Binomial Distribution through Tree Diagrams

Concept of binomial distribution using tree diagrams to represent various scenarios of rolling a fair dice multiple times and calculating probabilities. The content covers examples, calculations, and explanations to help you grasp the underlying principles easily.

• Binomial Distribution
• Tree Diagrams
• Probability
• Dice Rolling
• Mathematics

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Uploaded on Feb 27, 2025 | 0 Views

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Presentation Transcript

1. The Binomial Distribution The Binomial Distribution Through tree diagrams Through tree diagrams

2. Expand (a+b)2

3. Now let us consider the following: We roll a fair dice and record the number of 6 s we roll Represent this as a tree diagram

4. Now let us consider the following: We roll a fair dice and record the number of 6 s we roll, and call this X 2 1 6 1 6 Find P(X=2) Success + Success 1 6 Find P(X=1) 5 6 1 6 5 + Failure 6 Find P(X=0) 1 6 1 6 5 Success 6 5 6 Failure + 2 5 6 5 6 Failure 2 Note that these will sum to 1 5 6 2 2 1 6 5 1 6 + + 1 6

5. Now let us consider the following: We roll a fair dice twice and record the number of 6 s we roll, and call this X Success 0 times Success 1 time Success 2 times Does this look familiar? (a+b)2 = a2 + 2 x a x b + b2 2 5 6 2 2 1 6 5 1 6 + + 6

6. (a+b)2 = a2 + 2 x a x b + b2 2 5 6 2 2= What binomial would this have come from? 2 1 6 5 1 6 1 6+5 + + 6 6 Success 0 times Success 1 time Success 2 times 2 1 6+5 By writing probability outcomes as binomials we can quickly find probabilities for higher numbers of trials, rather than draw out huge tree diagrams 6 Success Failure

7. Expand (a+b)3

8. Now let us consider the following: We roll a fair dice 3 times and record the number of 6 s we roll, and call this X Find P(X=3) 3 1 6 1 6 Success 1 6 2 1 6 5 Success Failure 5 6 6 Success 1 6 2 1 6 1 6 5 Find P(X=2) Success Failure 2 Failure 6 5 6 1 6 5 6 5 6 2 1 6 1 6 1 6 5 Success Failure Success 2 6 1 6 5 6 5 6 5 6 Failure 2 1 6 1 6 5 6 Find P(X=1) Success Failure 5 6 3 5 6 5 6 Failure Find P(X=0) 2 3 3 2 1 6 5 5 6 1 6 3 1 5 6 + 3 + + 6 1 6

9. (a+b)3 = a3 + 3 x a2 x b + 3 x a x b2 + b3 2 2 3 3 3= 1 6 5 3 1 5 6 1 6 5 6 1 6+5 What binomial would this have come from? + 3 + 6 + 6 6 Success 3 times Success 1 time Success 0 times Success 2 times 3 1 6+5 By writing probability outcomes as binomials we can quickly find probabilities for higher numbers of trials, rather than draw out huge tree diagrams 6 Success Failure

10. Now let us consider the following: We roll a fair dice 4 times. By expanding (a + b)4, find the probability of rolling exactly two 6 s. a4 + 4 x a3 x b + 6 x a2 x b2 + 4 x a x b3 + b4 (a + b) 4 = So . 4 3 2 2 4 3 4 1 6+5 1 6 5 1 6 5 6 1 6 + 4 1 5 6 5 6 + 6 + 4 = + 6 6 6 Which probability represents rolling exactly two 6 s? Success Failure 2 2 1 6 5 6 So P (X = 2) = 6 =0.1157047047 = 0.116 (3SF)

11. Still not convinced? This is the last diagram, and furthest I will go with a tree diagram 1 6 Success 1 6 Success Find P(X=2) 5 6 Failure Success Failure Success Failure Success Failure Success Failure Success Failure 1 6 Success 1 6 Failure 5 6 2 2 1 6 5 6 Success 5 6 1 6 1 6 1 6 Success 2 2 1 6 2 5 6 2 5 6 Failure 5 6 1 6 1 6 5 6 Failure 5 6 5 6 1 6 1 6 Success 2 2 1 6 5 6 So P (X = 2) 5 6 1 6 Success 1 6 2 2 5 6 1 6 5 6 Failure 5 6 2 2 5 6 2 2 Failure 1 6 5 6 1 6 5 6 1 6 1 6 Success Failure = 6 Success 5 6 Failure 5 6 1 6 =0.1157407407 Success Failure 5 6 Failure 5 6 = 0.116 (3SF)

12. In General if P(Success) = p, then to find P(X = a) from n trials ? ? P (X = a) = ?? 1 ?? ? From the expansion where ? + (1 ?)? Failure Success