossible event has zero favourable outcomes. The probability is calculated as $\frac{0}{\text{Total outcomes}} = 0$.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure event or a certain event. Reason: A sure event includes all possible outcomes, so the number of favourable outcomes is equal to the total number of outcomes. The probability is $\frac{\text{Total outcomes}}{\text{Total outcomes}} = 1$.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1. Reason: Elementary events are the single outcomes of an experiment. The sum of their probabilities covers all possibilities, hence it must be 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1. Reason: The number of favourable outcomes can range from 0 (impossible event) to the total number of outcomes (sure event). Therefore, for any event E, $0 \leq P(E) \leq 1$.

(i) A driver attempts to start a car. The car starts or does not start. This experiment does not have equally likely outcomes. The car starting depends on many factors like the condition of the engine, amount of fuel, etc. A well-maintained car is much more likely to start than not.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. This experiment does not have equally likely outcomes. The outcome depends on the player's skill. A professional player is more likely to make the shot than miss it, while a novice is more likely to miss.

(iii) A trial is made to answer a true-false question. The answer is right or wrong. This experiment has equally likely outcomes, assuming the person is guessing. There are only two possibilities, right or wrong, and each has a chance of $\frac{1}{2}$. If the person knows the answer, it is not a random experiment.

(iv) A baby is born. It is a boy or a girl. This experiment has equally likely outcomes. From a biological standpoint, the probability of a baby being a boy or a girl is very close to $\frac{1}{2}$ for each.

Tossing a coin is considered a fair way of making a decision because a fair coin has two possible outcomes: a head or a tail. These two outcomes are equally likely. This means there is an equal chance for either team to win the toss. The result of a coin toss is unpredictable and not biased towards any particular outcome, ensuring impartiality.

Answer: (B) -1.5

Explanation: The probability of any event E must be a number between 0 and 1, inclusive. This is represented as $0 \leq P(E) \leq 1$.

Let's check the given options: (A) $\frac{2}{3} \approx 0.67$. This is between 0 and 1. (B) -1.5 is a negative number, which is less than 0. Probability cannot be negative. (C) $15\% = \frac{15}{100} = 0.15$. This is between 0 and 1. (D) 0.7 is between 0 and 1.

Therefore, -1.5 cannot be the probability of an event.

Given: The probability of an event E is $P(E) = 0.05$.

To Find: The probability of 'not E', which is denoted as $P(\overline{E})$.

Formula: We know that for any event E, the sum of the probability of the event and its complement is 1. $P(E) + P(\overline{E}) = 1$

Solution: Substituting the given value of $P(E)$ into the formula: $0.05 + P(\overline{E}) = 1$ $P(\overline{E}) = 1 - 0.05$ $P(\overline{E}) = 0.95$

Final Answer: The probability of 'not E' is 0.95.

Given: Let E be the event that 2 students have the same birthday. Then, 'not E' or $\overline{E}$ is the event that 2 students do not have the same birthday. The probability of 2 students not having the same birthday is $P(\overline{E}) = 0.992$.

To Find: The probability that the 2 students have the same birthday, which is $P(E)$.

Formula: We know that for any event E and its complement $\overline{E}$: $P(E) + P(\overline{E}) = 1$

Solution: Substituting the given value into the formula: $P(E) + 0.992 = 1$ $P(E) = 1 - 0.992$ $P(E) = 0.008$

Final Answer: The probability that the 2 students have the same birthday is 0.008.

Given: Number of male fish in the tank = 5 Number of female fish in the tank = 8

Total Number of Outcomes: The total number of fish in the tank is: Total number of fish = $5 + 8 = 13$

To Find: The probability that the fish taken out is a male fish.

Solution: Number of favourable outcomes (male fish) = 5

The probability of an event is given by the formula: $P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

Final Answer: The probability that the fish taken out is a male fish is $\frac{5}{13}$.

Given: Number of defective pens = 12 Number of good pens = 132

Total Number of Outcomes: The total number of pens in the lot is the sum of defective and good pens. Total number of pens = $12 + 132 = 144$

To Find: The probability that the pen taken out is a good one.

Solution: Number of favourable outcomes (good pens) = 132

$P(\text{good pen}) = \frac{\text{Number of good pens}}{\text{Total number of pens}}$ $P(\text{good pen}) = \frac{132}{144}$

To simplify the fraction, we can divide both numerator and denominator by their greatest common divisor, which is 12. $P(\text{good pen}) = \frac{132 \div 12}{144 \div 12} = \frac{11}{12}$

Final Answer: The probability that the pen taken out is a good one is $\frac{11}{12}$.

