PROBABILITY-Notes

Probability is a mathematical framework used to measure and reason about uncertainty. It assigns a numerical value to the likelihood of an event occurring when the outcome of an experiment cannot be predicted with absolute certainty. This numerical measure lies between 0 and 1, where 0 represents impossibility and 1 represents certainty.

In elementary probability, as studied in Class X, probability is grounded in logical reasoning rather than randomness alone. It is based on the assumption that all possible outcomes of a given experiment are equally likely, allowing probability to be calculated as a rational comparison between favourable outcomes and total possible outcomes.

Mathematically, if an experiment has a finite number of equally likely outcomes, the probability of an event is defined as:

\[\scriptsize\text{Probability of an event (P(E)) } = \dfrac{\text{Number of favourable outcomes​}}{\text{Total number of possible outcomes}​}\] This definition emphasises fairness, objectivity, and countability, making it suitable for structured experiments such as tossing coins, rolling dice, or drawing cards.

• Random Experiment

  A random experiment is a process that yields one of several possible outcomes, even though the experiment is performed under identical conditions each time. The key feature is uncertainty of outcome, not lack of control. Examples include tossing a coin or rolling a die.

• Sample Space

  The sample space is the complete set of all possible outcomes of a random experiment. It provides the universal reference against which all events are defined. Every probability calculation begins with a correct identification of the sample space.

• Event

  An event is a specific outcome or a group of outcomes drawn from the sample space. Events can be:
  • Simple events (single outcome)
  • Compound events (multiple outcomes)
  Probability always refers to an event, not to the experiment as a whole.

• Equally Likely Outcomes

  Outcomes are said to be equally likely if each has the same chance of occurring. This assumption is central to the classical definition of probability used at this level. Without equal likelihood, the standard probability formula does not apply directly.

• Classical (Theoretical) Probability

  he probability studied in Class X is theoretical, not experimental. It relies on logical analysis of outcomes rather than repeated trials. This approach assumes ideal conditions and focuses on reasoned calculation instead of observed frequency.

• Range of Probability Values

  The value of probability is always constrained as: \[\boxed{\;\boldsymbol{0\le P(E)\le 1}\;}\]
  • \(P(E) =0\) implies the event is impossible
  • \(P(E) =1\) implies the event is certain
  Any calculated probability outside this range indicates an error in reasoning or counting.

• Probability of an Impossible Event

  An impossible event is one that cannot occur under any circumstances within the defined experiment.
  Its probability is always zero, reinforcing the idea that probability quantifies feasibility.

• Probability of a Sure Event

  A sure (certain) event is one that must occur whenever the experiment is performed.
  Its probability is always one, representing complete certainty.

• Complementary Events

  If an event \(E\) occurs, its complement (denoted by \(\overline{E}\)) does not occur. Together, an event and its complement exhaust the sample space. Their probabilities satisfy the relationship: \[\boxed{\;\boldsymbol{P(E)+P(\overline{E})=1}\;}\]

• Logical Counting and Systematic Listing

  Accurate probability calculation depends heavily on systematic counting, such as listing outcomes, using tables, or applying logical classification.
  Errors in probability usually arise not from formulas, but from incorrect identification of outcomes.

• Role of Probability in Mathematics

  Probability acts as a bridge between pure mathematics and real-life reasoning. It introduces students to decision-making under uncertainty and prepares the foundation for advanced topics such as statistics, data analysis, and risk assessment.

Find the probability of getting a head when a coin is tossed once. Also, find the probability of getting a tail.

Solution

In the experiment of tossing the coin number of possible outcomes is 2 H or T
No of favourable outcomes of getting H is 1, so \[\begin{aligned} P(E)&=\dfrac{\text{No. of favourable outcome}}{\text{No. of all Possible outcome}}\\\\ &=\dfrac{1}{2} \end{aligned} \] Similarly, No of favourable outcomes of getting T is 1, so, \[P(F)=\dfrac{1}{2}\]

A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking at it. What is the probability that she takes out the
(i) yellow ball?
(ii) red ball?
(iii) blue ball?

Solution

Kritika takes out a ball from the bag without looking into it, so it is equally likely that she takes out any of them, therefore,
Total Possible outcome is 3.
(i)Favourable outcome to get the yellow ball out of 3 is =1 \[\begin{aligned}P(E)&=\dfrac{\text{No. of Favourable outcome}}{\text{No of total outcome}}\\\\ &=\dfrac{1}{3}\end{aligned}\] (ii)Favourable outcome to get the red ball out of 3 is =1 \[\begin{aligned}P(E)&=\dfrac{\text{No. of Favourable outcome}}{\text{No of total outcome}}\\\\ &=\dfrac{1}{3}\end{aligned}\] (iii)Favourable outcome to get blue ball out of 3 is =1 \[\begin{aligned}P(E)&=\dfrac{\text{No. of Favourable outcome}}{\text{No of total outcome}}\\\\ &=\dfrac{1}{3}\end{aligned}\]

Suppose we throw a die once.
(i) What is the probability of getting a number greater than 4 ?
(ii) What is the probability of getting a number less than or equal to 4 ?

Solution

Outcome of dice=(1),(2),(3),(4),(5) and (6)
Total number of possible outcomes = 6
(i) Possibility of getting a number greater than 4 is (5),(6), hence, No. of favourable events =2 \[ \begin{aligned} P(E)&=\dfrac{\text{No of favourable events}}{\text{No of all possible outcome}}\\\\ &= \dfrac{2}{6}\\\\ &=\dfrac{1}{3} \end{aligned} \] (ii) Possibility of getting a number less than or equal to 4 is {1},{2},{3},{4}, hence, No. of favourable events =4 \[ \begin{aligned} P(F)&=\dfrac{\text{No of favourable events}}{\text{No of all possible outcome}}\\\\ &= \dfrac{4}{6}\\\\ &=\dfrac{2}{3} \end{aligned} \]

One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace,
(ii) not be an ace.

