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The Mathematics of Maybe: Introduction to Probability - Tree diagrams

Grade 9CBSE

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

🔑Concepts

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A Tree Diagram is a visual representation used to list all possible outcomes of a sequence of events and calculate their probabilities.

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Each branch in a tree diagram represents a possible outcome. For a single trial, the sum of probabilities of all branches originating from a single point must be 11.

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To find the probability of a combined outcome (a specific path), multiply the probabilities along the branches: P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B).

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If an event can happen in multiple ways (multiple paths), the total probability is the sum of the probabilities of those individual paths.

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The sample space SS is the set of all possible outcomes. The probability of an event EE is given by the ratio of favorable outcomes to total outcomes: P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}.

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For any event EE, the probability of its complement (the event not happening) is P(E′)=1−P(E)P(E') = 1 - P(E).

📐Formulae

P(E)=Number of outcomes favorable to ETotal number of possible outcomesP(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}}

P(A∩B)=P(A)×P(B) (for independent events along branches)P(A \cap B) = P(A) \times P(B) \text{ (for independent events along branches)}

∑P(outcomes)=1\sum P(\text{outcomes}) = 1

P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1

💡Examples

Problem 1:

A coin is tossed twice. Use a tree diagram to find the probability of getting at least one head.

Solution:

  1. First toss: HH with P(H)=12P(H) = \frac{1}{2} and TT with P(T)=12P(T) = \frac{1}{2}.
  2. Second toss: From each branch, draw two more branches (HH and TT) each with probability 12\frac{1}{2}.
  3. Possible outcomes (paths):
  • HH:12×12=14HH: \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
  • HT:12×12=14HT: \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
  • TH:12×12=14TH: \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
  • TT:12×12=14TT: \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
  1. 'At least one head' includes HH,HT,THHH, HT, TH.
  2. P(at least one head)=14+14+14=34P(\text{at least one head}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}.

Explanation:

By multiplying the probabilities along the branches, we find the probability of each specific path. Since 'at least one head' can happen in three mutually exclusive ways, we sum those probabilities.

Problem 2:

A bag contains 33 Red balls and 22 Blue balls. A ball is drawn, its color is recorded, and then it is replaced before a second ball is drawn. Find the probability that both balls are of the same color.

Solution:

  1. Probability of Red (RR) = 35\frac{3}{5}; Probability of Blue (BB) = 25\frac{2}{5}.
  2. Tree branches:
  • Path 1 (RR then RR): P(R,R)=35×35=925P(R,R) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}
  • Path 2 (BB then BB): P(B,B)=25×25=425P(B,B) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}
  1. Probability of same color: P(Same Color)=P(R,R)+P(B,B)P(\text{Same Color}) = P(R,R) + P(B,B) P(Same Color)=925+425=1325P(\text{Same Color}) = \frac{9}{25} + \frac{4}{25} = \frac{13}{25}

Explanation:

Since the ball is replaced, the probabilities remain the same for the second draw. We identify the paths that satisfy the condition (both Red or both Blue) and sum their calculated probabilities.