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Introduction to Probability - Solve probability problems using tree diagrams and tabular representations

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The theoretical probability of an event EE, written as P(E)P(E), is defined as P(E)=Number of outcomes favorable to ETotal number of all possible outcomes of the experimentP(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of all possible outcomes of the experiment}}, assuming all outcomes are equally likely.

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The probability of any event EE is a number between 00 and 11 inclusive, i.e., 0≤P(E)≤10 \le P(E) \le 1.

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An event that cannot happen has a probability of 00 and is called an impossible event. An event that is certain to happen has a probability of 11 and is called a sure event.

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The sum of the probabilities of all the elementary events of an experiment is 11. For any event EE, P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1, where Eˉ\bar{E} represents 'not EE'.

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Tabular representations (Two-way tables) are useful for representing the sample space of two independent actions, such as rolling two dice. The outcomes are written as ordered pairs (x,y)(x, y). For two dice, there are 6×6=366 \times 6 = 36 total outcomes.

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Tree diagrams are visual tools used to list all possible outcomes of a sequence of events, such as tossing three coins or drawing balls from a bag without replacement. Each path from the root to a leaf represents one possible outcome.

📐Formulae

P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1

0≤P(E)≤10 \le P(E) \le 1

Total outcomes for n coins=2n\text{Total outcomes for } n \text{ coins} = 2^n

Total outcomes for n dice=6n\text{Total outcomes for } n \text{ dice} = 6^n

💡Examples

Problem 1:

Two fair dice are rolled simultaneously. Use a tabular representation to find the probability that the sum of the numbers on the top faces is 88.

Solution:

In a 6×66 \times 6 table, the outcomes where the sum is 88 are: E={(2,6),(3,5),(4,4),(5,3),(6,2)}E = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}. Total number of outcomes n(S)=36n(S) = 36. Number of favorable outcomes n(E)=5n(E) = 5. P(Sum is 8)=536P(\text{Sum is } 8) = \frac{5}{36}

Explanation:

A tabular representation lists all 36 pairs. We identify the diagonal where the sum of coordinates equals 8.

Problem 2:

Three coins are tossed simultaneously. Use a tree diagram to find the probability of getting exactly two heads.

Solution:

The sample space SS generated by the tree diagram is: S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}. Total outcomes n(S)=8n(S) = 8. Let EE be the event of getting exactly two heads: E={HHT,HTH,THH}E = \{HHT, HTH, THH\}. Number of favorable outcomes n(E)=3n(E) = 3. P(E)=38P(E) = \frac{3}{8}

Explanation:

A tree diagram branches first into H and T, then each of those branches into H and T again, and once more for the third coin, resulting in 23=82^3 = 8 paths.

Problem 3:

A bag contains 55 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.

Solution:

Let the number of blue balls be xx. Total balls =5+x= 5 + x. P(Red)=55+xP(\text{Red}) = \frac{5}{5+x} P(Blue)=x5+xP(\text{Blue}) = \frac{x}{5+x} According to the question: P(Blue)=2×P(Red)P(\text{Blue}) = 2 \times P(\text{Red}) x5+x=2×55+x\frac{x}{5+x} = 2 \times \frac{5}{5+x} x=10x = 10

Explanation:

We set up an equation based on the given ratio of probabilities and solve for the unknown variable xx representing the count of blue balls.