Solution

(i) Probability that the card will be an ace
Number of Aces in a pack of 52 cards = 4, therefore, the total possible favourable outcome that the card is an Ace is =4
Total possible outcome =52, hence \[ \begin{aligned} P(E)&=\dfrac{\text{No of Favourable events}}{\text{No of total events}}\\\\ &=\dfrac{4}{52}\\\\ &=\dfrac{1}{13} \end{aligned} \]

(ii) Probability that the card will not be an ace implies that the event is the complement of \(P(E)\)

\[ \begin{aligned} P(E) + P(\overline{E})&=1\\ P(\overline{E})&=1-P(E)\\ &=1-\dfrac{1}{13}\\\\ &=\dfrac{13-1}{13}\\\\ &=\dfrac{12}{13} \end{aligned} \]

Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?

Solution

Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
Probability of Sangeeta's winning is \(P(S)) and
Probability of Reshma's winning is \(R(S)) Both events S and R are complementary to each other, hence \[ \begin{aligned} P(S)+P(R)=1\\ 0.62+P(R)=1\\ P(R)=1-0.62\\ P(R)=0.38 \end{aligned} \]

There are 40 students in Class X of a school, of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts the cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?

Solution

Total number of posiible events = 40
Favourable event for a girl's name=25, hence \[ \begin{aligned} P(G)&=\dfrac{\text{No of favourable event}}{\text{No of total possible events}}\\\\ &=\dfrac{25}{40}\\\\ &=\dfrac{5}{8} \end{aligned} \] Favourable event for a boy's name=15, hence \[ \begin{aligned} P(B)&=\dfrac{\text{No of favourable event}}{\text{No of total possible events}}\\\\ &=\dfrac{15}{40}\\\\ &=\dfrac{3}{8} \end{aligned} \]

A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) white?
(ii) blue?
(iii) red?

Solution

Total outcome = 3+2+4=9
(i) possible outcome for a white marble is drawn=3, hence \[ \begin{aligned} P(W)&=\dfrac{\text{No of possible outcome}}{\text{No of total outcome}}\\\\ &=\dfrac{2}{9} \end{aligned} \] (ii) possible outcome for a blue marble is drawn=3, hence \[ \begin{aligned} P(B)&=\dfrac{\text{No of possible outcome}}{\text{No of total outcome}}\\\\ &=\dfrac{3}{9}\\\\ &=\dfrac{1}{3} \end{aligned} \] (iii) possible outcome for a red marble is drawn=4, hence \[ \begin{aligned} P(R)&=\dfrac{\text{No of possible outcome}}{\text{No of total outcome}}\\\\ &=\dfrac{4}{9}\\\\ \end{aligned} \]

Harpreet tosses two different coins simultaneously (say, one is of ₹ 1 and other of ₹ 2). What is the probability that she gets at least one head?

Solution

When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are equally likely.
Favourable outcome for gettingat least one head is (H, H), (H, T), (T, H) = 3, hence
\[ \begin{aligned} P(H)&=\dfrac{\text{No. of Favourable outcome}}{\text{No. of Total Outcome}}\\\\ &=\dfrac{3}{4} \end{aligned} \]

Probability that Harpreet gets at least one head is \(\frac{3}{4}\)

A carton consists of 100 shirts, of which 88 are good, 8 have minor defects, and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?

Solution

Total possible outcome =100 (i) Favourable outcome that Jimmy accepts the shirt=88, hence \[ \begin{aligned} P(J)&=\dfrac{\text{No of favourable outcome}}{\text{No. of total outcome}}\\\\ &=\dfrac{88}{100}\\\\ &=0.88 \end{aligned} \] (i) Favourable outcome that Sujatha accept the shirt=88 + 8=96, hence \[ \begin{aligned} P(S)&=\dfrac{\text{No of favourable outcome}}{\text{No. of total outcome}}\\\\ &=\dfrac{96}{100}\\\\ &=0.96 \end{aligned} \]

Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8?
(ii) 13?
(iii) less than or equal to 12?

Solution

Total Possible outcome =36 as arranged below \[ \begin{array}{|c|c|} \hline(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\\hline \hline(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\\hline \hline(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\\hline \hline(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\\hline \hline(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\\hline \hline(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\\hline \end{array} \] (i) Favourable event of getting sum of 8 is (2,6), (3,5), (4,4), (5,3), (6,2)= 5
(Here (2,6) & (6,2) and (3,5) & (5,3) are different due to differences in colour), hence \[ \begin{aligned} P(E)&=\frac{\text{No of Favourable Events}}{\text{No of Possible Events}}\\\\ &=\dfrac{5}{36} \end{aligned} \] (ii) Favourable event of getting a sum of 13 is = 0, hence \[ \begin{aligned} P(F)&=\frac{\text{No of Favourable Events}}{\text{No of Possible Events}}\\\\ &=\dfrac{0}{36}\\\\ &=0 \end{aligned} \] (iii) Favourable event of getting a sum less than or equal to 13 = 36, hence \[ \begin{aligned} P(G)&=\frac{\text{No of Favourable Events}}{\text{No of Possible Events}}\\\\ &=\dfrac{36}{36}\\\\ &=1 \end{aligned} \]