Given: A die is dropped on a rectangular region. Length of the rectangle = 3 m Breadth of the rectangle = 2 m Diameter of the circle inside the rectangle = 1 m

This is a problem of geometric probability, where probability is the ratio of favourable area to the total area.

Total Area (Total possible outcomes): The total area is the area of the rectangular region. Area of rectangle = Length $\times$ Breadth Area of rectangle = $3 \text{ m} \times 2 \text{ m} = 6 \text{ m}^2$

Favourable Area (Favourable outcomes): The favourable area is the area where the die can land, which is inside the circle. Radius of the circle = $\frac{\text{Diameter}}{2} = \frac{1}{2} \text{ m} = 0.5 \text{ m}$ Area of circle = $\pi r^2$ Area of circle = $\pi (0.5)^2 = 0.25\pi \text{ m}^2 = \frac{\pi}{4} \text{ m}^2$

To Find: The probability that the die will land inside the circle.

Formula: $P(\text{Event}) = \frac{\text{Area of favourable region}}{\text{Area of total region}}$

Solution: $P(\text{die lands inside the circle}) = \frac{\text{Area of the circle}}{\text{Area of the rectangle}}$ $P(\text{die lands inside the circle}) = \frac{\pi/4}{6} = \frac{\pi}{24}$

Final Answer: The probability that the die will land inside the circle is $\frac{\pi}{24}$.

(i) Completing the table: When two dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. The outcomes are pairs (die 1, die 2).

(ii) Evaluating the student's argument: Argument: 'There are 11 possible outcomes 2, 3, ..., 12. Therefore, each has a probability $\frac{1}{11}$.'

Conclusion: I do not agree with this argument.

Justification: The argument is incorrect because it assumes that the 11 possible sums (2, 3, ..., 12) are equally likely outcomes. However, they are not. As shown in the table above, the number of ways to obtain each sum is different. For example, there is only one way to get a sum of 2 (1,1), but there are six ways to get a sum of 7. The probability of an event depends on the number of elementary outcomes favourable to it. Since the elementary outcomes (the 36 pairs) are equally likely, but the sums are not, the probabilities for each sum are different.

Given: A coin is tossed 3 times. Let H denote Heads and T denote Tails.

Total Number of Outcomes: The possible outcomes of tossing a coin 3 times are: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Total number of possible outcomes = 8

Condition for Hanif to Win: Hanif wins if all tosses give the same result. The winning outcomes are: {HHH, TTT} Number of outcomes where Hanif wins = 2

Probability of Hanif Winning: $P(\text{Hanif wins}) = \frac{\text{Number of winning outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = \frac{1}{4}$

To Find: The probability that Hanif will lose the game.

Solution: Hanif loses if the outcome is not one of the winning outcomes. The event 'Hanif loses' is the complement of the event 'Hanif wins'.

Using the complement rule: $P(\text{Hanif loses}) = 1 - P(\text{Hanif wins})$ $P(\text{Hanif loses}) = 1 - \frac{1}{4} = \frac{3}{4}$

Alternative Method: List the outcomes where Hanif loses: {HHT, HTH, THH, HTT, THT, TTH} Number of outcomes where Hanif loses = 6

Final Answer: The probability that Hanif will lose the game is $\frac{3}{4}$.

Conclusion: The argument is not correct.

Reason: The argument lists three outcomes: 'two heads', 'two tails', and 'one of each'. However, these three outcomes are not equally likely. The actual elementary outcomes when two coins are tossed are:

There are 4 equally likely elementary outcomes.

Since the probabilities are not equal to $\frac{1}{3}$, the argument is incorrect.

Conclusion: The argument is correct.

Reason: When a die is thrown, the possible elementary outcomes are {1, 2, 3, 4, 5, 6}. There are 6 equally likely outcomes.

The event 'getting an odd number' corresponds to the outcomes {1, 3, 5}. Number of favourable outcomes for 'odd' = 3. $P(\text{odd number}) = \frac{3}{6} = \frac{1}{2}$

The event 'getting an even number' corresponds to the outcomes {2, 4, 6}. Number of favourable outcomes for 'even' = 3. $P(\text{even number}) = \frac{3}{6} = \frac{1}{2}$

The two outcomes, 'an odd number' and 'an even number', are indeed equally likely, and the probability for each is $\frac{1}{2}$. Therefore, the argument is correct.