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Combinatorics - Visualizing the Principle: The Tree Diagram-advanced

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, where each branch represents a possible choice at a particular stage.

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The Fundamental Principle of Counting states that if an event E1E_1 can occur in n1n_1 ways, followed by event E2E_2 in n2n_2 ways, and so on, then the total number of outcomes is given by the product n1×n2×⋯×nkn_1 \times n_2 \times \dots \times n_k.

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In advanced tree diagrams, the number of branches at a node may change based on previous choices, which is particularly useful for 'without replacement' scenarios or conditional constraints.

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The total number of outcomes in the sample space SS is equal to the total number of 'leaves' or endpoints at the final stage of the tree diagram.

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For multi-stage experiments involving kk trials with nn outcomes each, the total outcomes are nkn^k. This is represented by a perfectly symmetrical tree.

📐Formulae

N=n1×n2×n3×⋯×nkN = n_1 \times n_2 \times n_3 \times \dots \times n_k

P(A)=Number of favorable outcomes (paths)Total number of outcomes (total leaves)P(A) = \frac{\text{Number of favorable outcomes (paths)}}{\text{Total number of outcomes (total leaves)}}

Total outcomes for n objects taken r at a time with replacement=nr\text{Total outcomes for } n \text{ objects taken } r \text{ at a time with replacement} = n^r

💡Examples

Problem 1:

A bag contains 3 balls: one Red (RR), one Blue (BB), and one Green (GG). Two balls are drawn one after another without replacement. Use the concept of a tree diagram to find the total number of outcomes and the probability that the Blue ball is selected.

Solution:

  1. At the first stage, there are 3 possible outcomes: R,B,GR, B, G.
  2. Since the drawing is 'without replacement', if RR is drawn first, the second draw can only be BB or GG. If BB is drawn first, the second can be RR or GG. If GG is drawn first, the second can be RR or BB.
  3. The paths are: (R,B),(R,G),(B,R),(B,G),(G,R),(G,B)(R,B), (R,G), (B,R), (B,G), (G,R), (G,B).
  4. Total outcomes = 3×2=63 \times 2 = 6.
  5. Outcomes containing Blue (BB): (R,B),(B,R),(B,G),(G,B)(R,B), (B,R), (B,G), (G,B). Total = 4.
  6. Probability P(Blue)=46=23P(\text{Blue}) = \frac{4}{6} = \frac{2}{3}.

Explanation:

The tree diagram starts with 3 branches. Each of these branches then splits into 2 further branches (because one ball is removed). The total number of endpoints is the product of the choices at each step.

Problem 2:

A student has 2 shirts (White WW, Blue BB), 2 trousers (Black T1T_1, Grey T2T_2), and 2 pairs of shoes (Formal S1S_1, Casual S2S_2). How many different outfits can be formed? Show the calculation using the Fundamental Principle of Counting represented by a tree.

Solution:

Using the Fundamental Principle of Counting: Number of Shirt choices n1=2n_1 = 2 Number of Trouser choices n2=2n_2 = 2 Number of Shoe choices n3=2n_3 = 2

Total Outfits = n1×n2×n3n_1 \times n_2 \times n_3 2×2×2=82 \times 2 \times 2 = 8

The outcomes (leaves of the tree) are: W−T1−S1,W−T1−S2,W−T2−S1,W−T2−S2,B−T1−S1,B−T1−S2,B−T2−S1,B−T2−S2W-T_1-S_1, W-T_1-S_2, W-T_2-S_1, W-T_2-S_2, B-T_1-S_1, B-T_1-S_2, B-T_2-S_1, B-T_2-S_2.

Explanation:

Each shirt choice leads to two trouser choices, and each trouser choice leads to two shoe choices, creating a branching factor of 2 at every level.

Problem 3:

Three coins are tossed simultaneously. Using a tree diagram approach, find the probability of getting exactly two heads.

Solution:

Each coin toss has 2 outcomes: Head (HH) or Tail (TT). Stage 1 (Coin 1): H,TH, T (2 branches) Stage 2 (Coin 2): Each branch splits into H,TH, T (4 branches total) Stage 3 (Coin 3): Each branch splits into H,TH, T (8 branches total)

Sample Space S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} Total outcomes n(S)=23=8n(S) = 2^3 = 8. Outcomes with exactly two heads: E={HHT,HTH,THH}E = \{HHT, HTH, THH\} Number of favorable outcomes n(E)=3n(E) = 3. P(E)=n(E)n(S)=38P(E) = \frac{n(E)}{n(S)} = \frac{3}{8}

Explanation:

A tree diagram helps systematically list all 2×2×22 \times 2 \times 2 combinations without missing any, ensuring the calculation of the probability is accurate.